JEE Laws of Thermodynamics Questions

Melting takes place isothermally at $$T = 273\;\text{K}$$ and constant atmospheric pressure $$P = 1\;\text{atm}$$ (≈ $$1.013\times10^{5}\;\text{Pa}$$).

At constant pressure the heat absorbed equals the molar enthalpy of fusion:
$$q_{p}= \Delta H_{fus} \;\;(\text{positive})$$

The first-law relation connecting enthalpy and internal energy is
$$\Delta H = \Delta U + P\,\Delta V \quad -(1)$$

Rearranging, the change in internal energy is
$$\Delta U = \Delta H - P\,\Delta V \quad -(2)$$

When ice converts to water its volume decreases because water is denser than ice:
density of ice ≈ $$0.917\;\text{g cm}^{-3}$$, density of water ≈ $$1.00\;\text{g cm}^{-3}$$.
Hence $$\Delta V \lt 0$$ (final volume < initial volume).

Work done by the system on the atmosphere is
$$w = P\,\Delta V$$.
Since $$\Delta V \lt 0$$, $$w$$ is negative: the system does negative work, that is, the atmosphere does positive work on the system.

Substituting $$\Delta V \lt 0$$ in $$(2)$$:
$$\Delta U = \Delta H - P\,(\text{negative}) = \Delta H + (\text{positive term})$$.
Therefore $$\Delta U \gt 0$$; the internal energy increases, it certainly does not remain unchanged or decrease.

Summarising:
• Internal energy increases ⇒ Options A and C are incorrect.
• Work done on the system is positive ⇒ Option B is incorrect, Option D is correct.

Hence the correct choice is Option D: Positive work is done on the ice-water system by the atmosphere.

Assertion (A): With increase in pressure, volume falls off more rapidly in an isothermal process compared to an adiabatic process.

For isothermal: $$PV = C$$, so $$\frac{dV}{dP} = -\frac{V}{P}$$

For adiabatic: $$PV^\gamma = C$$, so $$\frac{dV}{dP} = -\frac{V}{\gamma P}$$

Since $$\gamma > 1$$: $$\left|\frac{dV}{dP}\right|_{isothermal} = \frac{V}{P} > \frac{V}{\gamma P} = \left|\frac{dV}{dP}\right|_{adiabatic}$$

So the volume decreases more rapidly (larger magnitude of dV/dP) in the isothermal process. Assertion A is true.

Reason (R): States the equations PV = constant (isothermal) and $$PV^\gamma$$ = constant (adiabatic). This is true and directly explains why the assertion holds (the extra factor of $$\gamma$$ in the adiabatic case makes the volume change slower).

Both A and R are true, and R is the correct explanation of A.

The correct answer is Option 1.

Let the number of moles, the pressure and the volume in the left compartment be $$n_1,\;P_L,\;V_L$$ and in the right compartment be $$n_2,\;P_R,\;V_R$$.

Given
$$n_1 = \frac32,\qquad n_2 = 1$$
length of each compartment at equilibrium: $$\frac{L}{2}$$
cross-sectional area of the piston: $$A$$

Therefore
$$V_L = A \left(\frac{L}{2}\right)=V_R = A\left(\frac{L}{2}\right)=V$$ where $$V = \frac{AL}{2}\,.$$

The piston is thermally conducting, hence the two gases finally attain the same temperature $$T$$. For an ideal gas, $$PV=nRT$$; therefore

$$P_LV = n_1 RT,\qquad P_RV = n_2 RT$$

Dividing the first relation by the second,

$$\frac{P_L}{P_R} = \frac{n_1}{n_2}= \frac{3/2}{1}= \frac32 \quad\Longrightarrow\quad P_L = \frac32\,P_R\;.$$  -(1)

Next, write the force-balance for the movable piston. Choose the rightward direction as positive.

• Force on piston by left gas (to the right): $$P_LA$$
• Force on piston by right gas (to the left): $$P_RA$$
• Force by the spring (to the left): $$F_s$$

At mechanical equilibrium, net force is zero:

$$P_LA - P_RA - F_s = 0 \quad\Longrightarrow\quad (P_L-P_R)A = F_s\;.$$  -(2)

The spring is fixed to the left wall; its natural length is $$\dfrac{2L}{5}$$ and its actual length equals the left-hand length $$\dfrac{L}{2}$$. Hence the extension is

$$\Delta x = \frac{L}{2} - \frac{2L}{5} = \frac{L}{10}\;.$$

Thus the spring force is

$$F_s = k\,\Delta x = k\left(\frac{L}{10}\right)=\frac{kL}{10}\;.$$  -(3)

Insert $$(1)$$ and $$(3)$$ into $$(2)$$:

$$(P_L-P_R)A = \frac{kL}{10}$$ $$\left(\frac32\,P_R - P_R\right)A = \frac{kL}{10}$$ $$\left(\frac12\,P_R\right)A = \frac{kL}{10}$$ $$P_R = \frac{kL}{10A}\times 2 = \frac{kL}{5A}\;.$$

Therefore the pressure in the right compartment is

$$P = P_R = \frac{kL}{A}\left(\frac15\right)\,.$$

Comparing with the required form $$P = \dfrac{kL}{A}\,\alpha$$ gives

$$\alpha = \frac15 = 0.2\;.$$

Hence, the desired value is 0.2.

Initial state:

For one complete cycle,

So net heat exchanged equals net work done:

$$Q=W$$

Now calculate work in each process.

1. Isothermal expansion from $$V_0\ ​to\ 4V_0$$

For isothermal process,

$$W_1=nRT\ln\frac{4V_0}{V_0}$$

Since initially

$$nRT=P_0V_0$$

So

$$W_1=P_0V_0\ln4$$

1. Isobaric compression

After isothermal expansion pressure becomes

$$P=\frac{P_0V_0}{4V_0}=\frac{P_0}{4}$$

Compression occurs at constant pressure $$\frac{P_0}{4}\ from\ 4V_0​\ to\ V_0​$$.

Work done:

1. Isochoric heating

$$W_3=0$$

Net work:

$$W=P_0V_0\ln4-\frac{3}{4}P_0V_0$$

For a cyclic process,

$$ΔU=0$$

So net heat exchanged equals net work done:

$$Q=W$$

And work done in a cycle is area enclosed in P-V diagram.

The path ABCA encloses the upper semicircle.

From figure:

• Center at

$$(300\text{ cc},300\text{ kPa})$$

• Radius:

$$r=100\text{ cc}$$

Area of semicircle:

$$W=\frac{1}{2}\pi r^2$$

$$=\frac{1}{2}\pi(100)^2$$

$$=5000\pi$$

Units are

$$(\text{kPa})(\text{cc})$$

Convert to SI:

$$1(\text{kPa})(\text{cc})=10^3\times10^{-6}=10^{-3}\text{ J}$$

So

$$Q=5000\pi\times10^{-3}$$

$$=5\pi J$$

When water falls freely, its gravitational potential energy is converted into internal energy (heat) on reaching the pool.
Ignoring any heat loss to the surroundings, the gain in internal energy raises the water’s temperature.

For a mass $$m$$ of water falling through height $$h$$, the potential energy lost is
$$\text{Potential energy} = m g h$$

This energy appears as heat: $$m c \Delta T$$, where $$c$$ is the specific heat capacity and $$\Delta T$$ is the rise in temperature.

Equating the two energies:
$$m g h = m c \Delta T$$

Cancel the common factor $$m$$ (mass of water):
$$g h = c \Delta T$$

Hence the temperature rise is
$$\Delta T = \frac{g h}{c}$$

Insert the given values: $$g = 10 \text{ m s}^{-2}$$, $$h = 200 \text{ m}$$, $$c = 4200 \text{ J kg}^{-1} \text{ K}^{-1}$$.

$$\Delta T = \frac{10 \times 200}{4200}$$

$$\Delta T = \frac{2000}{4200} = 0.476 \text{ K} \approx 0.48 \text{ K}$$

Therefore, the rise in temperature of the water is about $$0.48 \text{ K}$$.

Correct option: Option D (0.48 K)

Use ideal gas law:

$$PV=nRT$$

Given pressure increases linearly with temperature,

$$P∝T$$

or

$$\frac{P}{T}=\text{constant}$$

From ideal gas law,

$$\frac{P}{T}=\frac{nR}{V}$$

Since P/T is constant,

$$V=\text{constant}$$

So process is isochoric.

Hence statement E is true.

For isochoric process,

$$W=\int PdV=0$$

So A is true.

First law:

$$Q=\Delta U+W$$

Since

W=0

$$Q=ΔU$$

So B ("heat added is different from change in internal energy") is false.

Volume does not increase (constant), so C is false.

Pressure increases linearly with temperature, so temperature increases.

For ideal gas internal energy depends only on temperature, so internal energy increases.

D is true.

Correct statements:

A, D, E

This question is about thermoelectric materials (Seebeck effect), where a temperature difference is converted to electrical energy.

For efficient thermoelectric energy harvesting, the material should:

1. Have high electrical conductivity - to efficiently transport the generated electrical current with minimal resistive losses.

2. Have low thermal conductivity - to maintain the temperature difference across the material (if thermal conductivity is high, heat flows quickly and the temperature gradient disappears).

This is characterized by a high figure of merit $$ZT = \frac{S^2\sigma T}{\kappa}$$, where $$S$$ is the Seebeck coefficient, $$\sigma$$ is electrical conductivity, $$\kappa$$ is thermal conductivity, and $$T$$ is temperature.

A high ZT requires high electrical conductivity ($$\sigma$$) and low thermal conductivity ($$\kappa$$).

The correct answer is Option 4: low thermal conductivity and high electrical conductivity.

For reversible heating,

$$dS=\frac{dQ}{T}$$

For water,

$$dQ=mcdT$$

so

$$dS=\frac{mcdT}{T}$$

Integrating from $$T_1\ ​to\ T_2​$$,

$$ΔS=\int^{_{ }}\frac{mcdT}{T}$$

$$\Delta S=mc\ln\left(\frac{T_2}{T_1}\right)$$

Given specific heat

$$c=1\ \text{J kg}^{-1}\text{K}^{-1}$$

Thus

$$\Delta S=m\ln\left(\frac{T_2}{T_1}\right)$$

Concept: Extensive vs intensive thermodynamic variables.

Definitions:

Extensive variables depend on the size/amount of the system (they scale with the quantity of matter). Examples: internal energy, volume, mass, enthalpy, entropy.

Intensive variables are independent of the size of the system. Examples: pressure, temperature, density.

Evaluating each statement:

(A) Internal energy, volume (V), and mass (M) are extensive variables.

All three depend on the amount of matter. Statement A is correct.

(B) Pressure (P), temperature (T), and density ($$\rho$$) are intensive variables.

None of these depend on the system size. Statement B is correct.

(C) Volume (V), temperature (T), and density ($$\rho$$) are intensive variables.

Volume is an extensive variable (it doubles when the system size doubles). Statement C is incorrect.

(D) Mass (M), temperature (T), and internal energy are extensive variables.

Temperature is an intensive variable. Statement D is incorrect.

Correct statements: (A) and (B) only, which corresponds to Option 4.

The efficiency of a Carnot engine is given by the formula:

$$\eta = 1 - \frac{T_c}{T_h}$$

where $$T_c$$ is the cold reservoir temperature and $$T_h$$ is the hot reservoir temperature, both in Kelvin.

For the original engine $$E$$ working between $$T_1 = 473 \, \text{K}$$ and $$T_2 = 273 \, \text{K}$$:

$$\eta_{12} = 1 - \frac{T_2}{T_1} = 1 - \frac{273}{473} = \frac{473 - 273}{473} = \frac{200}{473}$$

For engine $$E_1$$ working between $$T_1 = 473 \, \text{K}$$ and $$T_3 = 373 \, \text{K}$$:

$$\eta_1 = 1 - \frac{T_3}{T_1} = 1 - \frac{373}{473} = \frac{473 - 373}{473} = \frac{100}{473}$$

For engine $$E_2$$ working between $$T_3 = 373 \, \text{K}$$ and $$T_2 = 273 \, \text{K}$$:

$$\eta_2 = 1 - \frac{T_2}{T_3} = 1 - \frac{273}{373} = \frac{373 - 273}{373} = \frac{100}{373}$$

Now, evaluate the options:

Option A: $$\eta_{12} = \eta_1 \eta_2$$

Compute $$\eta_1 \eta_2$$:

$$\eta_1 \eta_2 = \left( \frac{100}{473} \right) \times \left( \frac{100}{373} \right) = \frac{10000}{473 \times 373}$$

Compare with $$\eta_{12} = \frac{200}{473}$$:

$$\frac{200}{473} \neq \frac{10000}{473 \times 373}$$

Since they are not equal, option A is incorrect.

Option B: $$\eta_{12} \geq \eta_1 + \eta_2$$

Compute $$\eta_1 + \eta_2$$:

$$\eta_1 + \eta_2 = \frac{100}{473} + \frac{100}{373} = 100 \left( \frac{1}{473} + \frac{1}{373} \right) = 100 \left( \frac{373 + 473}{473 \times 373} \right) = 100 \times \frac{846}{473 \times 373} = \frac{84600}{473 \times 373}$$

Now, express $$\eta_{12}$$ with the same denominator:

$$\eta_{12} = \frac{200}{473} = \frac{200 \times 373}{473 \times 373} = \frac{74600}{473 \times 373}$$

Compare $$\frac{74600}{473 \times 373}$$ and $$\frac{84600}{473 \times 373}$$:

Since $$74600 < 84600$$, it follows that $$\eta_{12} < \eta_1 + \eta_2$$.

Thus, $$\eta_{12} \geq \eta_1 + \eta_2$$ is false, so option B is incorrect.

Option C: $$\eta_{12} = \eta_1 + \eta_2$$

From above:

$$\eta_{12} = \frac{200}{473} \approx 0.4228$$

$$\eta_1 + \eta_2 = \frac{100}{473} + \frac{100}{373} \approx 0.2114 + 0.2681 = 0.4795$$

Since $$0.4228 \neq 0.4795$$, they are not equal. Thus, option C is incorrect.

Option D: $$\eta_{12} \leq \eta_1 + \eta_2$$

From the comparison in option B, $$\eta_{12} < \eta_1 + \eta_2$$, which implies $$\eta_{12} \leq \eta_1 + \eta_2$$ is true.

Therefore, option D is correct.

The correct answer is option D.

For an adiabatic process of an ideal gas, the heat exchanged with the surroundings is zero: $$Q = 0$$.

First law of thermodynamics: $$\Delta U = Q + W$$, where $$W$$ is the work done on the gas.
Because $$Q = 0$$, we have

$$\Delta U = W \quad -(1)$$

For a monoatomic ideal gas, the internal energy is $$U = \dfrac{3}{2}nRT$$. Hence

$$\Delta U = \dfrac{3}{2}nR\,(T_2 - T_1) \quad -(2)$$

Using $$(1)$$ and $$(2)$$, the work done on the gas is

$$W = \dfrac{3}{2}nR\,(T_2 - T_1) \quad -(3)$$

Thus, $$W$$ is directly proportional to $$(T_2 - T_1)$$.

For an adiabatic change of an ideal gas, the pressure-volume relation is $$PV^{\gamma} = \text{constant}$$, where $$\gamma = \dfrac{C_P}{C_V}$$. For a monoatomic gas, $$\gamma = \dfrac{5}{3}$$. Consequently

$$TV^{\gamma - 1} = TV^{2/3} = \text{constant}$$,
which is not simply $$TV = \text{constant}$$, and $$PV$$ is also not constant.

Now examine each statement:

Statement (A): “Internal energy is constant.”
From $$(2)$$, $$\Delta U \neq 0$$ when temperature changes, so (A) is false.

Statement (B): “Work done in the process is equal to the change in internal energy.”
Equation $$(1)$$ shows $$W = \Delta U$$ (with the ‘work on the gas’ sign convention). Hence (B) is true.

Statement (C): “The product of temperature and volume is a constant.”
We have $$TV^{2/3} = \text{constant}$$, not $$TV$$. Therefore (C) is false.

Statement (D): “The product of pressure and volume is a constant.”
For adiabatic change, $$PV^{5/3} = \text{constant}$$, so $$PV$$ is not constant. (D) is false.

Statement (E): “The work done to change the temperature from $$T_1$$ to $$T_2$$ is proportional to $$(T_2 - T_1)$$.”
Equation $$(3)$$ confirms this proportionality. Hence (E) is true.

Therefore, only statements (B) and (E) are correct.

Option C (B, E only) is the correct choice.

The process is sudden and the container is thermally insulated, so no heat enters or leaves the gas. Such a process is adiabatic.

For an adiabatic change in an ideal gas, the equation is
$$P V^{\gamma} = \text{constant}$$

Let the initial state be $$(P_1 , V_1)$$ and the final state be $$(P_2 , V_2)$$. Writing the adiabatic relation for the two states gives
$$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$$

Re-arrange to obtain the pressure ratio:
$$\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^{\gamma}$$ $$-(1)$$

The gas is compressed to $$\left(\frac{1}{8}\right)^{\text{th}}$$ of its original volume, so
$$V_2 = \frac{V_1}{8}$$ ⇒ $$\frac{V_1}{V_2} = 8$$

Given $$\gamma = \frac{5}{3}$$ for a monoatomic gas, substitute these values in $$(1)$$:
$$\frac{P_2}{P_1} = 8^{\,\frac{5}{3}}$$

Write 8 as $$2^3$$ and simplify:
$$8^{\,\frac{5}{3}} = \left(2^3\right)^{\frac{5}{3}} = 2^{\,3 \times \frac{5}{3}} = 2^5 = 32$$

Hence the ratio of the final pressure to the initial pressure is $$32$$.

Therefore, Option C is correct.

Since the gas is heated isothermally,

$$ΔU=0$$

(for ideal gas internal energy depends only on temperature)

So heat supplied by filament equals work done by gas:

$$Q=W$$

Total work done by gas has two parts:

1. Work done against atmosphere/piston weight
2. Work done in stretching spring

At equilibrium initially,

$$P_0A=Mg$$

where A is piston area.

Gas expands from height $$L_0$$ to $$L_1$$​

Change in volume:

Work against piston weight:

$$W_1=Mg(L_1-L_0)$$

Spring force law:

$$F=kx^3$$

Initially spring is natural length, so initial extension

x=0

Final extension:

$$x=L_1-L_0$$

Work stored in spring:

$$W_2​=\int kx^3dx\ \ from\ 0\ to\ L_{1\ }-L_0$$

$$\frac{k}{4}(L_1-L_0)^4$$

Also because process is reversible isothermal,

$$W=nRT\ln\frac{V_1}{V_0}$$

with

so

$$W_3=nRT\ln\frac{L_1}{L_0}$$

This is the expansion work of ideal gas.

Hence total energy supplied by filament

A. Pressure varies inversely with volume of an ideal gas.

For an isothermal process,

So

$$P\propto\frac{1}{V}$$

A → III

B. Heat absorbed goes partly to increase internal energy and partly to do work.

That is true for isobaric process:

Both $$ΔU$$ and W are generally nonzero.

B → IV

C. Heat is neither absorbed nor released.

That is adiabatic process:

Q=0

C → I

D. No work is done on or by gas.

In isochoric process volume is constant:

$$W=P\Delta V=0$$

D → II

For an adiabatic process involving an ideal gas, the relation between temperature and volume is
$$T\,V^{\gamma-1}= \text{constant}$$

Given data:
Initial volume $$V_1 = 800\;\text{cm}^3$$
Final volume $$V_2 = 200\;\text{cm}^3$$
Initial temperature $$T_1 = 27^\circ\text{C} = 27 + 273 = 300\;\text{K}$$
Specific-heat ratio $$\gamma = 1.5$$

Applying the adiabatic relation for the final temperature $$T_2$$:
$$T_1\,V_1^{\gamma-1}=T_2\,V_2^{\gamma-1}$$

Solve for $$T_2$$:
$$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$

Compute the volume ratio and exponent:
$$\frac{V_1}{V_2} = \frac{800}{200} = 4$$
$$\gamma - 1 = 1.5 - 1 = 0.5$$

Therefore,
$$T_2 = 300\,\text{K}\;\times\;4^{0.5} = 300\,\text{K}\;\times\;2 = 600\,\text{K}$$

The change in temperature is
$$\Delta T = T_2 - T_1 = 600\,\text{K} - 300\,\text{K} = 300\,\text{K}$$

Hence, the temperature rises by $$300\;\text{K}$$.
Option D is correct.

For any thermodynamic process the first law of thermodynamics states
$$dQ = dU + dW$$ where

$$dQ$$ = heat supplied to the system, $$dU$$ = change in internal energy and $$dW$$ = work done by the system.

Adiabatic process: By definition no heat is exchanged with the surroundings, hence
$$dQ = 0$$ $$-(1)$$

Putting $$dQ = 0$$ in the first-law equation $$-(1)$$ gives
$$0 = dU + dW \;\;\Longrightarrow\;\; dU = -\,dW$$ $$-(2)$$

Equation $$-(2)$$ shows that the work done by the gas is numerically equal to the decrease (negative) in its internal energy, or equivalently, the increase in internal energy is numerically equal to the negative of the work done.

Now recall the definition of molar heat capacity for any path:
$$C = \frac{1}{n}\,\frac{dQ}{dT}$$ where $$n$$ is the number of moles.

For an adiabatic process $$dQ = 0$$, therefore
$$C = \frac{1}{n}\,\frac{0}{dT} = 0$$ $$-(3)$$

Hence the molar heat capacity along an adiabatic path is zero.

Let us test each option:

Option A The molar heat capacity is infinite - contradicts $$-(3)$$ (false).

Option B Work done by the gas equals the increase in internal energy - according to $$-(2)$$ the signs are opposite (false).

Option C The molar heat capacity is zero - matches $$-(3)$$ (true).

Option D The internal energy of the gas decreases as the temperature increases - no such rule exists; internal energy is directly related to temperature for an ideal gas (false).

Therefore the correct statement is Option C: the molar heat capacity in an adiabatic process is zero.

Path is:

$$A(2V_0,P_0)→B(3V_0,P_0)→C(3V_0,2P_0)→D(V_0,2P_0)$$

Work done is

$$W=\int PdV$$

Calculate segment-wise.

Along AB (constant pressure $$P_0$$​):

$$W_{AB}=P_0(3V_0-2V_0)=P_0V_0$$

Along BC (constant volume):

$$W_{BC}=0$$

Along CD (constant pressure $$2P_0$$ compression):

$$W_{CD}=2P_0(V_0-3V_0)$$

$$=2P_0(-2V_0)=-4P_0V_0$$

Total work:

$$W=W_{AB}+W_{BC}+W_{CD}$$

$$P_0V_0-4P_0V_0$$

$$=-3P_0V_0$$​

Using Newton's law of cooling in the approximate form:

$$ \frac{\Delta T}{t} = k\left(\bar{T} - T_s\right) $$

where $$\bar{T}$$ is the average temperature and $$T_s$$ is the surrounding temperature.

First case: Cooling from 90° to 80° in time $$t$$:

$$\bar{T} = 85°$$, $$\Delta T = 10°$$

$$ \frac{10}{t} = k(85 - 20) = 65k \quad \cdots (1) $$

Second case: Cooling from 80° to 60° in time $$t'$$:

$$\bar{T} = 70°$$, $$\Delta T = 20°$$

$$ \frac{20}{t'} = k(70 - 20) = 50k \quad \cdots (2) $$

Dividing (2) by (1):

$$ \frac{20/t'}{10/t} = \frac{50k}{65k} $$

$$ \frac{2t}{t'} = \frac{10}{13} $$

$$ t' = \frac{2t \times 13}{10} = \frac{26t}{10} = \frac{13t}{5} $$

The correct answer is Option 4: $$\frac{13}{5}t$$.

For an adiabatic process, $$Q = 0$$ (no heat exchange with the surroundings).

From the first law of thermodynamics:

$$\Delta U = Q - W = -W$$

$$W = -\Delta U$$

For an ideal gas, the internal energy depends only on temperature:

$$\Delta U = nC_v\Delta T$$

Therefore:

$$W = -nC_v\Delta T$$

This shows that the work done in an adiabatic process depends only on the change in temperature (for a given amount of ideal gas with fixed $$C_v$$).

The answer is Option A: change in its temperature.

The first law of thermodynamics gives the relation $$\Delta Q = \Delta U + \Delta W$$ where:
  • $$\Delta Q$$ is the heat supplied to the system,
  • $$\Delta U$$ is the change in internal energy,
  • $$\Delta W$$ is the work done by the system.

For each thermodynamic process we check which of these three quantities is zero or non-zero.

Case A: Isothermal process

Definition: Temperature remains constant.
For an ideal gas, internal energy depends only on temperature, so when temperature is constant $$\Delta U = 0$$.
Heat supplied equals work done, but both are generally non-zero.
Hence the correct description is $$\Delta U = 0$$, i.e. List-II item (IV).

Case B: Adiabatic process

Definition: No heat is exchanged with the surroundings.
Therefore $$\Delta Q = 0$$.
Neither $$\Delta U$$ nor $$\Delta W$$ is automatically zero; they are related by $$\Delta U = -\Delta W$$.
Thus the matching statement is $$\Delta Q = 0$$, i.e. List-II item (II).

Case C: Isobaric process

Definition: Pressure remains constant.
At constant pressure, the gas expands or compresses, so work is done: $$\Delta W \ne 0$$.
Temperature usually changes, so internal energy changes: $$\Delta U \ne 0$$.
Only $$\Delta U \ne 0$$ is listed among the four given statements, therefore we match with List-II item (III).

Case D: Isochoric process

Definition: Volume remains constant.
Work done for a volume change is $$\Delta W = P\,\Delta V$$. With $$\Delta V = 0$$ we get $$\Delta W = 0$$.
Heat added goes entirely into changing internal energy: $$\Delta Q = \Delta U$$, so neither of them is necessarily zero.
Thus the matching statement is $$\Delta W = 0$$, i.e. List-II item (I).

Collecting the matches:
(A) Isothermal → (IV) $$\Delta U = 0$$
(B) Adiabatic → (II) $$\Delta Q = 0$$
(C) Isobaric → (III) $$\Delta U \ne 0$$
(D) Isochoric → (I) $$\Delta W = 0$$

This corresponds to Option C.

Final Answer: Option C   [ (A)-(IV), (B)-(II), (C)-(III), (D)-(I) ]

From the graph,$$A→B$$ is adiabatic, so no heat is exchanged in this part:

$$Q_{AB}=0$$

Hence net heat absorbed by the gas comes only during $$B→C$$, which is an isothermal expansion.

For one mole of an ideal gas in an isothermal process,

$$Q=W=RT\ln\frac{V_C}{V_B}$$

From the diagram, the path $$B→C$$ lies on the $$450K$$ isotherm, so

$$T=450K$$

Also,

$$\frac{V_C}{V_B}=\frac{8\times10^{-6}}{6\times10^{-6}}=\frac{4}{3}$$

Therefore,

$$Q=450R\ln\frac{4}{3}$$

The first law of thermodynamics for a quasi-static process is written as $$\Delta Q = \Delta U + \Delta W$$, where $$\Delta W$$ is the work done by the gas.

For expansion/compression work at constant external pressure, $$\Delta W = P\,\Delta V$$, so the first law becomes $$\Delta Q = \Delta U + P\,\Delta V$$ $$-(1)$$.

Now analyse each process in List-I.

Case A: Isobaric (constant pressure)

Pressure stays constant, therefore $$P = \text{constant}$$. Substituting in $$(1)$$ gives
$$\Delta Q = \Delta U + P\,\Delta V$$.
This is exactly the expression in List-II (IV). Hence A → IV.

Case B: Isochoric (constant volume)

Here $$\Delta V = 0$$, so no work is done: $$\Delta W = P\,\Delta V = 0$$.
Using the first law: $$\Delta Q = \Delta U + 0 \; \Rightarrow \; \Delta Q = \Delta U$$.
This matches List-II (II). Hence B → II.

Case C: Adiabatic (no heat exchange)

By definition, $$\Delta Q = 0$$ for an adiabatic process.
Thus the relation in List-II (III) fits. Hence C → III.

Case D: Isothermal (constant temperature)

For an ideal gas, internal energy depends only on temperature. Since temperature is constant,
$$\Delta U = 0$$.
First law then gives $$\Delta Q = \Delta W$$.
This is relation (I) in List-II. Hence D → I.

Collecting the matches: (A-IV), (B-II), (C-III), (D-I).

The option with this sequence is Option C.

Final Answer: Option C

We need to evaluate the Assertion and Reason about adiabatic processes.

Assertion (A): "In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases."

Analysis: The term "shrunk to half its volume" in an insulated container describes a free expansion or compression scenario. However, "adiabatically shrunk" means the gas is compressed (volume decreases). For adiabatic compression of an ideal gas, using the relation $$TV^{\gamma - 1} = \text{constant}$$:

$$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$$

Since $$V_2 = V_1/2$$ (volume halved) and $$\gamma > 1$$:

$$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = T_1 \cdot 2^{\gamma - 1} > T_1$$

The temperature increases during adiabatic compression, not decreases. Assertion (A) is FALSE.

Reason (R): "Free expansion of an ideal gas is an irreversible and an adiabatic process."

Analysis: Free expansion occurs when a gas expands into a vacuum. It is indeed both irreversible (spontaneous, cannot be reversed without external work) and adiabatic (no heat exchange since $$Q = 0$$; also $$W = 0$$ against vacuum). For an ideal gas, free expansion produces no temperature change ($$\Delta U = 0$$). Reason (R) is TRUE.

The correct answer is Option 2: (A) is false but (R) is true.

For an adiabatic process (sudden compression), the relation between temperature and volume is $$ TV^{\gamma - 1} = \text{constant} $$. In this case, $$\gamma = \frac{C_p}{C_v} = \frac{3}{2}$$, the initial temperature is $$T_1 = 0°C = 273$$ K, and the gas is compressed to one-fourth of its original volume, so $$V_2 = \frac{V_1}{4}$$.

Applying $$ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} $$ gives $$ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = 273 \times (4)^{3/2 - 1} = 273 \times 4^{1/2} = 273 \times 2 = 546 \text{ K} $$.

Therefore, the change in temperature is $$ \Delta T = T_2 - T_1 = 546 - 273 = 273 \text{ K} $$. The temperature rise is 273 K.

For an ideal gas,

$$PV=nRT$$

For an isobaric process (P= constant),

$$V=\frac{nR}{P}T$$

This is a straight line on a V-T graph, with slope

$$slope=\frac{nR}{P}$$

Since same mass of gas is used, n is constant, so slope is inversely proportional to pressure:

$$\text{slope}\propto\frac{1}{P}$$

Line $$P_2$$​ is steeper than line $$P_1$$, so

$$\frac{nR}{P_2}>\frac{nR}{P_1}$$

which gives

$$P_1>P_2$$

For a real gas, the relation between the molar specific heats is

$$C_P - C_V \;=\; T\Bigl(\frac{\partial P}{\partial T}\Bigr)_V \Bigl(\frac{\partial V}{\partial T}\Bigr)_P$$

This identity is obtained from the first & second laws together with exact-differential criteria, and it is valid for one mole of any substance.

The gas in the question obeys the equation of state

$$P\,V^2 \;=\; R\,T \;\;-(1)$$

Step 1: Find $$\Bigl(\dfrac{\partial P}{\partial T}\Bigr)_V$$.

From $$(1)$$: $$P \;=\; \dfrac{R\,T}{V^{2}}$$. Keeping $$V$$ constant and differentiating with respect to $$T$$:

$$\Bigl(\frac{\partial P}{\partial T}\Bigr)_V = \frac{R}{V^{2}} \;\;-(2)$$

Step 2: Find $$\Bigl(\dfrac{\partial V}{\partial T}\Bigr)_P$$.

Rewrite $$(1)$$ to express $$V$$ in terms of $$T$$ at fixed $$P$$:

$$V^{2} = \frac{R\,T}{P} \quad\Longrightarrow\quad 2V\,dV = \frac{R}{P}\,dT$$ so

$$\Bigl(\frac{\partial V}{\partial T}\Bigr)_P = \frac{R}{2\,P\,V} \;\;-(3)$$

Step 3: Compute $$C_P - C_V$$.

Substitute $$(2)$$ and $$(3)$$ into the heat-capacity relation:

$$\begin{aligned} C_P - C_V &= T\left(\frac{R}{V^{2}}\right)\left(\frac{R}{2\,P\,V}\right)\\[4pt] &= \frac{T\,R^{2}}{2\,P\,V^{3}} \;\;-(4) \end{aligned}$$

Step 4: Eliminate $$P$$ using the equation of state.

From $$(1)$$: $$P = \dfrac{R\,T}{V^{2}}$$. Insert this in $$(4)$$:

$$\begin{aligned} C_P - C_V &= \frac{T\,R^{2}}{2\,(\tfrac{R\,T}{V^{2}})\,V^{3}}\\[6pt] &= \frac{T\,R^{2}}{2\,R\,T\,V}\\[4pt] &= \frac{R}{2\,V} \end{aligned}$$

Therefore

$$C_P = C_V + \frac{R}{2\,V}$$

Among the given options this matches Option C.

Answer: Option C

For an ideal gas,

$$PV=nRT$$

Also density,

$$\rho=\frac{m}{V}$$

and

$$n=\frac{m}{M}$$

So ideal gas law becomes

$$P=\frac{\rho}{M}RT$$

or

$$P=\left(\frac{\rho R}{M}\right)T$$

This is the equation of a straight line in a P-T graph, whose slope is

$$slope=\frac{\rho R}{M}$$

Since R and molar mass M are constant, slope is directly proportional to density:

$$slope∝ρ$$

From the graph, line corresponding to $$\rho_1$$​ has greatest slope, then

$$\rho_2\ then\ \rho_3$$

Hence,

$$\rho_1>\rho_2>\rho_3$$

For a cyclic process,

ΔU=0

So from first law,

Q=W

Thus heat absorbed equals net work done, which is area enclosed by the cycle on the P-V diagram.

The loop is a circle touching

P=340,  P=60 and

V=340,  V=60

So diameter is

$$340−60=280$$

Hence radius is

r=140

Area enclosed

$$W=\pi r^2$$

$$=\pi(140)^2$$

$$=19600\pi$$

Now units:

Pressure is in cc and volume in kPa (as labeled), so

$$1(\text{kPa})(\text{cc})=10^3\times10^{-6}=10^{-3}J$$

Thus

$$Q=19600\pi\times10^{-3}$$

$$=19.6\pi\ \text{J}=61.6J$$

The diagram (not shown here) is a rectangle in the $$P$$-$$T$$ plane with the numerical labels

$$J:(P_0,T_0),\;K:(P_0,3T_0),\;L:(2P_0,3T_0),\;M:(2P_0,T_0).$$

One mole of a monatomic ideal gas moves along the four sides JK, KL, LM, MJ in that order.

For one mole the ideal-gas equation is $$PV = RT$$ and the internal energy is $$U = \tfrac32 RT.$$ Work done by the gas in any step is $$W = \int P\,dV,$$ and the first law gives $$Q = \Delta U + W.$$ We treat every leg separately.

Temperatures: $$T_0 \rightarrow 3T_0 \;(\Delta T = 2T_0).$$
Change in internal energy $$\Delta U_{JK} = \tfrac32 R\,(2T_0) = 3RT_0.$$

Volume rises from $$V_J = \frac{RT_0}{P_0}$$ to $$V_K = \frac{3RT_0}{P_0}.$$ Work done $$W_{JK} = P_0\,(V_K-V_J) = P_0 \left(\frac{2RT_0}{P_0}\right)=2RT_0.$$

Pressure $$P_0 \rightarrow 2P_0,$$ so volume $$V_K = \frac{3RT_0}{P_0}$$ compresses to $$V_L = \frac{3RT_0}{2P_0} = \tfrac12 V_K.$$ Work done $$W_{KL}=RT\ln\!\left(\frac{V_L}{V_K}\right)=3RT_0\ln\!\left(\tfrac12\right)=-3RT_0\ln2.$$

Because the temperature is constant, $$\Delta U_{KL}=0,$$ hence $$Q_{KL}=W_{KL}=-3RT_0\ln2.$$

Temperature $$3T_0 \rightarrow T_0 \;(\Delta T = -2T_0).$$
Change in internal energy $$\Delta U_{LM}=\tfrac32 R(-2T_0)=-3RT_0.$$

Volumes: $$V_L=\frac{3RT_0}{2P_0},\;V_M=\frac{RT_0}{2P_0}$$ so $$\Delta V = -\frac{RT_0}{P_0}.$$ Work done $$W_{LM}=2P_0\left(-\frac{RT_0}{P_0}\right)=-2RT_0.$$

Pressure $$2P_0 \rightarrow P_0,$$ volume doubles: $$V_M=\frac{RT_0}{2P_0},\;V_J=\frac{RT_0}{P_0}=2V_M.$$ Work done $$W_{MJ}=RT_0\ln\!\left(\frac{V_J}{V_M}\right)=RT_0\ln2.$$

Again $$\Delta U_{MJ}=0$$ (temperature constant), so $$Q_{MJ}=W_{MJ}=RT_0\ln2.$$

$$W_{\text{cycle}} = 2RT_0 + (-3RT_0\ln2) + (-2RT_0) + RT_0\ln2$$ $$\phantom{W_{\text{cycle}}}=0 - 2RT_0\ln2 = -2RT_0\ln2.$$

Now we can match each quantity with List-II:

(P) Work in the entire cycle: $$-2RT_0\ln2$$ → (4)
(Q) $$\Delta U_{JK}=3RT_0$$ → (3)
(R) Heat in KL: $$Q_{KL}=-3RT_0\ln2$$ → (5)
(S) $$\Delta U_{MJ}=0$$ → (2)

Therefore the correct set of matches is P → 4; Q → 3; R → 5; S → 2, which is Option B.

or a cyclic process, net work done is the sum of work in each path.

Cycle is

$$C→A→B→C$$

For C→A, volume is constant, so

$$W_{CA}=0$$

For A→B, given

$$PV^3=RT$$

Temperature is

$$27^{\circ}C=300K$$

and

R=8

So

$$PV^3=8(300)=2400$$

Thus

$$P=\frac{2400}{V^3}$$

Work done from A to B is

$$W_{AB}=\int PdV$$

$$=\int_2^4\frac{2400}{V^3}dV$$

$$=2400\int_2^4V^{-3}dV$$

$$=2400\left(\frac{-1}{2V^2}\right)_2^4$$

$$2400\left(\frac{1}{8}-\frac{1}{32}\right)$$

$$=2400\cdot\frac{3}{32}$$

=225J

For B→C, pressure is constant at

P=10

So

=10(2−4)

=−20J

Hence net work done in the complete cycle is

W=0+225−20W

=205J

Match each definition carefully.

(A) “A force that restores an elastic body of unit area to its original state”

Force per unit area developed internally to restore original shape is definition of stress.

So

A→(III)

(B) “Two equal and opposite forces parallel to opposite faces”

Parallel tangential forces produce shear deformation, associated with shear modulus.

So

B→(IV)

(C) “Forces perpendicular everywhere to the surface per unit area same everywhere”

This describes uniform normal pressure acting in all directions, associated with bulk modulus.

So

C→(I)

(D) “Two equal and opposite forces perpendicular to opposite faces”

Normal tensile/compressive stress produces longitudinal strain, associated with Young’s modulus.

So

D→(II)

To just reach infinity with zero final speed, total mechanical energy at infinity should be zero.

At Earth’s surface:

Potential energy is

$$U=-\frac{GMm}{R_E}$$

Required kinetic energy = escape energy so that

$$K+U=0$$

Thus

$$K=\frac{GMm}{R_E}$$

Now use

$$g=\frac{GM}{R_E^2}$$

So

$$GM=gR_E^2$$

Substitute:

$$K=\frac{(gR_E^2)m}{R_E}$$

$$K=mgR_E$$

For heating at constant volume, the change in internal energy equals the heat added:

$$\Delta U = m c_v \Delta T$$

Given: $$m = 0.08$$ kg, $$c_v = 0.17$$ kcal kg$$^{-1}$$ °C$$^{-1}$$, $$\Delta T = 5°C$$.

$$\Delta U = 0.08 \times 0.17 \times 5 = 0.068 \text{ kcal}$$

Converting to joules ($$1 \text{ cal} = 4.18 \text{ J}$$, so $$1 \text{ kcal} = 4180 \text{ J}$$):

$$\Delta U = 0.068 \times 4180 = 284.24 \text{ J} \approx 284 \text{ J}$$

The answer is approximately $$284$$ J, which corresponds to Option (3).

For an isobaric process (constant pressure), the work done by the gas is given by:

$$W = P \Delta V$$

Using the ideal gas law, $$PV = nRT$$, we can express the work as:

$$W = P \Delta V = nR \Delta T$$

Given that the work done by the gas is 200 J:

$$nR \Delta T = 200 \text{ J} \quad \text{(Equation 1)}$$

The heat supplied to the gas in an isobaric process is:

$$Q = n C_p \Delta T$$

where $$C_p$$ is the molar specific heat at constant pressure.

For a diatomic gas with $$\gamma = 1.4$$, the ratio of specific heats is $$\gamma = \frac{C_p}{C_v} = \frac{7}{5}$$. The molar specific heat at constant volume is $$C_v = \frac{5}{2}R$$ (since diatomic gases have 5 degrees of freedom). Therefore:

$$C_p = \gamma C_v = \frac{7}{5} \times \frac{5}{2}R = \frac{7}{2}R$$

Substituting into the expression for Q:

$$Q = n \left( \frac{7}{2}R \right) \Delta T = \frac{7}{2} (nR \Delta T)$$

Using Equation 1, where $$nR \Delta T = 200 \text{ J}$$:

$$Q = \frac{7}{2} \times 200 = 7 \times 100 = 700 \text{ J}$$

Thus, the heat given to the gas is 700 J.

Verification using the first law of thermodynamics:

The change in internal energy is:

$$\Delta U = n C_v \Delta T = n \left( \frac{5}{2}R \right) \Delta T = \frac{5}{2} (nR \Delta T) = \frac{5}{2} \times 200 = 500 \text{ J}$$

From the first law, $$\Delta U = Q - W$$:

$$500 = Q - 200 \implies Q = 700 \text{ J}$$

This confirms the result.

The correct option is D. 700 J.

Find the work done when a gas at temperature $$T$$ is adiabatically expanded to double its volume with $$\gamma = 3/2$$ and $$\mu = 1$$ mole.

For an adiabatic process with an ideal gas, the relation $$TV^{\gamma-1} = \text{constant}$$ implies

$$ T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1} $$

With $$V_f = 2V_i$$ and $$\gamma - 1 = 3/2 - 1 = 1/2$$, we have

$$ T \cdot V_i^{1/2} = T_f \cdot (2V_i)^{1/2} $$

$$ T_f = T \cdot \frac{V_i^{1/2}}{(2V_i)^{1/2}} = T \cdot \frac{1}{\sqrt{2}} = \frac{T}{\sqrt{2}} $$

For an adiabatic process, $$Q = 0$$, so the work done by the gas is $$W = -\Delta U$$. Equivalently, integrating $$W = \int P\,dV$$ with $$PV^\gamma = \text{const}$$ yields

$$W = \int P\,dV = \frac{P_iV_i - P_fV_f}{\gamma-1} = \frac{nRT_i - nRT_f}{\gamma-1} = \frac{nR(T_i-T_f)}{\gamma-1}$$

Substituting $$n = 1$$, $$T_i = T$$, $$T_f = T/\sqrt{2}$$, and $$\gamma - 1 = 1/2$$ gives

$$ W = \frac{1 \times R \times (T - T/\sqrt{2})}{1/2} = 2R \cdot T\left(1 - \frac{1}{\sqrt{2}}\right) = 2RT\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) = RT(2 - \sqrt{2}) $$

The correct answer is Option C: $$RT[2 - \sqrt{2}]$$.

Work done is the area under the P-V curve.

From A→BA\to BA→B, pressure changes linearly, so work done is area of trapezium:

$$W_{AB}=\frac{P_A+P_B}{2}(V_B-V_A)$$

Given

$$P_A=8000\ \text{dyne/cm}^2,\quad P_B=4000\ \text{dyne/cm}^2$$

$$V_A=3m^3\ and\ V_B=7m^3$$

So

$$W_{AB}=\frac{8000+4000}{2}(7-3)$$

$$=6000\times4=24000$$

From B→C, process is isobaric compression, so

$$W_{BC}=P(V_C-V_B)$$

$$=4000(3−7)$$

=−16000

Therefore total work done,

$$W=W_{AB}+W_{BC}$$

$$=24000−16000=8000$$

Now converting units:

$$1\ \text{dyne/cm}^2=0.1\ \text{Pa}$$

So

$$8000\ \text{dyne/cm}^2=800\ \text{Pa}$$

Thus,

$$W=800(m^3)=800J$$

We need to find the work done by one mole of helium gas when 48 J of heat is supplied and the temperature increases by 2°C.

We know that according to the First Law of Thermodynamics $$Q = \Delta U + W$$, where $$Q$$ is heat supplied, $$\Delta U$$ is the change in internal energy, and $$W$$ is the work done by the gas. For an ideal gas, the change in internal energy depends only on temperature change, giving $$\Delta U = n C_v \Delta T$$. Since helium is a monoatomic ideal gas with three translational degrees of freedom, its molar heat capacity at constant volume is $$C_v = \frac{3}{2}R$$.

Substituting $$n = 1$$ mol, $$\Delta T = 2$$ K (because 2°C corresponds to 2 K), and $$R = 8.3 \, \text{J K}^{-1}\text{mol}^{-1}$$ into the expression for the change in internal energy yields $$\Delta U = n C_v \Delta T = 1 \times \frac{3}{2} \times 8.3 \times 2$$.

This gives $$\Delta U = 1 \times 1.5 \times 8.3 \times 2 = 1 \times 12.45 \times 2 = 24.9 \, \text{J}$$.

Applying the First Law then results in the work done by the gas as $$W = Q - \Delta U = 48 - 24.9 = 23.1 \, \text{J}$$.

The correct answer is Option (4): 23.1 J.

Yes — this question was asking to identify the processes, not compare work/heat.

From the graph, process A is the less steep hyperbola, corresponding to an isothermal process:

$$PV=k$$

For process BBB, the curve is steeper than an isotherm, so it represents an adiabatic process:

$$PV^{\gamma}=k$$

because adiabatic curves are steeper than isotherms on a P-V diagram.

For an adiabatic process,

$$PV^{\gamma}=\text{constant}$$

Also ideal gas law gives

PV=nRT

or

$$V=\frac{nRT}{P}$$

Substitute into adiabatic equation:

$$P\left(\frac{nRT}{P}\right)^{\gamma}=\text{constant}$$

$$P\cdot\frac{(nR)^{\gamma}T^{\gamma}}{P^{\gamma}}=\text{constant}$$

$$P^{1-\gamma}T^{\gamma}=\text{constant}$$

Given pressure is proportional to cube of absolute temperature,

$$P\propto T^3$$

So

$$P=kT^3$$

Comparing with

$$P^{1-\gamma}T^{\gamma}=\text{constant}$$

write

$$P\propto T^{\frac{\gamma}{\gamma-1}}$$

Thus,

$$=3\frac{\gamma}{\gamma-1}$$

Solving,

$$γ=3γ−3$$

$$2γ=3$$

$$γ=\frac{3}{2}$$

But for an adiabatic process so,

$$\gamma=\frac{C_P}{C_V}$$

For an adiabatic process: $$PV^\gamma = \text{constant}$$ and $$TV^{\gamma-1} = \text{constant}$$.

From the ideal gas law: $$PV = nRT$$, so $$V = \frac{nRT}{P}$$.

Substituting: $$P\left(\frac{nRT}{P}\right)^\gamma = \text{const}$$

$$P^{1-\gamma}T^\gamma = \text{const}$$

$$\frac{T^\gamma}{P^{\gamma-1}} = \text{const}$$

$$P^{\gamma-1} \propto T^\gamma$$

$$P \propto T^{\gamma/(\gamma-1)}$$

Given $$P \propto T^3$$:

$$\frac{\gamma}{\gamma - 1} = 3$$

$$\gamma = 3\gamma - 3$$

$$2\gamma = 3$$

$$\gamma = \frac{3}{2}$$

The answer is $$\frac{C_p}{C_v} = \frac{3}{2}$$, which corresponds to Option (2).

Average translational KE: $$K = \frac{3}{2}k_BT$$.

Initial: $$T_1 = -78°C = 195$$ K, $$K_1 = K$$.

For $$K_2 = 2K$$: $$T_2 = 2T_1 = 390$$ K = $$117°C$$.

The correct answer is Option 2: $$117°C$$.

We need to find the total energy of 10 non-rigid diatomic molecules at temperature $$T$$ by applying the equipartition theorem.

According to the equipartition theorem, each degree of freedom contributes $$\frac{1}{2}k_B T$$ of energy per molecule, and the total energy per molecule is therefore $$E = \frac{f}{2} k_B T$$, where $$f$$ is the number of degrees of freedom and $$k_B$$ is Boltzmann's constant.

In a non-rigid diatomic molecule, there are three translational degrees of freedom (motion along the x, y, and z axes), two rotational degrees of freedom (rotation about the two axes perpendicular to the bond axis, since rotation about the bond axis is negligible), and two vibrational degrees of freedom (one for kinetic energy and one for potential energy). Hence, the total number of degrees of freedom is $$f = 3 + 2 + 2 = 7$$.

Substituting $$f = 7$$ into the expression for the energy per molecule gives $$E_{\text{per molecule}} = \frac{7}{2} k_B T$$.

For 10 molecules, this becomes $$E_{\text{total}} = 10 \times \frac{7}{2} k_B T = \frac{70}{2} k_B T = 35 \, k_B T$$.

The correct answer is Option (2): $$35 K_B T$$.

PV^(3/2) = K. W = ∫PdV = ∫K/V^(3/2)dV = K[-2/V^(1/2)] = -2K(1/√V₂ - 1/√V₁)

= 2K(1/√V₁ - 1/√V₂) = 2(P₁V₁^(3/2)/√V₁ - P₂V₂^(3/2)/√V₂) = 2(P₁V₁ - P₂V₂).

The correct answer is Option 1.

For an adiabatic process: $$PV^{\gamma} = \text{constant}$$.

$$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$$

$$\frac{P_1}{P_2} = \left(\frac{V_2}{V_1}\right)^{\gamma} = \left(\frac{4}{5}\right)^{3/2}$$

$$= \frac{4^{3/2}}{5^{3/2}} = \frac{8}{5\sqrt{5}}$$

The correct answer is Option 3: $$\frac{8}{5\sqrt{5}}$$.

For an adiabatic process,

$$PV^{\gamma}=\text{constant}$$

There are two adiabatic curves:

For adiabatic through a and c:

$$P_aV_a^{\gamma}=P_cV_c^{\gamma}$$

So,

$$\frac{P_a}{P_c}=\frac{V_c^{\gamma}}{V_a^{\gamma}}$$

For adiabatic through b and d:

$$P_bV_b^{\gamma}=P_dV_d^{\gamma}$$

So,

$$\frac{P_b}{P_d}=\frac{V_d^{\gamma}}{V_b^{\gamma}}$$

Now a and b lie on same isotherm, so

$$P_aV_a=P_bV_b$$

$$\frac{P_a}{P_b}=\frac{V_b}{V_a}$$

Also c and d lie on same isotherm, so

$$P_cV_c=P_dV_d$$

$$\frac{P_c}{P_d}=\frac{V_d}{V_c}$$

Now divide the two adiabatic equations:

$$\frac{P_a/P_c}{P_b/P_d}=\left(\frac{\frac{V_C}{V_a}}{\frac{V_d}{V_b}}\right)^{\gamma}$$

Using isothermal relations,

$$\frac{(V_c/V_a)}{(V_d/V_b)}=\left(\frac{V_bV_c}{V_aV_d}\right)^{\gamma}$$

This simplifies only if

$$V_bV_c=V_aV_d$$

Hence,

$$\frac{V_a}{V_d}=\frac{V_b}{V_c}$$

In the graph, the lines have positive slope in the (log⁡V,log⁡P) plane, so they represent

$$\log P=n\log V$$

or

$$P=V^n$$

which can be written as

This is a polytropic process with index

m=−n

For a polytropic process, molar heat capacity is

Substituting m=−n,

$$C=C_V+\frac{R}{1+n}$$

For process A, slope is given by

$$\tan⁡θ=γ$$

so

$$n=γ$$

Hence

$$C_A=C_V+\frac{R}{1+\gamma}$$

For process BBB, angle is $$45^{\circ}$$, so

n=1

and

$$C_B=C_V+\frac{R}{2}$$

Also,

$$C_P=C_V+R$$

Now comparing:

$$C_P>C_B$$

because

$$R>\frac{R}{2}$$

Also since γ>1,

$$\frac{R}{2}>\frac{R}{1+\gamma}$$

so

$$C_B>C_A$$

and clearly

$$C_A>C_V$$

Therefore,

$$CP>CB>CA>CV$$

We need to evaluate two statements about thermodynamics.

Statement I: "If heat is added to a system, its temperature must increase."

This is FALSE. Consider the following counterexamples:

- During a phase change (e.g., melting of ice at 0°C), heat is added but the temperature remains constant.

- In an isothermal expansion of an ideal gas, heat is added to do work, but the temperature stays constant.

Statement II: "If positive work is done by a system in a thermodynamic process, its volume must increase."

This is FALSE. Consider these counterexamples:

- In a cyclic process, the system can perform net positive work while the volume returns to its original value.

- A system may do non-expansion (non-PdV) work, such as electrical work, without any change in volume.

Since both statements are false, the correct answer is Option B: Both Statement I and Statement II are false.

For an ideal gas undergoing an adiabatic process, the relation between temperature and volume is
$$T\,V^{\gamma-1}= \text{constant} \quad -(1)$$

Case 1: Adiabatic expansion
Initial state $$\left(T_A,V_0\right)$$ → final state $$\left(T_f,5V_0\right)$$.
Using $$(1):$$
$$T_A\,V_0^{\gamma-1}=T_f\,(5V_0)^{\gamma-1}$$
$$\Longrightarrow\; T_A=T_f\,5^{\gamma-1} \quad -(2)$$

Case 2: Isothermal expansion
Initial state $$\left(T_B,V_0\right)$$ → final state $$\left(T_f,5V_0\right)$$.
For an isothermal process, temperature is constant, so
$$T_B=T_f \quad -(3)$$

Dividing $$(2)$$ by $$(3):$$
$$\frac{T_A}{T_B}= \frac{T_f\,5^{\gamma-1}}{T_f}=5^{\gamma-1}$$

Hence the required ratio is $$T_A/T_B = 5^{\gamma-1}$$.

Option A which is: $$5^{\gamma-1}$$

For an ideal gas kept at constant pressure, the work done is related to the temperature rise as
$$W = P\Delta V = nR\Delta T$$

The mixture contains
- 2 moles of monatomic gas with $$\gamma_1 = \dfrac{5}{3}$$,
- 1 mole of diatomic gas with $$\gamma_2 = \dfrac{7}{5}$$.
Total moles
$$n = 2 + 1 = 3$$

Given work done
$$W = 66 \text{ J}$$, hence from the relation above
$$\Delta T = \frac{W}{nR} = \frac{66}{3R} = \frac{22}{R}$$ $$-(1)$$

For each component we first evaluate its molar specific heat at constant volume $$C_v$$.
The relation between $$\gamma$$, $$C_p$$ and $$C_v$$ is
$$\gamma = \frac{C_p}{C_v}, \qquad C_p - C_v = R$$ $$-(2)$$

Monatomic gas: $$\gamma_1 = \dfrac{5}{3}$$.
Using $$(2)$$,
$$C_{v1} = \frac{R}{\gamma_1 - 1} = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3}{2}R$$

Diatomic gas: $$\gamma_2 = \dfrac{7}{5}$$.
Similarly,
$$C_{v2} = \frac{R}{\gamma_2 - 1} = \frac{R}{\frac{7}{5} - 1} = \frac{R}{\frac{2}{5}} = \frac{5}{2}R$$

The change in internal energy of the mixture is the sum over all moles:
$$\Delta U = n_1 C_{v1}\Delta T + n_2 C_{v2}\Delta T$$
$$\Delta U = 2\left(\frac{3}{2}R\right)\Delta T \;+\, 1\left(\frac{5}{2}R\right)\Delta T$$
$$\Delta U = \left(3R + \frac{5}{2}R\right)\Delta T = \frac{11}{2}R\Delta T$$ $$-(3)$$

Substitute $$\Delta T$$ from $$(1)$$ into $$(3)$$:
$$\Delta U = \frac{11}{2}R \left(\frac{22}{R}\right) = \frac{11 \times 22}{2} = 121 \text{ J}$$

Hence, the change in internal energy of the gas mixture is 121 Joule.

We use the first law of thermodynamics: $$\Delta U = Q - W$$.

The heat absorbed during vaporisation is:

$$Q = mL = 1 \times 2257 = 2257 \text{ kJ}$$

Now, the work done during expansion at constant pressure is:

$$W = P\Delta V = P(V_2 - V_1)$$

$$W = 1 \times 10^5 \times (1.671 - 1.00 \times 10^{-3})$$

$$W = 1 \times 10^5 \times 1.670 = 1.670 \times 10^5 \text{ J} = 167 \text{ kJ}$$

So the change in internal energy is:

$$\Delta U = Q - W = 2257 - 167 = 2090 \text{ kJ}$$

Hence, the answer is $$+2090$$ kJ.

The process involves a phase change from liquid to vapour at constant pressure. The heat supplied (Q) is used to increase the internal energy (ΔU) and to perform work against the external pressure (W). The first law of thermodynamics states:

$$Q = \Delta U + W$$

The work done by the system is given by $$W = P \Delta V$$, where P is the pressure and ΔV is the change in volume.

Given that 10% of the heat supplied is used for increasing the volume, this means the work done W is 10% of Q:

$$W = 0.1 Q$$

Substituting the expression for work:

$$P \Delta V = 0.1 Q$$

Now, plug in the known values:

Pressure, $$P = 3 \times 10^5$$ Pa
Change in volume, $$\Delta V = 1600 \text{cm}^3 = 1600 \times 10^{-6} \text{m}^3 = 0.0016 \text{m}^3$$

Calculate W:

$$W = P \Delta V = (3 \times 10^5) \times (0.0016) = 480 \text{J}$$

Using the relation $$W = 0.1 Q$$:

$$480 = 0.1 Q$$

Solve for Q:

$$Q = \frac{480}{0.1} = 4800 \text{J}$$

Now apply the first law to find ΔU:

$$Q = \Delta U + W$$
$$4800 = \Delta U + 480$$
$$\Delta U = 4800 - 480 = 4320 \text{J}$$

Therefore, the increase in internal energy is 4320 J.

The correct option is A. 4320 J.

We need to identify which statement is NOT true for adiabatic compression of a gas. An adiabatic process is one in which no heat is exchanged between the system and its surroundings, meaning $$Q = 0$$. The First Law of Thermodynamics states:
$$\Delta U = Q - W$$

For an adiabatic process ($$Q = 0$$):
$$\Delta U = -W$$

Option 1: "There is no heat supplied to the system." This is TRUE. By definition, an adiabatic process has $$Q = 0$$, meaning no heat enters or leaves the system.

Option 2: "There is no change in the internal energy." This is NOT TRUE. From the First Law with $$Q = 0$$, we have $$\Delta U = -W$$. During compression, work is done ON the gas, so the gas does negative work ($$W < 0$$). Therefore:
$$\Delta U = -W = -(\text{negative value}) > 0$$
The internal energy INCREASES during adiabatic compression. The statement $$\Delta U = 0$$ would only hold if $$W = 0$$ (no work done), which contradicts the fact that the gas is being compressed. Note: $$\Delta U = 0$$ is characteristic of an isothermal process for an ideal gas, not an adiabatic one.

Option 3: "The temperature of the gas increases." This is TRUE. Since $$\Delta U > 0$$, and for an ideal gas the internal energy depends only on temperature via $$U = nC_vT$$, an increase in internal energy means an increase in temperature. This is why adiabatic compression heats up a gas, as seen in diesel engines where adiabatic compression raises the air temperature enough to ignite fuel.

Option 4: "The change in internal energy is equal to the work done on the gas." This is TRUE. From $$\Delta U = -W$$, and noting that work done ON the gas is $$W_{on} = -W$$, we get $$\Delta U = W_{on}$$. Since no heat escapes ($$Q = 0$$), all the work done on the gas goes entirely into increasing its internal energy.

The statement that is NOT true is Option 2.
The correct answer is Option 2.

If $$dQ$$ and $$dW$$ represent the heat supplied to the system and the work done on the system respectively, then according to the first law of thermodynamics $$dQ = dU - dW$$. The first law states that the change in internal energy equals heat added plus work done on the system: $$dU = dQ + dW_{\text{on}}$$ where $$dW_{\text{on}}$$ is the work done on the system (IUPAC convention). Rearranging gives $$dQ = dU - dW_{\text{on}}$$, matching the assertion since $$dW$$ denotes the work done on the system. Assertion A is correct.

The first law of thermodynamics is indeed a statement of the law of conservation of energy applied to thermodynamic systems — energy can neither be created nor destroyed, only converted from one form to another. Reason R is correct.

The mathematical form $$dQ = dU - dW$$ follows directly from energy conservation: the heat supplied equals the change in internal energy minus the work done on the system, so R provides the fundamental basis for A. R is the correct explanation of A.

Both A and R are correct, and R is the correct explanation of A.

For an ideal gas, the internal energy depends only on temperature: $$U = nC_vT$$.

In an isothermal process, the temperature remains constant, and therefore the internal energy remains constant ($$\Delta U = 0$$).

By the first law of thermodynamics: $$\Delta Q = \Delta U + \Delta W = 0 + \Delta W = \Delta W$$

So in an isothermal process, all the heat supplied to the system is used to do work, while the internal energy stays constant.

Note: In an isochoric process, volume is constant ($$W = 0$$, but $$\Delta U$$ can change). In an adiabatic process, $$Q = 0$$ but $$U$$ changes. In an isobaric process, pressure is constant but $$U$$ can change.

A Carnot engine operates between two reservoirs with efficiency $$\frac{1}{3}$$. When the cold reservoir temperature is raised by $$x$$, the efficiency decreases to $$\frac{1}{6}$$. The hot reservoir temperature is $$99°$$C.
In Kelvin this gives $$T_H = 99 + 273 = 372$$ K.

The initial efficiency satisfies the relation $$\eta = 1 - \frac{T_C}{T_H}$$. Substituting $$\frac{1}{3} = 1 - \frac{T_C}{372}$$ leads to $$\frac{T_C}{372} = \frac{2}{3}$$ and hence $$T_C = 248$$ K.

After increasing the cold reservoir temperature by $$x$$, the efficiency becomes $$\frac{1}{6} = 1 - \frac{T_C + x}{372}$$. Rearranging gives $$\frac{T_C + x}{372} = \frac{5}{6}$$, so $$T_C + x = 310$$.

Solving for $$x$$ yields $$x = 310 - 248 = 62$$ K, so the answer is Option D: 62 K.

A Carnot engine has 50% efficiency at a source temperature of 600 K. Find the new source temperature needed to achieve 70% efficiency while keeping the sink temperature unchanged.

Find the sink temperature.

Carnot efficiency: $$\eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}$$

$$0.50 = 1 - \frac{T_{\text{cold}}}{600}$$

$$T_{\text{cold}} = 600 \times 0.50 = 300 \text{ K}$$

Find the new source temperature.

$$0.70 = 1 - \frac{300}{T_{\text{new}}}$$

$$\frac{300}{T_{\text{new}}} = 0.30$$

$$T_{\text{new}} = \frac{300}{0.30} = 1000 \text{ K}$$

The correct answer is 1000 K.

We have a gas at temperature $$T$$ that is adiabatically expanded to double its volume, with $$\gamma = \frac{3}{2}$$.

For an adiabatic process, $$TV^{\gamma - 1} = \text{constant}$$. With $$\gamma - 1 = \frac{1}{2}$$:

$$T_1 V_1^{1/2} = T_2 V_2^{1/2}$$

$$T \cdot V^{1/2} = T_2 \cdot (2V)^{1/2}$$

so $$T_2 = \frac{T}{\sqrt{2}}$$.

Now the work done by the gas in an adiabatic process (for 1 mole) is

$$W = \frac{R(T_1 - T_2)}{\gamma - 1}$$

Substituting:

$$W = \frac{R\left(T - \frac{T}{\sqrt{2}}\right)}{\frac{1}{2}} = 2RT\left(1 - \frac{1}{\sqrt{2}}\right) = 2RT \cdot \frac{\sqrt{2} - 1}{\sqrt{2}}$$

Simplifying (by multiplying numerator and denominator by $$\sqrt{2}$$):

$$W = \frac{2RT(\sqrt{2} - 1) \cdot \sqrt{2}}{2} = RT(\sqrt{2})(\sqrt{2} - 1) = RT(2 - \sqrt{2})$$

Hence, the correct answer is $$W = RT(2 - \sqrt{2})$$.

Given: Both gases start at pressure P, volume V. Monoatomic gas ($$\gamma = 5/3$$).

Container A (isothermal compression):

$$P_A V_A = PV$$, with $$V_A = V/8$$:

$$P_A = 8P$$

Container B (adiabatic compression):

$$P_B V_B^\gamma = PV^\gamma$$, with $$V_B = V/8$$:

$$P_B (V/8)^{5/3} = PV^{5/3}$$

$$P_B = P \times 8^{5/3} = P \times (2^3)^{5/3} = P \times 2^5 = 32P$$

Ratio:

$$\frac{P_B}{P_A} = \frac{32P}{8P} = 4$$

The correct answer is Option 2: 4.

We have a Carnot engine operating between a hot reservoir at $$T_H = 127°C = 400$$ K and a cold reservoir at $$T_C = 27°C = 300$$ K, producing work $$W = 2$$ kJ.

The efficiency of the Carnot engine is:

$$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$$

Now, since $$\eta = \frac{W}{Q_H}$$, the heat absorbed from the hot reservoir is:

$$Q_H = \frac{W}{\eta} = \frac{2}{0.25} = 8 \text{ kJ}$$

Hence, the correct answer is Option 1.

By the first law of thermodynamics, the rate of heat supply equals the rate of change of internal energy plus the rate of work done by the system:

$$\frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt}$$

We are given $$\frac{dQ}{dt} = 1000\;\text{W}$$ and $$\frac{dW}{dt} = 200\;\text{W}$$. Substituting:

$$1000 = \frac{dU}{dt} + 200$$

So $$\frac{dU}{dt} = 1000 - 200 = 800\;\text{W}$$. Hence, the rate at which internal energy increases is 800 W.

We need to identify the correct statements about heat given to an ideal gas in an isothermal process.

In an isothermal process:

Temperature remains constant, i.e., $$\Delta T = 0$$.

Statement A: Internal energy will decrease.

For an ideal gas, internal energy depends only on temperature: $$U = nC_vT$$. Since $$\Delta T = 0$$, we have $$\Delta U = 0$$. Statement A is INCORRECT.

Statement B: Internal energy will increase.

Since $$\Delta U = 0$$, Statement B is INCORRECT.

Statement C: Internal energy will not change.

Since $$\Delta U = 0$$, Statement C is CORRECT. ✓

Statement D: The gas will do positive work.

By first law: $$Q = \Delta U + W$$. Since $$\Delta U = 0$$ and $$Q > 0$$ (heat is given), we get $$W = Q > 0$$.

The gas does positive work (it expands). Statement D is CORRECT. ✓

Statement E: The gas will do negative work.

Since $$W > 0$$, Statement E is INCORRECT.

The correct answer is Option 4: C and D only.

For an isothermal process involving an ideal gas, we have $$PV = nRT = \text{constant}$$.

This gives the relation $$P = \frac{nRT}{V}$$, which is a rectangular hyperbola in the P-V plane.

Key properties of isothermal curves (isotherms) on a P-V diagram:

1. Each isotherm is a rectangular hyperbola.

2. For a higher temperature, the hyperbola is farther from the origin.

3. At any given volume, a higher temperature gives a higher pressure (since $$P = \frac{nRT}{V}$$).

Since $$T_3 > T_2 > T_1$$, the isotherm for $$T_3$$ should be the outermost curve (farthest from origin), $$T_2$$ in the middle, and $$T_1$$ closest to the origin.

The correct graph shows rectangular hyperbolas that do not intersect, with $$T_3$$ being the highest curve.

The answer is Option A.

Matching thermodynamic processes with their characteristics:

A. Isothermal Process: Temperature is constant, so internal energy doesn't change (for ideal gas). ΔU = 0, Q = W → II. No change in internal energy

B. Adiabatic Process: No heat exchange (Q = 0), so W = -ΔU. Work done by gas decreases internal energy → I. Work done by the gas decreases internal energy

C. Isochoric Process: Constant volume, so no work done (W = 0). All heat goes to change internal energy → IV. No work is done on or by the gas

D. Isobaric Process: Constant pressure, heat absorbed goes partly to increase internal energy and partly to do work (Q = ΔU + W) → III

Matching: A-II, B-I, C-IV, D-III

The correct answer is Option 2.

Since for an isobaric (constant pressure) process, the work done by the gas is $$W = P\Delta V = nR\Delta T$$.

Meanwhile, the heat supplied is $$Q = nC_p\Delta T$$.

Therefore, the ratio of heat to work is $$\frac{Q}{W} = \frac{nC_p\Delta T}{nR\Delta T} = \frac{C_p}{R}$$. For an ideal gas, $$C_p = \frac{\gamma R}{\gamma - 1}$$, which gives $$\frac{Q}{W} = \frac{\gamma}{\gamma - 1}$$.

Substituting the given values, $$\gamma = 1.4$$ and $$W = 400$$ J, we get $$Q = W \times \frac{\gamma}{\gamma - 1} = 400 \times \frac{1.4}{1.4 - 1}$$, hence $$Q = 400 \times \frac{1.4}{0.4} = 400 \times 3.5 = 1400 \text{ J}$$.

The answer is $$\boxed{1400}$$ J.

A monoatomic gas performs work $$W = \frac{Q}{4}$$ where $$Q$$ is the heat supplied. We need to find the molar heat capacity in terms of $$R$$.

Since by the first law of thermodynamics $$Q = \Delta U + W$$ and $$W = \frac{Q}{4}$$, we have

$$Q = \Delta U + \frac{Q}{4}$$

This gives

$$Q - \frac{Q}{4} = \Delta U$$

so that

$$\frac{3Q}{4} = \Delta U$$

and hence

$$Q = \frac{4}{3}\Delta U$$

Now, for a monoatomic ideal gas with $$n$$ moles, the change in internal energy can be expressed as

$$\Delta U = nC_v \Delta T = n \cdot \frac{3R}{2} \cdot \Delta T$$

Substituting this into the expression for $$Q$$ yields

$$Q = \frac{4}{3} \cdot n \cdot \frac{3R}{2} \cdot \Delta T = n \cdot 2R \cdot \Delta T$$

From the definition $$Q = nC\Delta T$$, it follows that

$$C = 2R$$

Therefore, the molar heat capacity of the gas is 2R.

We have a Carnot engine with reservoir temperature 527°C and sink temperature 200 K. The work done is 12000 kJ.

Converting the temperatures to Kelvin, we get $$T_1 = 527 + 273 = 800 \text{ K}$$ and $$T_2 = 200 \text{ K}.$$

The efficiency of the Carnot engine is given by $$\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{200}{800} = 1 - \frac{1}{4} = \frac{3}{4}.$$

Since efficiency is the ratio of work done to heat absorbed, $$\eta = \frac{W}{Q_1}.$$ Solving for the heat absorbed, $$Q_1 = \frac{W}{\eta} = \frac{12000}{\frac{3}{4}} = 12000 \times \frac{4}{3} = 16000 \text{ kJ}.$$

Converting to joules yields $$Q_1 = 16000 \text{ kJ} = 16 \times 10^6 \text{ J}.$$

The quantity of heat absorbed by the engine from the reservoir is 16 $$\times 10^6$$ J.

We have a diatomic gas with $$\gamma = \dfrac{7}{5}$$ undergoing an adiabatic process, where the density changes from $$d_1$$ to $$d_2$$ with $$\dfrac{d_2}{d_1} = 32$$, and the temperature becomes $$\alpha^2$$ times the initial temperature.

For an adiabatic process, we have the relation $$T \propto d^{\gamma - 1}$$, where $$d$$ is the density. This follows from combining the adiabatic law $$PV^{\gamma} = \text{constant}$$ with the ideal gas law $$PV = nRT$$ and the relation $$d = \dfrac{m}{V}$$.

Therefore, $$\dfrac{T_2}{T_1} = \left(\dfrac{d_2}{d_1}\right)^{\gamma - 1} = 32^{\gamma - 1}$$. Now $$\gamma - 1 = \dfrac{7}{5} - 1 = \dfrac{2}{5}$$, and $$32 = 2^5$$, so $$32^{2/5} = (2^5)^{2/5} = 2^2 = 4$$.

Since the temperature becomes $$\alpha^2$$ times the initial temperature, we have $$\alpha^2 = 4$$. The question asks for the value that the temperature ratio equals, which is $$\alpha^2 = 4$$.

Hence, the correct answer is 4.

We need to find the minimum temperature of the hot reservoir for a heat engine.

Since the cold reservoir temperature is $$T_C = 324$$ K, the heat absorbed from the hot reservoir is $$Q_H = 300$$ J, and the heat delivered to the cold reservoir is $$Q_C = 180$$ J, the engine must operate at maximum (Carnot) efficiency to achieve the minimum hot reservoir temperature.

For a Carnot engine, the relationship between heat exchanged and reservoir temperatures is given by

$$\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$$

Substituting the given values into this expression, we have

$$\frac{180}{300} = \frac{324}{T_H}$$

From this,

$$\frac{3}{5} = \frac{324}{T_H}$$

This gives

$$T_H = \frac{324 \times 5}{3}$$

Next, simplifying the right-hand side yields

$$T_H = \frac{1620}{3} = 540$$ K

The minimum temperature of the hot reservoir is 540 K.

The mass of steam is m = 50 g with an initial temperature of 100 °C and a final temperature of 20 °C. The latent heat of vaporization is L = 540 cal/g and the specific heat of water is c = 1 cal/(g·°C).

Condensation of steam to water at 100 °C releases $$Q_1 = mL = 50 \times 540 = 27000 \text{ cal},$$ and cooling this water from 100 °C to 20 °C releases $$Q_2 = mc\Delta T = 50 \times 1 \times (100 - 20) = 4000 \text{ cal}.$$ Therefore, the total heat rejected per minute is $$Q = Q_1 + Q_2 = 27000 + 4000 = 31000 \text{ cal} = 31 \times 10^3 \text{ cal}.$$

The heat rejected by the steam engine per minute is 31 $$\times 10^3$$ cal.

We are given that the degrees of freedom per molecule of a gas is $$f = 8$$, and the gas performs $$W = 150 \text{ J}$$ of work when expanding under constant pressure. We need to find the heat absorbed.

For a gas with $$f$$ degrees of freedom, the molar heat capacity at constant volume is $$C_v = \frac{f}{2}R = \frac{8}{2}R = 4R$$, and the molar heat capacity at constant pressure is $$C_p = C_v + R = 4R + R = 5R$$.

At constant pressure, the work done by the gas is $$W = nR\Delta T$$, and the heat absorbed is $$Q = nC_p\Delta T$$.

Taking the ratio, $$\frac{Q}{W} = \frac{nC_p\Delta T}{nR\Delta T} = \frac{C_p}{R} = \frac{5R}{R} = 5$$.

So $$Q = 5W = 5 \times 150 = 750 \text{ J}$$.

Hence, the amount of heat absorbed by the gas is $$\textbf{750}$$ J.

Process I - Water ⟶ Steam at 100 °C, constant pressure $$P = 10^{5}\;{\rm Pa}$$

Heat supplied equals the enthalpy change: $$Q = \Delta H = mL$$.
Given mass $$m = 10^{-3}\;{\rm kg}$$ and latent heat $$L = 2250\;{\rm kJ\,kg^{-1}}$$,

$$\Delta H = 10^{-3}\times 2250\;{\rm kJ}=2.25\;{\rm kJ}$$

The work done on the surroundings is $$W = P\Delta V$$ with
$$\Delta V = 10^{-3}-10^{-6}=9.99\times 10^{-4}\;{\rm m^{3}}$$, hence

$$W = 10^{5}\times 9.99\times 10^{-4}\;{\rm J}\approx 1.0\times 10^{2}\;{\rm J}=0.10\;{\rm kJ}$$

By the first law, $$\Delta U = \Delta H - P\Delta V = 2.25-0.10 = 2.15\;{\rm kJ}\approx 2\;{\rm kJ}$$.
Thus I → (P).

Process II - 0.2 mol rigid diatomic gas, isobaric, $$V\;{\rm to}\;3V$$ at $$T_i = 500\;{\rm K}$$

At constant pressure, $$\dfrac{T_f}{T_i} = \dfrac{V_f}{V_i}=3\; \Rightarrow\; T_f = 1500\;{\rm K}$$, so $$\Delta T = 1000\;{\rm K}$$.

For a rigid diatomic gas, $$C_v = \dfrac{5}{2}R = 20\;{\rm J\,mol^{-1}K^{-1}}$$.
Hence

$$\Delta U = nC_v\Delta T = 0.2\times 20\times 1000 = 4000\;{\rm J}=4\;{\rm kJ}$$

Therefore II → (R).

Process III - 1 mol monatomic gas, adiabatic compression from $$V_i=\dfrac13\;{\rm m^{3}}$$, $$P_i = 2\;{\rm kPa}$$ to $$V_f = \dfrac{V_i}{8}$$

Initial temperature: $$T_i=\dfrac{P_iV_i}{R}= \dfrac{2000\times\dfrac13}{8}=83.3\;{\rm K}$$.

For a monatomic gas, $$\gamma=\dfrac53$$ and $$TV^{\gamma-1}={\rm const}$$.
Thus $$T_f = T_i\left(\dfrac{V_i}{V_f}\right)^{\gamma-1}=83.3\times 8^{2/3}=83.3\times 4 = 333\;{\rm K}$$.

$$\Delta T = 333-83.3 = 250\;{\rm K}$$, and $$C_v = \dfrac32 R = 12\;{\rm J\,mol^{-1}K^{-1}}$$, so

$$\Delta U = nC_v\Delta T = 1\times 12\times 250 = 3000\;{\rm J}=3\;{\rm kJ}$$

Hence III → (T).

Process IV - 3 mol vibrating diatomic gas, isobaric; heat supplied $$Q = 9\;{\rm kJ}$$

For a diatomic gas with vibrations active:
$$C_v = \dfrac{7}{2}R,\quad C_p=C_v+R=\dfrac{9}{2}R$$

At constant pressure, $$Q = nC_p\Delta T \Rightarrow \Delta T = \dfrac{Q}{nC_p}$$.
Internal-energy change: $$\Delta U = nC_v\Delta T = \dfrac{C_v}{C_p}\,Q = \dfrac{7/2}{9/2}\,Q = \dfrac79\,Q$$

$$\Delta U = \dfrac79 \times 9\;{\rm kJ}=7\;{\rm kJ}$$

Thus IV → (Q).

Collecting the matches:
I → P, II → R, III → T, IV → Q

Hence the correct option is Option C.

The heat released by the copper block as it cools from $$500°$$C to $$0°$$C (temperature of ice) is:

$$Q = mc\Delta T$$

where $$m = 5.0$$ kg $$= 5000$$ g, $$c = 0.39$$ J g$$^{-1}$$ °C$$^{-1}$$, and $$\Delta T = 500°$$C.

$$Q = 5000 \times 0.39 \times 500 = 975000 \text{ J}$$

The mass of ice that can melt using this heat is:

$$m_{ice} = \frac{Q}{L_f}$$

where $$L_f = 335$$ J g$$^{-1}$$.

$$m_{ice} = \frac{975000}{335} = 2910.4 \text{ g} \approx 2.9 \text{ kg}$$

Hence, the correct answer is Option C.

We need to find the increase in internal energy when $$7$$ moles of a monoatomic ideal gas undergoes a temperature increase of $$\Delta T = 40 \text{ K}$$ at constant pressure.

For an ideal gas, the change in internal energy depends only on temperature change (not on the process). For a monoatomic ideal gas:

$$\Delta U = n C_V \Delta T$$

where $$C_V = \dfrac{3}{2}R$$ for a monoatomic ideal gas.

$$\Delta U = n \times \frac{3}{2}R \times \Delta T$$

$$\Delta U = 7 \times \frac{3}{2} \times 8.3 \times 40$$

$$\Delta U = 7 \times 1.5 \times 8.3 \times 40$$

$$\Delta U = 7 \times 1.5 \times 332$$

$$\Delta U = 7 \times 498$$

$$\Delta U = 3486 \text{ J}$$

The correct answer is Option B: $$3486 \text{ J}$$.

We need to find the temperature of the source of a Carnot engine. The efficiency of a Carnot engine is given by $$\eta = 1 - \frac{T_2}{T_1}$$ and it is stated that initially $$\eta = 50\% = 0.5$$. Therefore $$0.5 = 1 - \frac{T_2}{T_1}$$ implies $$\frac{T_2}{T_1} = 0.5$$ and hence $$T_2 = 0.5 \, T_1 \quad \cdots (1)$$.

When the sink temperature is reduced by 40°C (equal to 40 K), the efficiency increases by 30%. If this increase were additive, the new efficiency would be $$0.5 + 0.3 = 0.8$$ so that $$0.8 = 1 - \frac{T_2 - 40}{T_1}$$ giving $$\frac{T_2 - 40}{T_1} = 0.2$$. Substituting from (1) yields $$\frac{0.5 \, T_1 - 40}{T_1} = 0.2$$, leading to $$0.5 \, T_1 - 40 = 0.2 \, T_1$$ and hence $$0.3 \, T_1 = 40$$ so that $$T_1 = \frac{40}{0.3} = 133.3 \text{ K}$$, which does not match any of the given options.

Interpreting the 30% increase as a relative increase instead gives a new efficiency of $$0.5 \times 1.3 = 0.65$$. In that case $$0.65 = 1 - \frac{T_2 - 40}{T_1}$$ implies $$\frac{T_2 - 40}{T_1} = 0.35$$. Using $$T_2 = 0.5 \, T_1$$ again, we have $$\frac{0.5 \, T_1 - 40}{T_1} = 0.35$$, so $$0.5 \, T_1 - 40 = 0.35 \, T_1$$, giving $$0.15 \, T_1 = 40$$ and hence $$T_1 = \frac{40}{0.15} = 266.67 \text{ K} \approx 266.7 \text{ K}$$.

Therefore, the correct answer is Option C: 266.7 K.

A gas at volume $$V$$, temperature $$27°C$$, and pressure $$P_1 = 2 \times 10^7 \text{ N/m}^2$$ first expands isothermally to volume $$2V$$, then adiabatically to volume $$4V$$. We need to find the final pressure. Given $$\gamma = 1.5$$.

In the isothermal expansion from $$V$$ to $$2V$$, we use the relation $$P_1 V_1 = P_2 V_2$$, so that $$2 \times 10^7 \times V = P_2 \times 2V$$ and thus $$P_2 = 1 \times 10^7 \text{ Pa}$$.

During the subsequent adiabatic expansion from $$2V$$ to $$4V$$, the condition $$P_2 V_2^\gamma = P_3 V_3^\gamma$$ leads to $$P_3 = P_2 \left(\frac{V_2}{V_3}\right)^\gamma = 10^7 \times \left(\frac{2V}{4V}\right)^{1.5}$$, and so $$P_3 = 10^7 \times \left(\frac{1}{2}\right)^{1.5}$$, $$P_3 = 10^7 \times \frac{1}{2\sqrt{2}}$$, $$P_3 = \frac{10^7}{2.828}$$, giving $$P_3 \approx 3.536 \times 10^6 \text{ Pa}$$.

Hence, the correct answer is Option B.

We need to verify both statements about an adiabatic process for an ideal gas.

Checking Statement I:

For an adiabatic process, $$Q = 0$$.

From the first law of thermodynamics: $$W = -\Delta U = -\mu C_v(T_2 - T_1)$$

Since $$C_v = \frac{R}{\gamma - 1}$$:

$$W = -\mu \times \frac{R}{\gamma - 1} \times (T_2 - T_1)$$

$$W = \frac{\mu R(T_1 - T_2)}{\gamma - 1} = \frac{\mu R(T_2 - T_1)}{1 - \gamma}$$

This matches the given formula. Statement I is TRUE.

Checking Statement II:

When work is done ON the gas in an adiabatic process, $$W < 0$$ (work done by gas is negative, meaning work is done on the gas).

From the first law (adiabatic, $$Q = 0$$): $$\Delta U = -W_{\text{by gas}}$$

When work is done on the gas, $$W_{\text{on gas}} > 0$$, so $$\Delta U > 0$$.

Since internal energy increases, $$T_2 > T_1$$, meaning the temperature rises.

Statement II is TRUE.

Both statements are true. The correct answer is Option A.

Heat absorbed $$Q_H = 5000$$ kcal, source temperature $$T_H = 727°$$C, sink temperature $$T_C = 127°$$C. Converting the temperatures to Kelvin gives $$T_H = 727 + 273 = 1000 \text{ K}$$ and $$T_C = 127 + 273 = 400 \text{ K}$$.

The efficiency of a Carnot engine is $$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{400}{1000} = 1 - 0.4 = 0.6$$.

The heat input in joules is $$Q_H = 5000 \text{ kcal} = 5000 \times 4200 \text{ J} = 21 \times 10^6 \text{ J}$$.

The work done by the engine is $$W = \eta \times Q_H = 0.6 \times 21 \times 10^6$$ and therefore $$= 12.6 \times 10^6 \text{ J}$$. The work done by the Carnot engine is $$12.6 \times 10^6$$ J.

The correct answer is Option D.

A monoatomic gas at pressure $$P$$ and volume $$V$$ is suddenly compressed to $$\dfrac{V}{8}$$ (one-eighth of its original volume). We need to find the final pressure given constant entropy (adiabatic process).

Constant entropy means the process is adiabatic (isentropic). For an adiabatic process:

$$PV^{\gamma} = \text{constant}$$

For a monoatomic ideal gas:

$$\gamma = \frac{C_P}{C_V} = \frac{5/2 \cdot R}{3/2 \cdot R} = \frac{5}{3}$$

$$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$$

$$P \cdot V^{5/3} = P_2 \cdot \left(\frac{V}{8}\right)^{5/3}$$

$$P_2 = P \times \left(\frac{V}{V/8}\right)^{5/3} = P \times 8^{5/3}$$

Now, $$8^{5/3} = (2^3)^{5/3} = 2^5 = 32$$.

$$P_2 = 32P$$

The correct answer is Option C: $$32P$$.

In the first case, a single Carnot engine operates between 300 K and 100 K. In the second case, a combination of two Carnot engines is used between the same temperature limits.

Since the efficiency of a Carnot engine is given by $$\eta = 1 - \frac{T_C}{T_H}$$, for the single engine we have $$\eta_1 = 1 - \frac{T_C}{T_H} = 1 - \frac{100}{300} = 1 - \frac{1}{3} = \frac{2}{3}$$.

Next, consider the series combination of two Carnot engines: Engine 1 operates between the hot reservoir at $$T_H = 300\text{ K}$$ and an intermediate temperature $$T$$, while Engine 2 operates between $$T$$ and the cold reservoir at $$T_C = 100\text{ K}$$. The heat rejected by Engine 1 becomes the heat absorbed by Engine 2.

Let $$Q_1$$ be the heat absorbed by Engine 1. Then the work done by Engine 1 is $$W_1 = Q_1\left(1 - \frac{T}{300}\right)$$, and the heat rejected by Engine 1 is $$Q_2 = Q_1 \cdot \frac{T}{300}$$, which is absorbed by Engine 2.

For Engine 2, the work done is $$W_2 = Q_2\left(1 - \frac{100}{T}\right) = Q_1 \cdot \frac{T}{300}\left(1 - \frac{100}{T}\right)$$. Simplifying this gives $$W_2 = Q_1\left(\frac{T}{300} - \frac{100}{300}\right) = Q_1\left(\frac{T - 100}{300}\right)$$.

Therefore, the total work output from the combination is $$W_{\text{total}} = W_1 + W_2 = Q_1\left(1 - \frac{T}{300}\right) + Q_1\left(\frac{T - 100}{300}\right)$$. Substituting and combining terms yields $$W_{\text{total}} = Q_1\left(1 - \frac{100}{300}\right) = Q_1 \times \frac{2}{3}$$.

It follows that the overall efficiency of the combination is $$\eta_{\text{combination}} = \frac{W_{\text{total}}}{Q_1} = \frac{2}{3}$$, which is exactly the same as the efficiency of the single Carnot engine between 300 K and 100 K.

This result is consistent with the principle that two Carnot engines arranged in series have the same overall efficiency as a single Carnot engine operating between the same extreme temperatures, regardless of the choice of intermediate temperature.

Answer: Option A: same as the 1st case

An ideal gas expands from $$V_1$$ to $$V_2$$ via three processes starting from the same initial conditions. We compare the work done: $$W_1$$ (isothermal), $$W_2$$ (adiabatic), $$W_3$$ (isobaric).

All three processes start from the same point $$(V_1, P_1)$$ on the P-V diagram and end at $$V_2$$.

For expansion from $$V_1$$ to $$V_2$$:

- Isobaric process: Pressure stays at $$P_1$$ throughout. The P-V curve is a horizontal line at $$P = P_1$$.

- Isothermal process: $$PV = P_1V_1 = \text{const}$$, so $$P = P_1V_1/V$$. The pressure drops as $$1/V$$ — the curve lies below the isobaric line.

- Adiabatic process: $$PV^\gamma = P_1V_1^\gamma = \text{const}$$, with $$\gamma > 1$$. The pressure drops faster than in the isothermal case, so this curve lies below the isothermal curve.

Work done equals the area under the P-V curve from $$V_1$$ to $$V_2$$. Since the isobaric curve is highest, isothermal is in the middle, and adiabatic is lowest:

$$W_{\text{adiabatic}} < W_{\text{isothermal}} < W_{\text{isobaric}}$$

$$W_2 < W_1 < W_3$$

This ordering can be understood physically: in an isobaric process, the gas maintains constant pressure (maximum area under curve). In an isothermal process, the gas absorbs heat to maintain temperature, keeping the pressure relatively high. In an adiabatic process, no heat is absorbed, so the temperature and pressure drop the most, giving the smallest area under the curve.

Hence, the correct answer is Option A: $$W_2 < W_1 < W_3$$.

We are given the heat sink temperature $$T_{sink} = 27°C = 300$$ K, the initial efficiency $$\eta_1 = 25\% = 0.25$$, and we wish to increase the efficiency by 100% of the original, doubling it to $$50\%$$. The Carnot efficiency is given by $$\eta = 1 - \frac{T_{sink}}{T_{source}}$$.

Substituting $$\eta_1 = 0.25$$ leads to $$0.25 = 1 - \frac{300}{T_1}$$, which gives $$\frac{300}{T_1} = 0.75$$ and hence $$T_1 = \frac{300}{0.75} = 400 \text{ K}$$.

For the doubled efficiency $$\eta_2 = 0.50$$ we have $$0.50 = 1 - \frac{300}{T_2}$$, so $$\frac{300}{T_2} = 0.50$$, giving $$T_2 = \frac{300}{0.50} = 600 \text{ K}$$.

Therefore, the required increase in source temperature is $$\Delta T = T_2 - T_1 = 600 - 400 = 200 \text{ K} = 200°C$$, meaning the source temperature must be raised by $$200°C$$.

Final Answer: Option B.

Use first law:

$$Q=\Delta U+W$$

(where W is work done by gas)

For process BC:

Given no heat exchange,

$$Q_{BC}=0$$

Work done on gas = 50 J

So work done by gas

$$W_{BC}=-50J$$

Thus

$$\Delta U_{BC}-50=0$$

$$ΔU_{BC}=50$$

So

$$U_C-U_B=50$$

For process AB:

$$Q_{AB}=40$$

From graph AB is vertical (constant volume), so

$$W_{AB}=0$$

Therefore

$$40=\Delta U_{AB}$$

$$U_B-U_A=40$$

Since

$$U_A=1560$$

$$U_B=1600$$

Then

$$U_C=1650$$

For process CA:

$$Q_{CA}=-60$$

Change in internal energy:

$$\Delta U_{CA}=U_A-U_C$$

$$=1560−1650=−90$$

Now

$$Q_{CA}=\Delta U_{CA}+W_{CA}\\-60=-90+W_{CA}$$

$$W_{CA}=30\text{ J}$$

We need to find the ratio $$\frac{\eta_1}{\eta_2}$$ for two Carnot engines.

Convert temperatures to Kelvin.

For engine 1: $$T_1 = 447 + 273 = 720 \text{ K}$$, $$T_2 = 147 + 273 = 420 \text{ K}$$

For engine 2: $$T_1 = 947 + 273 = 1220 \text{ K}$$, $$T_2 = 47 + 273 = 320 \text{ K}$$

Calculate the efficiencies.

Calculate the ratio.

The correct answer is Option B: $$0.56$$.

We have $$n$$ moles of a perfect gas undergoing a cyclic process ABCA.

For $$A \to B$$ (isothermal expansion at temperature $$T$$ from $$V_1$$ to $$V_2 = 2V_1$$), the work done is $$W_{AB} = nRT\ln\frac{V_2}{V_1} = nRT\ln 2$$.

Since $$A \to B$$ is isothermal, we have $$P_1V_1 = P_2V_2 = nRT$$. With $$V_2 = 2V_1$$, this gives $$P_2 = \frac{P_1}{2}$$ and $$P_2V_1 = \frac{nRT}{2}$$.

For $$B \to C$$ (isobaric compression at pressure $$P_2$$ from $$V_2 = 2V_1$$ back to $$V_1$$), the work done is $$W_{BC} = P_2(V_1 - V_2) = P_2(V_1 - 2V_1) = -P_2V_1 = -\frac{nRT}{2}$$.

For $$C \to A$$ (isochoric process at constant volume $$V_1$$), no work is done since volume does not change. So $$W_{CA} = 0$$.

The total work done in the complete cycle is $$W = W_{AB} + W_{BC} + W_{CA} = nRT\ln 2 - \frac{nRT}{2} + 0 = nRT\left(\ln 2 - \frac{1}{2}\right)$$.

Hence, the correct answer is Option A.

For every Carnot engine the efficiency is given by the well-known relation

Engine A works between the temperatures $$T_1$$ (source) and $$T$$ (sink). Let the heat absorbed from the hot reservoir be $$Q_1$$. Using the above formula we have for engine A

Therefore the work obtained from engine A is

The heat rejected by engine A to the intermediate reservoir at temperature $$T$$ is found from the first law:

According to the statement of the problem, engine B absorbs only half of this rejected heat, so

Engine B now works between the temperatures $$T$$ (source) and $$T_3$$ (sink). Its Carnot efficiency is

Hence the work obtained from engine B is

The condition given is that both engines produce the same work, i.e.

Substituting the expressions for $$W_A$$ and $$W_B$$ we get

The factor $$Q_1$$ is present on both sides, so it cancels out. Multiplying the remaining equation by $$2T_1$$ we obtain

Rewriting the terms inside the brackets gives

Simplifying each side separately:

Expanding and collecting the temperature terms leads to

Now shift all terms involving $$T$$ to one side and the constants to the other:

Finally, solving for the unknown intermediate temperature $$T$$ yields

Hence, the correct answer is Option D.

We are given 1 mole of a rigid diatomic gas that performs work $$W = \frac{Q}{5}$$ when heat $$Q$$ is supplied to it. We need to find the molar heat capacity during this process.

For a rigid diatomic gas, there are no vibrational degrees of freedom, so $$C_v = \frac{5R}{2}$$.

By the first law of thermodynamics, $$Q = \Delta U + W$$. Substituting $$W = \frac{Q}{5}$$, we get $$Q = \Delta U + \frac{Q}{5}$$, which gives $$\Delta U = \frac{4Q}{5}$$.

For 1 mole of gas, $$\Delta U = n C_v \Delta T = \frac{5R}{2} \Delta T$$. So $$\frac{4Q}{5} = \frac{5R}{2} \Delta T$$.

The molar heat capacity of the process is defined as $$C = \frac{Q}{n \Delta T} = \frac{Q}{\Delta T}$$ (since $$n = 1$$). From the above equation, $$\Delta T = \frac{8Q}{25R}$$.

Therefore, $$C = \frac{Q}{\frac{8Q}{25R}} = \frac{25R}{8}$$.

Comparing with $$\frac{xR}{8}$$, we get $$x = 25$$.

For an adiabatic process of an ideal gas we always have the relation

$$P\,V^{\gamma}= \text{constant}.$$

The work done by the gas (positive for expansion, negative for compression) in a reversible adiabatic process is given by the formula

$$W=\frac{P_2V_2-P_1V_1}{1-\gamma}=\frac{P_1V_1-P_2V_2}{\gamma-1}.$$

Here the given data are

$$\gamma = 1.5, \quad V_1 = 1200\ \text{cm}^3, \quad V_2 = 300\ \text{cm}^3, \quad P_1 = 200\ \text{kPa}.$$

First we convert the volumes into SI units (m$$^3$$):

$$V_1 = 1200\ \text{cm}^3 = 1200\times10^{-6}\ \text{m}^3 = 1.2\times10^{-3}\ \text{m}^3,$$

$$V_2 = 300\ \text{cm}^3 = 300\times10^{-6}\ \text{m}^3 = 3.0\times10^{-4}\ \text{m}^3.$$

Using $$P_1V_1^{\gamma}=P_2V_2^{\gamma},$$ we find the final pressure:

$$P_2=P_1\left(\frac{V_1}{V_2}\right)^{\gamma}.$$

The volume ratio is

$$\frac{V_1}{V_2}= \frac{1.2\times10^{-3}}{3.0\times10^{-4}} = 4,$$ so

$$P_2 = 200\ \text{kPa}\;\bigl(4\bigr)^{1.5}.$$

Because $$4^{1.5}=4^{3/2}=(4^1)(4^{1/2})=4\times2=8,$$ we obtain

$$P_2 = 200\ \text{kPa}\times 8 = 1600\ \text{kPa}.$$

Next we calculate the products $$P_1V_1$$ and $$P_2V_2,$$ remembering that $$1\ \text{kPa}=10^{3}\ \text{Pa}$$ and $$1\ \text{Pa}\cdot\text{m}^3=1\ \text{J}.$$

$$P_1V_1 = 200\ \text{kPa}\times1.2\times10^{-3}\ \text{m}^3 = 200\times10^{3}\ \text{Pa}\times1.2\times10^{-3}\ \text{m}^3$$

$$\phantom{P_1V_1} = 200\times1.2\times(10^{3}\times10^{-3})\ \text{J} = 240\ \text{J}.$$

$$P_2V_2 = 1600\ \text{kPa}\times3.0\times10^{-4}\ \text{m}^3 = 1600\times10^{3}\ \text{Pa}\times3.0\times10^{-4}\ \text{m}^3$$

$$\phantom{P_2V_2} = 1600\times3.0\times(10^{3}\times10^{-4})\ \text{J} = 480\ \text{J}.$$

Substituting these values into the work formula that is convenient for magnitude,

$$W = \left|\frac{P_1V_1-P_2V_2}{\gamma-1}\right|,$$

we have

$$P_1V_1-P_2V_2 = 240\ \text{J}-480\ \text{J} = -240\ \text{J},$$

and since $$\gamma-1 = 1.5-1 = 0.5,$$

$$W = \left|\frac{-240\ \text{J}}{0.5}\right| = \left|-480\ \text{J}\right| = 480\ \text{J}.$$

The negative sign merely indicated that the gas was compressed; the question asks for the absolute value, which is 480 J.

So, the answer is $$480\ \text{J}.$$

We have a heat engine that absorbs heat $$Q_1 = 300\ \text{J}$$ from a hot reservoir at temperature $$T_1$$ and rejects heat $$Q_2 = 240\ \text{J}$$ to a cold reservoir at temperature $$T_2 = 400\ \text{K}$$.

First, we find the work done in one complete cycle. The First Law of Thermodynamics for a cyclic engine states that the net work output $$W$$ equals the net heat absorbed:

$$W = Q_1 - Q_2.$$

Substituting the given heats,

$$W = 300\ \text{J} - 240\ \text{J} = 60\ \text{J}.$$

Now we determine the actual thermal efficiency $$\eta_{\text{actual}}$$ of this engine. By definition, thermal efficiency is the ratio of work output to heat absorbed from the hot reservoir:

$$\eta_{\text{actual}} = \frac{W}{Q_1}.$$

Substituting $$W = 60\ \text{J}$$ and $$Q_1 = 300\ \text{J},$$ we get

$$\eta_{\text{actual}} = \frac{60}{300} = 0.20.$$

According to the Second Law of Thermodynamics, no real engine can be more efficient than a reversible (Carnot) engine operating between the same two temperature reservoirs. For a Carnot engine the efficiency is

$$\eta_{\text{Carnot}} = 1 - \frac{T_2}{T_1},$$

where $$T_1$$ is the temperature of the hot reservoir and $$T_2 = 400\ \text{K}$$ is the temperature of the cold reservoir.

To allow the given engine to operate without violating the Second Law, its efficiency must not exceed the Carnot efficiency. Hence we require

$$\eta_{\text{actual}} \le \eta_{\text{Carnot}}.$$

Substituting the expressions,

$$0.20 \le 1 - \frac{T_2}{T_1}.$$

Rewriting the right-hand side,

$$0.20 \le 1 - \frac{400}{T_1}.$$

Now we isolate the fraction:

$$\frac{400}{T_1} \le 1 - 0.20 = 0.80.$$

Next, we solve for $$T_1$$ by inverting and rearranging:

$$\frac{1}{T_1} \le \frac{0.80}{400} \quad\Longrightarrow\quad T_1 \ge \frac{400}{0.80}.$$

Carrying out the division,

$$T_1 \ge 500\ \text{K}.$$

This value represents the minimum possible temperature of the hot reservoir that ensures the engine’s efficiency does not exceed the Carnot limit.

So, the answer is $$500\ \text{K}$$.

We are given that the pressure of a gas depends on its volume as $$P = kV^3$$. For an ideal gas, $$PV = nRT$$, so substituting gives $$kV^3 \cdot V = nRT$$, which means $$kV^4 = nRT$$.

To find the work done, we use $$W = \int P\,dV$$. Since $$P = kV^3$$, we have $$dP = 3kV^2\,dV$$, so $$V\,dP = 3kV^3\,dV = 3P\,dV$$. From the product rule, $$d(PV) = P\,dV + V\,dP = P\,dV + 3P\,dV = 4P\,dV$$. Since $$PV = nRT$$, we have $$d(PV) = nR\,dT$$. Therefore, $$4P\,dV = nR\,dT$$, giving $$P\,dV = \frac{nR}{4}\,dT$$.

Integrating from $$T_1 = 100°C = 373\,\text{K}$$ to $$T_2 = 300°C = 573\,\text{K}$$:

$$W = \int_{T_1}^{T_2} \frac{nR}{4}\,dT = \frac{nR}{4}(573 - 373) = \frac{nR}{4} \times 200 = 50\,nR$$

Comparing with $$W = xnR$$, we get $$x = 50$$.

We are given a monoatomic gas with $$V = KT^{2/3}$$ and need to find the work done when the temperature changes by 90 K.

For 1 mole of an ideal gas, $$PV = RT$$, so $$P = \frac{RT}{V} = \frac{RT}{KT^{2/3}} = \frac{R}{K} T^{1/3}$$.

The work done is $$W = \int P \, dV$$. Since $$V = KT^{2/3}$$, we have $$dV = K \cdot \frac{2}{3} T^{-1/3} \, dT$$.

Substituting, $$W = \int \frac{R}{K} T^{1/3} \cdot K \cdot \frac{2}{3} T^{-1/3} \, dT = \int \frac{2R}{3} \, dT$$.

For a temperature change of $$\Delta T = 90$$ K, $$W = \frac{2R}{3} \times 90 = 60R$$.

Therefore, $$x = 60$$.

For a reversible (Carnot) heat engine, the efficiency is $$\eta = 1 - \frac{T_2}{T_1}$$, where $$T_1$$ is the source temperature and $$T_2$$ is the sink temperature.

Given that the engine converts one-fourth of heat input into work, the initial efficiency is $$\eta_1 = \frac{1}{4}$$. So $$1 - \frac{T_2}{T_1} = \frac{1}{4}$$, which gives $$\frac{T_2}{T_1} = \frac{3}{4}$$, hence $$T_2 = \frac{3}{4}T_1$$.

When the sink temperature is reduced by 52 K, the new sink temperature is $$T_2' = T_2 - 52$$ and the new efficiency is doubled: $$\eta_2 = \frac{1}{2}$$. So $$1 - \frac{T_2 - 52}{T_1} = \frac{1}{2}$$, which gives $$\frac{T_2 - 52}{T_1} = \frac{1}{2}$$, hence $$T_2 - 52 = \frac{T_1}{2}$$.

Substituting $$T_2 = \frac{3}{4}T_1$$: $$\frac{3}{4}T_1 - 52 = \frac{1}{2}T_1$$, so $$\frac{3}{4}T_1 - \frac{1}{2}T_1 = 52$$, giving $$\frac{1}{4}T_1 = 52$$.

Therefore $$T_1 = 208$$ K.

For an ideal heat engine (Carnot engine), the efficiency is given by $$\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$$, where temperatures are in Kelvin.

The source temperature is $$T_{\text{source}} = 127°\text{C} = 127 + 273 = 400$$ K. The required efficiency is $$\eta = 60\% = 0.6$$.

Substituting: $$0.6 = 1 - \frac{T_{\text{sink}}}{400}$$, which gives $$\frac{T_{\text{sink}}}{400} = 0.4$$, so $$T_{\text{sink}} = 160$$ K.

Converting back to Celsius: $$T_{\text{sink}} = 160 - 273 = -113$$°C.

The temperature of the sink is $$\boxed{-113}$$°C.

We have an ideal diatomic gas that can translate in three directions and rotate about two perpendicular axes, but it cannot oscillate. Hence the total number of active degrees of freedom is five. For an ideal gas the molar heat capacities are related to the degrees of freedom by the formula

$$C_V = \frac{f}{2}\,R \qquad\text{and}\qquad C_P = C_V + R,$$

where $$f$$ is the number of degrees of freedom and $$R$$ is the universal gas constant. Substituting $$f = 5$$ we obtain

$$C_V = \frac{5}{2}R,$$

and therefore

$$C_P = \frac{5}{2}R + R = \frac{7}{2}R.$$

The number of moles present is $$n = 3.00\,\text{mol}$$ and the rise in temperature is $$\Delta T = 40.0^{\circ}\text{C} = 40.0\,\text{K}.$$

The process occurs at constant pressure. First, let us evaluate the change in internal energy. For an ideal gas

$$\Delta U = n\,C_V\,\Delta T.$$

Substituting the known values gives

$$\Delta U = 3.00 \left(\frac{5}{2}R\right)(40.0) = 3.00 \times \frac{5}{2} \times 40.0\,R = 3.00 \times 100\,R = 300\,R.$$

Next, we calculate the work done by the gas at constant pressure. The formula for the work in an isobaric (constant-pressure) process is

$$W = n\,R\,\Delta T.$$

Substituting once more we get

$$W = 3.00\,R\,(40.0) = 120\,R.$$

Now we form the ratio of the change in internal energy to the work done:

$$\frac{\Delta U}{W} = \frac{300\,R}{120\,R} = \frac{300}{120} = 2.5 = \frac{25}{10}.$$

The problem states that this ratio equals $$\dfrac{x}{10},$$ so by direct comparison we have $$x = 25.$$

So, the answer is $$25$$.

For the complete cycle,

$$W_{\text{net}}=W_{AB}+W_{BC}+W_{CA}$$

For A→B, expansion is isothermal.

For one mole,

Since at A,

and

$$V_B=2V_1$$

so

$$W_{AB}=P_1V_1\ln2$$

For B→CB \to CB→C, volume is constant, so

$$W_{BC}=0$$

For C→A, process is adiabatic compression.

For adiabatic process,

Q=0

and

For compression from C to A,

$$W_{CA}=\frac{P_CV_C-P_AV_A}{\gamma-1}$$

Now

and

$$P_AV_A=P_1V_1$$

Thus

So net work done is

$$W_{\text{net}}=RT\ln2-\frac{P_1V_1}{2(\gamma-1)}$$

Given data:

Heat supplied to the system: $$\dot Q = 6000 \text{ J min}^{-1}$$
Useful power (work done by the system): $$P = 90 \text{ W} = 90 \text{ J s}^{-1}$$
Required rise in internal energy: $$\Delta U = 2.5 \times 10^3 \text{ J}$$

Step 1: Convert the heat-input rate to SI units (joules per second).

$$6000 \text{ J min}^{-1} = \frac{6000 \text{ J}}{60 \text{ s}} = 100 \text{ J s}^{-1}$$

Step 2: Write the First Law of Thermodynamics in its rate form.

The first law states $$\Delta Q = \Delta U + \Delta W$$. Dividing by $$\Delta t$$ gives the power form
$$\frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} + \frac{\Delta W}{\Delta t}$$
or $$\dot Q = \dot U + \dot W$$ $$-(1)$$

Step 3: Substitute the known rates into equation $$-(1)$$.

Heat input rate: $$\dot Q = 100 \text{ J s}^{-1}$$
Work output rate: $$\dot W = 90 \text{ J s}^{-1}$$
Therefore,
$$\dot U = \dot Q - \dot W = 100 - 90 = 10 \text{ J s}^{-1}$$

Step 4: Find the time needed to raise the internal energy by $$\Delta U = 2.5 \times 10^3 \text{ J}$$.

Using $$\dot U = \frac{\Delta U}{\Delta t}$$, we get
$$\Delta t = \frac{\Delta U}{\dot U} = \frac{2.5 \times 10^3}{10} = 250 \text{ s}$$

Answer: $$\boxed{2.5 \times 10^2 \text{ s}}$$   (Option B)

In thermodynamics, state functions (or point functions) are properties that depend only on the current state of the system, such as internal energy, pressure, temperature, and entropy. Their values are independent of the path taken to reach that state.

Heat and work, on the other hand, are not properties of the system — they are energy transfers that occur during a process. Their values depend on the specific path or process by which the system transitions from one state to another. For example, an ideal gas expanding isothermally does different amounts of work compared to an adiabatic expansion between the same two states.

Therefore, heat and work are path functions.

We need to identify the correct graphical representations for two different thermodynamic processes—isothermal and adiabatic—across different state variables ($$P-V$$, $$V-T$$, and $$P-T$$ diagrams).

1. Core Thermodynamic Principles

For an ideal gas undergoing isothermal and adiabatic expansions:

• Isothermal Process: Temperature remains constant ($$T = \text{constant}$$).
  • Governing equation: $$PV = \text{constant}$$
• Adiabatic Process: No heat exchange occurs ($$Q = 0$$).
  • Governing equation: $$PV^\gamma = \text{constant}$$ (where $$\gamma > 1$$ is the adiabatic index)

2. Analyzing the Slopes on a $$P-V$$ Diagram (Graph a)

Differentiating both governing equations with respect to volume ($$V$$) gives the slopes of the curves on a pressure-volume graph:

• Isothermal slope: $$\frac{dP}{dV} = -\frac{P}{V}$$
• Adiabatic slope: $$\frac{dP}{dV} = -\gamma \frac{P}{V}$$

Since $$\gamma > 1$$, the adiabatic curve is steeper than the isothermal curve during expansion. In graph (a), the label pointing to the steeper curve is incorrectly labeled as isothermal, making graph (a) incorrect.

3. Analyzing the $$V-T$$ Diagram (Graph c)

• Isothermal: Since $$T$$ is constant, the curve must be a straight vertical line parallel to the volume axis.
• Adiabatic: From $$PV^\gamma = \text{constant}$$ and the ideal gas law ($$P = \frac{nRT}{V}$$), we get:

  $$T V^{\gamma - 1} = \text{constant}$$

  As volume ($$V$$) increases during an expansion, temperature ($$T$$) must decrease. This produces a downward-sloping curve. Graph (c) correctly displays both of these behaviors.

4. Analyzing the $$P-T$$ Diagram (Graph d)

• Isothermal: Since $$T$$ is constant, the curve must be a straight vertical line parallel to the pressure axis.
• Adiabatic: From $$PV^\gamma = \text{constant}$$ and the ideal gas law ($$V = \frac{nRT}{P}$$), we get:

  $$P^{1-\gamma} T^\gamma = \text{constant} \implies P \propto T^{\frac{\gamma}{\gamma - 1}}$$

  Since $$\gamma > 1$$, as temperature ($$T$$) decreases during an expansion, pressure ($$P$$) must also decrease. This forms a curved path dropping toward the origin. Graph (d) correctly displays both of these behaviors.

Conclusion

Graphs (c) and (d) accurately represent the characteristics of the isothermal and adiabatic processes under their respective thermodynamic variables.

Correct Answer: (c) and (d) 
HENCE ; Option B would be the right answer

The efficiency of a Carnot engine operating between a cold reservoir at temperature $$T_C = 400$$ K and a hot reservoir at $$T_H = 800$$ K is $$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{400}{800} = \frac{1}{2}$$.

Since efficiency is also defined as $$\eta = \frac{W}{Q_H}$$, where $$W = 1200$$ J is the work output and $$Q_H$$ is the heat absorbed from the source, we have $$\frac{1}{2} = \frac{1200}{Q_H}$$, giving $$Q_H = 2400$$ J.

The correct answer is option 4: 2400 J.

For a diatomic gas heated at constant pressure, we need to find the ratio $$dU : dQ : dW$$.

The change in internal energy at constant pressure is $$dU = nC_V\,dT = n \cdot \frac{5}{2}R\,dT$$.

The heat supplied at constant pressure is $$dQ = nC_P\,dT = n \cdot \frac{7}{2}R\,dT$$.

The work done by the gas at constant pressure is $$dW = dQ - dU = n(C_P - C_V)\,dT = nR\,dT$$.

Therefore the ratio is $$dU : dQ : dW = \frac{5}{2}R : \frac{7}{2}R : R = 5 : 7 : 2$$.

The correct answer is Option (3): $$5 : 7 : 2$$.

Let us denote the temperature of the hot reservoir (source) by $$T_1$$ and that of the cold reservoir (sink) by $$T_2$$. Throughout the calculation all temperatures will be expressed on the absolute (kelvin) scale; a change of 1 °C equals a change of 1 K, so differences can be handled in either unit without confusion.

For an ideal (Carnot) heat engine, the efficiency $$\eta$$ is given by the well-known relation

$$\eta \;=\;1-\dfrac{T_2}{T_1}.$$

We are first told that the efficiency is $$\dfrac16$$. Substituting this value into the formula we have

$$1-\dfrac{T_2}{T_1}\;=\;\dfrac16.$$

Rearranging term by term,

$$\dfrac{T_2}{T_1}\;=\;1-\dfrac16\;=\;\dfrac56,$$

and hence

$$T_2=\dfrac56\,T_1. \quad -(1)$$

Next, the problem states that the sink temperature is lowered by 62 °C (that is, by 62 K). Let the new sink temperature be $$T_2'$$. Then

$$T_2' = T_2 - 62.$$

With this altered sink temperature the efficiency doubles, becoming $$2\times\dfrac16=\dfrac13$$. Applying the efficiency formula once more, we get

$$1-\dfrac{T_2'}{T_1}\;=\;\dfrac13.$$

Solving for the ratio,

$$\dfrac{T_2'}{T_1}\;=\;1-\dfrac13\;=\;\dfrac23,$$

so

$$T_2'=\dfrac23\,T_1. \quad -(2)$$

Now we substitute $$T_2=\dfrac56\,T_1$$ from equation (1) into the definition of $$T_2'$$:

$$T_2' = T_2 - 62 \;=\;\dfrac56\,T_1\;-\;62.$$

But equation (2) tells us that $$T_2'=\dfrac23\,T_1$$. Equating the two expressions for $$T_2'$$ gives

$$\dfrac56\,T_1\;-\;62 \;=\;\dfrac23\,T_1.$$

To collect like terms, we bring the right-hand term to the left:

$$\dfrac56\,T_1 - \dfrac23\,T_1 \;=\;62.$$

Writing both fractions with a common denominator 6, $$\dfrac23\,T_1=\dfrac46\,T_1$$, so

$$\left(\dfrac56 - \dfrac46\right)T_1 \;=\;62.$$

The difference in the parentheses is $$\dfrac16$$, hence

$$\dfrac16\,T_1 \;=\;62.$$

Multiplying both sides by 6 we arrive at

$$T_1 \;=\;372\;\text{K}.$$

Finally, converting back to the Celsius scale (recall $$T(\text{°C}) = T(\text{K}) - 273$$), we obtain

$$T_1 = 372\;\text{K} - 273 = 99\;^\circ\text{C}.$$

Hence, the correct answer is Option D.

For a completely reversible (Carnot) heat engine, the efficiency $$\eta$$ is related to the source temperature $$T_1$$ and the sink (cold reservoir) temperature $$T_2$$ by the well-known Carnot relation:

$$\eta \;=\;1-\dfrac{T_2}{T_1}$$

We are first told that the efficiency is $$\dfrac14$$. Substituting this value, we have

$$\dfrac14 \;=\;1-\dfrac{T_2}{T_1}$$

Now we isolate the fraction $$\dfrac{T_2}{T_1}$$. Moving terms carefully,

$$\dfrac{T_2}{T_1}=1-\dfrac14=\dfrac34$$

Multiplying both sides by $$T_1$$ gives

$$T_2=\dfrac34\,T_1$$

Next, the problem says that the sink temperature is lowered by 58 °C, and the efficiency becomes double. Because temperature differences are identical in the Celsius and Kelvin scales, we can use the same numerical value in kelvin. Let the new sink temperature be $$T_2'$$ and the (unchanged) source temperature remain $$T_1$$. We therefore have

$$T_2'=T_2-58$$

The efficiency is now doubled, becoming $$2\times\dfrac14=\dfrac12$$. Using the Carnot formula again,

$$\dfrac12 \;=\;1-\dfrac{T_2'}{T_1}$$

Solving for $$T_2'$$,

$$\dfrac{T_2'}{T_1}=1-\dfrac12=\dfrac12\quad\Longrightarrow\quad T_2'=\dfrac12\,T_1$$

But we also have $$T_2' = T_2 - 58$$. Substituting $$T_2=\dfrac34\,T_1$$ obtained earlier,

$$\dfrac12\,T_1 \;=\;\dfrac34\,T_1 - 58$$

Bringing the terms containing $$T_1$$ to one side,

$$\dfrac34\,T_1 - \dfrac12\,T_1 = 58$$

The left-hand side simplifies since $$\dfrac34-\dfrac12=\dfrac14$$, giving

$$\dfrac14\,T_1 = 58$$

Multiplying both sides by 4, we find the source temperature:

$$T_1 = 4 \times 58 = 232\text{ K}$$

Finally, we determine the original sink temperature from $$T_2=\dfrac34\,T_1$$:

$$T_2=\dfrac34 \times 232 = \dfrac{696}{4}=174\text{ K}$$

Hence, the correct answer is Option C.

For an adiabatic process involving an ideal gas, we have $$PV^\gamma = \text{constant}$$.

Taking the differential of both sides: $$V^\gamma \, dP + P \cdot \gamma V^{\gamma - 1} \, dV = 0$$.

Dividing throughout by $$PV^\gamma$$: $$\frac{dP}{P} + \gamma \frac{dV}{V} = 0$$.

Therefore, the fractional change in pressure is $$\frac{dP}{P} = -\gamma \frac{dV}{V}$$.

We begin by converting the given Celsius temperatures into Kelvin, because the thermodynamic formulae for a Carnot refrigerator demand absolute temperatures.

We have
$$T_{\text{cold}} = -10^{\circ}{\rm C} = -10 + 273 = 263\ {\rm K}$$
and
$$T_{\text{hot}} = 25^{\circ}{\rm C} = 25 + 273 = 298\ {\rm K}.$$

For an ideal, perfectly reversible refrigerator the coefficient of performance (COP) is given by the Carnot expression

$$\text{COP} = \frac{Q_{\text{cold}}}{W} = \frac{T_{\text{cold}}}{T_{\text{hot}} - T_{\text{cold}}},$$

where $$Q_{\text{cold}}$$ is the heat absorbed per second from the cold chamber and $$W$$ is the mechanical work (power) supplied per second.

Substituting the numerical values, we obtain

$$\text{COP} = \frac{263}{298 - 263} = \frac{263}{35}.$$

Carrying out the division gives

$$\text{COP} = 7.514.$$

Now, the power input is given as $$W = 35\ {\rm W} = 35\ {\rm J\,s^{-1}}.$$
Using the relation $$Q_{\text{cold}} = \text{COP} \times W$$ we can find the average heat drawn from the cold compartment each second.

So

$$Q_{\text{cold}} = 7.514 \times 35\ {\rm J\,s^{-1}}.$$

Multiplying, we get

$$Q_{\text{cold}} = 263\ {\rm J\,s^{-1}}\;(\text{approximately}).$$

This quantity represents the average heat that the refrigerator transfers per second from the low-temperature space when there are no energy losses.

Hence, the correct answer is Option C.

Entropy is an extensive thermodynamic property, which means it scales with the size of the system and is additive over subsystems. If a composite system is made up of two independent parts with entropies $$S_1$$ and $$S_2$$, the total entropy of the composite system is $$S_{\text{total}} = S_1 + S_2$$.

When the piston separating the two parts of the ideal gas is removed, the gas from both compartments mixes. Since this is an irreversible process (free expansion into each other's space), the entropy of the system can only increase or stay the same according to the second law of thermodynamics. However, for an ideal gas where both sides are at the same temperature and pressure, the mixing does not generate additional entropy.

Regardless of whether additional entropy is generated during mixing, the total entropy is at minimum $$S_1 + S_2$$. Among the given options — $$S_1 \times S_2$$, $$S_1 - S_2$$, $$\frac{S_1}{S_2}$$, and $$S_1 + S_2$$ — the physically meaningful and correct answer is $$S_1 + S_2$$, since entropy is additive.

We need to match each thermodynamic process with the quantity that remains constant.

(a) Isothermal process: The prefix "iso" means equal and "thermal" refers to temperature. So in an isothermal process, temperature is constant. This matches with (ii).

(b) Isochoric process: "Choric" comes from the Greek word for volume. So in an isochoric process, volume is constant. This matches with (iii).

(c) Adiabatic process: In an adiabatic process, no heat is exchanged with the surroundings. So the heat content is constant ($$\Delta Q = 0$$). This matches with (iv).

(d) Isobaric process: "Baric" refers to pressure. So in an isobaric process, pressure is constant. This matches with (i).

The correct matching is (a)→(ii), (b)→(iii), (c)→(iv), (d)→(i).

Hence, the correct answer is Option A.

First, we change the given temperatures into the absolute (kelvin) scale because all thermodynamic formulas use kelvin.

We have $$T_{1}=27^{\circ}\text{C}=27+273=300\ \text{K}$$ and $$T_{2}=37^{\circ}\text{C}=37+273=310\ \text{K}.$$

Now we must find the molar heat capacity at constant volume, $$C_{V},$$ of the gas. According to the classical equipartition theorem every quadratic degree of freedom contributes $$\tfrac12kT$$ (per molecule) to the internal energy. For one mole the contribution becomes $$\tfrac12RT.$$ Hence we first count the quadratic degrees of freedom (d.o.f.).

For a non-linear polyatomic molecule we have

• 3 translational d.o.f.
• 3 rotational d.o.f.
• 4 vibrational modes, and each vibrational mode supplies two quadratic d.o.f. (one kinetic + one potential), so $$4\times2=8$$ vibrational d.o.f.

Therefore the total number of quadratic degrees of freedom is

$$f = 3 + 3 + 8 = 14.$$

Stating the equipartition formula for one mole, the internal energy is

$$U = \frac{f}{2}RT.$$

Consequently the molar heat capacity at constant volume is

$$C_{V} = \frac{\partial U}{\partial T} = \frac{f}{2}R = \frac{14}{2}R = 7R.$$

We now calculate the change in internal energy when the temperature changes by $$\Delta T = T_{2}-T_{1} = 310\ \text{K} - 300\ \text{K} = 10\ \text{K}.$$

Hence (for one mole)

$$\Delta U = C_{V}\,\Delta T = 7R\,(10\ \text{K}).$$

Substituting $$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1},$$ we get

$$\Delta U = 7 \times 8.314 \times 10 = 581.98\ \text{J} \approx 582\ \text{J}.$$

The process is specified to be adiabatic, so we write the first law in the form

$$\Delta U = Q - W,$$

and because $$Q = 0$$ for an adiabatic change, this simplifies to

$$\Delta U = -W.$$

Rearranging, we have

$$W = -\Delta U.$$

Since $$\Delta U$$ is positive (the gas has gained internal energy), $$W$$ is negative, meaning that the work is done on the gas. Its magnitude equals $$|\Delta U| \approx 582\ \text{J}.$$

Therefore, the work done on the gas is close to 582 J.

Hence, the correct answer is Option 2.

We start from the first law of thermodynamics, which in differential form is stated as $$\Delta Q=\Delta U+\Delta W,$$ where $$\Delta Q$$ is the heat supplied to the system, $$\Delta U$$ is the change in internal energy and $$\Delta W$$ is the work done by the system.

Now we examine every thermodynamic process one by one and translate its defining property into conditions on $$\Delta Q,\;\Delta U,\;\Delta W.$

For an adiabatic process the defining statement is “no heat enters or leaves the system.” Mathematically this means $$\Delta Q=0.$$ Substituting $$\Delta Q=0$$ in the first-law expression we get $$0=\Delta U+\Delta W,$$ so generally neither $$\Delta U$$ nor $$\Delta W$$ has to be zero, only their sum must vanish. Hence the sole obligatory condition is $$\Delta Q=0,$$ which is condition (B).

For an isothermal process carried out on an ideal gas the temperature stays constant. The internal energy of an ideal gas depends only on temperature, so constant temperature immediately gives $$\Delta U=0.$$ Putting $$\Delta U=0$$ into $$\Delta Q=\Delta U+\Delta W$$ yields $$\Delta Q=\Delta W,$$ but neither of these terms is forced to be zero. Thus the characteristic condition is $$\Delta U=0,$$ corresponding to condition (D).

For an isochoric process the volume remains fixed, so the system cannot do pressure-volume work. The work term is $$\Delta W = P\,\Delta V,$$ and since $$\Delta V=0,$$ we have $$\Delta W=0.$$ Substituting $$\Delta W=0$$ in the first law gives $$\Delta Q=\Delta U,$$ with both quantities in general non-zero. Therefore the key condition is $$\Delta W=0,$$ i.e. condition (A).

For an isobaric process the pressure stays constant but the volume can change. Hence $$\Delta V\neq 0,$$ so $$\Delta W=P\,\Delta V\neq 0.$$ Temperature usually changes, making $$\Delta U\neq 0,$$ and because both $$\Delta U$$ and $$\Delta W$$ are non-zero, the heat supplied $$\Delta Q=\Delta U+\Delta W$$ is also non-zero. Thus all three changes are generally different from zero: $$\Delta U\neq 0,\;\Delta W\neq 0,\;\Delta Q\neq 0,$$ matching condition (C).

Collecting the matches we have:

$$ \begin{aligned} \text{(I) Adiabatic} &\longrightarrow \text{(B)} \\[4pt] \text{(II) Isothermal} &\longrightarrow \text{(D)} \\[4pt] \text{(III) Isochoric} &\longrightarrow \text{(A)} \\[4pt] \text{(IV) Isobaric} &\longrightarrow \text{(C)} \end{aligned} $$

This pattern corresponds exactly to Option D.

Hence, the correct answer is Option D.

We are told that one mole of an ideal di-atomic gas is initially at temperature $$T_1 = 300\;{\rm K}$$ and volume $$V_1$$. Its ratio of specific heats is given as $$\gamma = 1.4$$. The gas is first compressed adiabatically to a smaller volume $$V_2$$ where $$V_2 = \dfrac{V_1}{16}$$.

For any reversible (quasi-static) adiabatic process of an ideal gas, we recall the relation

$$T\,V^{\gamma-1} = \text{constant}.$$

This means that for the initial state (1) and the state just after the adiabatic compression (2) we must have

$$T_1\,V_1^{\gamma-1} \;=\; T_2\,V_2^{\gamma-1}.$$

Solving for $$T_2$$ gives

$$T_2 \;=\; T_1 \left(\dfrac{V_1}{V_2}\right)^{\gamma-1}.$$

We know $$\dfrac{V_1}{V_2} = 16$$ and $$\gamma - 1 = 1.4 - 1 = 0.4$$, so

$$T_2 \;=\; 300\;{\rm K}\;\times\; 16^{0.4}.$$

We next evaluate $$16^{0.4}$$. Noting that $$16 = 2^4$$, we have

$$16^{0.4} = (2^4)^{0.4} = 2^{4 \times 0.4} = 2^{1.6}.$$

Using $$2^{1.6} \approx 3.03$$, we obtain

$$T_2 \;\approx\; 300 \times 3.03 \;=\; 909\;{\rm K}.$$

The second step in the problem is an isobaric (constant-pressure) expansion from volume $$V_2$$ to volume $$V_3 = 2V_2$$. For a constant-pressure process of an ideal gas we invoke the ideal-gas law in the form

$$\dfrac{T}{V} = \text{constant (when $P$ is constant)},$$

so that

$$\dfrac{T_2}{V_2} = \dfrac{T_3}{V_3}.$$

Solving for the final temperature $$T_3$$ we get

$$T_3 \;=\; T_2 \left(\dfrac{V_3}{V_2}\right).$$

Since $$V_3 = 2V_2$$, this reduces to

$$T_3 \;=\; T_2 \times 2.$$

Substituting the value of $$T_2$$ found above, we find

$$T_3 \;\approx\; 2 \times 909\;{\rm K} \;=\; 1818\;{\rm K}.$$

Rounding to the nearest integer, the final temperature becomes $$1819\;{\rm K}$$.

Hence, the correct answer is Option 1819.

We are given that 5 moles of air (treated as an ideal di-atomic gas with rigid molecules) are initially at a temperature of $$T_i = 20^{\circ}\text{C} = 293\ \text{K}$$ and a pressure of $$P_i = 1\ \text{atm} = 1.013\times10^{5}\ \text{Pa}$$. The gas is compressed adiabatically to one-tenth of its original volume, i.e. $$V_f = \dfrac{V_i}{10}$$. We have to find the change in internal energy $$\Delta U$$ produced by this adiabatic compression.

For an ideal gas, the internal energy depends only on temperature, and the change in internal energy is given by the formula

$$\Delta U = n\,C_v\,(T_f - T_i)$$

where $$n$$ is the number of moles, $$C_v$$ is the molar heat capacity at constant volume, $$T_i$$ is the initial absolute temperature and $$T_f$$ is the final absolute temperature. Hence our first task is to determine $$C_v$$ and $$T_f$$.

Because the molecules are rigid di-atomic, they possess three translational and two rotational degrees of freedom, so the total degrees of freedom are $$f = 5$$. The molar heat capacity at constant volume is therefore

$$C_v = \dfrac{f}{2}\,R = \dfrac{5}{2}\,R$$

where $$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$$ is the universal gas constant. Thus

$$C_v = \dfrac{5}{2}\,(8.314) = 20.785\ \text{J mol}^{-1}\text{K}^{-1}.$$

For an adiabatic (reversible) process in an ideal gas we have the relation

$$T\,V^{\gamma - 1} = \text{constant},$$

where $$\gamma = \dfrac{C_p}{C_v} = 1 + \dfrac{2}{f}.$$ Substituting $$f = 5$$ gives

$$\gamma = 1 + \dfrac{2}{5} = \dfrac{7}{5} = 1.4.$$

Using the adiabatic relation for the initial state $$(T_i,V_i)$$ and the final state $$(T_f,V_f)$$ we write

$$T_i\,V_i^{\gamma-1} = T_f\,V_f^{\gamma-1}.$$

Dividing both sides by $$T_i$$ and rearranging for $$T_f$$ we get

$$T_f = T_i\left(\dfrac{V_i}{V_f}\right)^{\gamma-1}.$$

Since $$V_f = \dfrac{V_i}{10}$$, the volume ratio is

$$\dfrac{V_i}{V_f} = 10.$$

Therefore

$$T_f = 293\ \text{K}\; \times\; 10^{\,\gamma-1} = 293\ \text{K}\; \times\;10^{\,1.4 -1} = 293\ \text{K}\; \times\;10^{0.4}.$$

Now, $$10^{0.4} = e^{\,0.4\ln 10} = e^{\,0.4 \times 2.302585} = e^{0.921034} \approx 2.512.$$ Hence

$$T_f \approx 293\ \text{K}\,\times\,2.512 \approx 737\ \text{K}.$$

The change in temperature is therefore

$$\Delta T = T_f - T_i = 737\ \text{K} - 293\ \text{K} = 444\ \text{K}.$$

Substituting $$n = 5$$, $$C_v = 20.785\ \text{J mol}^{-1}\text{K}^{-1}$$ and $$\Delta T = 444\ \text{K}$$ into the formula for $$\Delta U$$, we obtain

$$\Delta U = 5\ \text{mol}\;\times\;20.785\ \dfrac{\text{J}}{\text{mol K}}\;\times\;444\ \text{K}.$$ $$\Delta U = 5 \times (20.785 \times 444)\ \text{J}.$$

First compute the bracketed product:

$$20.785 \times 444 = 20.785 \times (400 + 44) = 20.785 \times 400 + 20.785 \times 44 = 8\,314 + 914.5 = 9\,228.5\ \text{J}.$$

Now multiply by 5 moles:

$$\Delta U = 5 \times 9\,228.5\ \text{J} = 46\,142.5\ \text{J} \approx 4.614\times10^{4}\ \text{J}.$$

Converting to kilojoules,

$$\Delta U \approx 46.1\ \text{kJ}.$$

Rounded to the nearest integer, $$\Delta U = 46\ \text{kJ}.$$ So, the answer is $$46\ \text{kJ}$$.

We have a Carnot heat engine working between two thermal reservoirs whose absolute temperatures are $$T_h = 900\,\text{K}$$ (hot reservoir) and $$T_c = 300\,\text{K}$$ (cold reservoir).

The thermal efficiency $$\eta$$ of an ideal Carnot engine is given by the well-known formula

$$\eta = 1 - \frac{T_c}{T_h}.$$

Substituting the given temperatures, we obtain

$$\eta = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3}.$$

By definition, the efficiency is also the ratio of the useful work output $$W$$ to the heat energy $$Q_h$$ absorbed from the hot reservoir:

$$\eta = \frac{W}{Q_h}.$$

The work done per cycle is given as $$W = 1200\,\text{J}.$$ Solving the above relation for $$Q_h$$, we have

$$Q_h = \frac{W}{\eta}.$$

Substituting the numerical values,

$$Q_h = \frac{1200}{\dfrac{2}{3}} = 1200 \times \frac{3}{2} = 1800\,\text{J}.$$

The heat rejected to the cold reservoir, denoted by $$Q_c$$, is found using the first law of thermodynamics for a complete cycle, which states that the net heat absorbed equals the work done:

$$Q_h - Q_c = W.$$

Rearranging gives

$$Q_c = Q_h - W.$$

Substituting $$Q_h = 1800\,\text{J}$$ and $$W = 1200\,\text{J},$$ we get

$$Q_c = 1800 - 1200 = 600\,\text{J}.$$

So, the answer is $$600\,\text{J}.$$

We have 100 g of water at 0 °C that must be converted to ice at the same temperature. During freezing the refrigerator must remove only the latent heat, so the heat taken out from the water-ice system (the cold reservoir) is

$$Q_C = m\,L = 100\;\text{g}\times 80\;\text{cal g}^{-1}=8000\;\text{cal}.$$

To obtain the minimum possible work input, the refrigerator has to operate reversibly, i.e. as an ideal Carnot refrigerator, between the cold‐reservoir temperature 0 °C and the surrounding (hot reservoir) temperature 27 °C.

First we convert both temperatures to the absolute (Kelvin) scale:

$$T_C = 0\;^\circ\text{C} = 273\;\text{K},\qquad T_H = 27\;^\circ\text{C} = 300\;\text{K}.$$

The Carnot coefficient of performance (COP) for a refrigerator is stated by the formula

$$\text{COP}_{\text{max}} = \frac{T_C}{T_H - T_C}.$$

Substituting the temperatures,

$$\text{COP}_{\text{max}} = \frac{273}{300-273} = \frac{273}{27} = 10.111\;.$$

The work that must be supplied to a refrigerator is related to the heat extracted and the COP by the equation

$$W_{\text{min}} = \frac{Q_C}{\text{COP}_{\text{max}}}.$$

So,

$$W_{\text{min}} = \frac{8000}{10.111} \approx 791.3\;\text{cal}.$$

The heat rejected to the surroundings (the hot reservoir) is the sum of the heat extracted from the cold side and the work put in:

$$Q_H = Q_C + W_{\text{min}} = 8000 + 791.3 = 8791.3\;\text{cal}.$$

Rounding this to the nearest integer gives

$$Q_H \approx 8791\;\text{cal}.$$

Hence, the correct answer is Option A.

We first recall the basic form of the First Law of Thermodynamics, which is written as

$$\Delta U \;=\; Q \;-\; W,$$

where $$\Delta U$$ is the change in internal energy of the system, $$Q$$ is the heat supplied to the system, and $$W$$ is the work done by the system on the surroundings.

Now consider what actually happens the moment a rubber balloon filled with helium suddenly bursts. The rubber skin tears so quickly that the gas inside finds itself in the open atmosphere almost instantaneously. Because the time interval of the tearing is extremely small, there is practically no opportunity for energy to flow as heat between the helium and the surrounding air. In thermodynamics, a process in which $$Q = 0$$ is called adiabatic. Substituting $$Q = 0$$ into the First Law gives

$$\Delta U \;=\; -W.$$

Next, let us decide whether the work term represents a reversible or an irreversible process. In a reversible expansion every intermediate state is in mechanical equilibrium, so the external pressure $$P_{\text{ext}}$$ always matches the internal pressure $$P_{\text{int}}$$ infinitesimally closely, and the system passes through well-defined equilibrium states. In the bursting of a balloon the internal pressure, initially $$1.7\ \text{atm}$$, is suddenly released against an external pressure of roughly $$1\ \text{atm}$$. That large pressure difference means the gas expands in one violent “jump,” far from equilibrium. Because the expansion occurs against a finite pressure difference, it cannot trace a sequence of equilibrium states, so it is by definition irreversible.

Putting the two observations together:

• Heat exchange: $$Q = 0 \;\Rightarrow\;$$ adiabatic.
• Mechanical path: large pressure difference, single rapid step $$\Rightarrow$$ irreversible.

Therefore the expansion of helium immediately after the balloon bursts is best described as an irreversible adiabatic process.

Hence, the correct answer is Option B.

For a heat engine we use the First Law of Thermodynamics which says that over one complete cycle, the net work done $$W$$ equals the algebraic sum of all heats exchanged:

$$W=\sum Q_i$$

The heats exchanged in the given cycle are

$$+1915\ \text{J},\; -40\ \text{J},\; +125\ \text{J},\; -Q\ \text{J}$$

(A positive sign means heat is absorbed by the engine, a negative sign means heat is rejected.)

So the net heat, and therefore the work done in one cycle, is

$$W = 1915 + (-40) + 125 + (-Q)$$

Adding the known numbers step by step:

$$1915 + 125 = 2040$$

$$2040 - 40 = 2000$$

Hence

$$W = 2000 - Q$$

Next, efficiency $$\eta$$ of a heat engine is defined as

$$\eta = \frac{W}{Q_{\text{in}}}$$

where $$Q_{\text{in}}$$ is the total heat absorbed (all positive terms only). The positive heats here are $$1915\ \text{J}$$ and $$125\ \text{J}$$, so

$$Q_{\text{in}} = 1915 + 125 = 2040\ \text{J}$$

The given efficiency is $$\eta = 0.500$$, therefore

$$0.500 = \frac{W}{2040}$$

Substituting $$W = 2000 - Q$$ into this relation gives

$$0.500 = \frac{2000 - Q}{2040}$$

Multiplying both sides by $$2040$$:

$$0.500 \times 2040 = 2000 - Q$$

$$1020 = 2000 - Q$$

Rearranging to isolate $$Q$$:

$$-Q = 1020 - 2000$$

$$-Q = -980$$

$$Q = 980\ \text{J}$$

Hence, the correct answer is Option C.

We start by writing all data in convenient SI units. The initial volume of air is $$V_1 = 1\;\text{L} = 1 \times 10^{-3}\;\text{m}^3$$ and the final volume is $$V_2 = 3\;\text{L} = 3 \times 10^{-3}\;\text{m}^3$$. At STP the initial pressure is $$P_1 = 1\;\text{atm} = 1.013 \times 10^{5}\;\text{Pa}$$ and the initial temperature is $$T_1 = 273\;\text{K}$$. For air the adiabatic index is $$\gamma = 1.40$$.

An adiabatic process for an ideal gas obeys the relation $$P V^{\gamma} = \text{constant}.$$ Using this, the final pressure $$P_2$$ can be obtained from

$$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}\,,$$

so

$$P_2 = P_1 \left(\dfrac{V_1}{V_2}\right)^{\gamma}.$$

Here $$\dfrac{V_1}{V_2} = \dfrac{1}{3}$$, hence

$$P_2 = 1.013 \times 10^{5}\;\text{Pa}\;\times \left(\dfrac{1}{3}\right)^{1.40}.$$

The question supplies $$3^{1.4} = 4.6555$$, therefore

$$\left(\dfrac{1}{3}\right)^{1.4} = \dfrac{1}{3^{1.4}} = \dfrac{1}{4.6555} = 0.2148.$$

Substituting, we get

$$P_2 = 1.013 \times 10^{5}\;\text{Pa} \times 0.2148 = 2.1768 \times 10^{4}\;\text{Pa}.$$

Now we recall the formula for the work done during an adiabatic process for an ideal gas:

$$W = \dfrac{P_1 V_1 - P_2 V_2}{\gamma - 1}.$$

First calculate the two pressure-volume products (remembering that $$1\;\text{Pa}\cdot\text{m}^3 = 1\;\text{J}$$):

$$P_1 V_1 = \bigl(1.013 \times 10^{5}\;\text{Pa}\bigr)\bigl(1 \times 10^{-3}\;\text{m}^3\bigr) = 101.3\;\text{J},$$

$$P_2 V_2 = \bigl(2.1768 \times 10^{4}\;\text{Pa}\bigr)\bigl(3 \times 10^{-3}\;\text{m}^3\bigr) = 65.304\;\text{J}.$$

The difference of these two terms is

$$P_1 V_1 - P_2 V_2 = 101.3\;\text{J} - 65.304\;\text{J} = 35.996\;\text{J}.$$

Now divide by $$\gamma - 1$$. Since $$\gamma = 1.40$$, we have $$\gamma - 1 = 0.40$$, so

$$W = \dfrac{35.996\;\text{J}}{0.40} = 89.99\;\text{J} \approx 90.5\;\text{J}.$$

Hence, the correct answer is Option B.

Let us denote the initial state of the di-atomic gas by the subscript 1 and the final state after the adiabatic change by the subscript 2.

So, the initial pressure and density are $$P_1$$ and $$\rho_1$$, while the final pressure and density are $$P_2$$ and $$\rho_2$$ respectively.

According to the statement of the problem, the density becomes $$32$$ times its initial value. Therefore

$$\rho_2 \;=\; 32\,\rho_1.$$

We are told that the final pressure is $$n$$ times the initial pressure. Hence

$$P_2 \;=\; n\,P_1.$$

Because the change is adiabatic, we must use the adiabatic relation that connects pressure and volume. First we recall the formula

$$P\,V^{\gamma} \;=\; \text{constant}$$

where $$\gamma$$ is the ratio of specific heats $$C_P/C_V$$. For a di-atomic ideal gas (with no vibrational modes excited in the temperature range considered) we have

$$\gamma \;=\;\frac{7}{5}.$$

To involve the density, we note that density is inversely proportional to volume for a fixed mass of gas. In symbols, if $$m$$ is the mass,

$$\rho \;=\;\frac{m}{V}\quad\Longrightarrow\quad V \;=\;\frac{m}{\rho} \;\propto\; \frac{1}{\rho}.$$

Substituting $$V = \dfrac{1}{\rho}$$ (ignoring the constant mass factor which cancels out when we form ratios) into the adiabatic relation, we obtain

$$P\Bigl(\frac{1}{\rho}\Bigr)^{\gamma} \;=\; \text{constant}.$$

Multiplying both sides by $$\rho^{\gamma}$$ gives

$$P \;\rho^{-\gamma}\;=\;\text{constant}\quad\Longrightarrow\quad P\;\rho^{-\gamma} = \text{constant}.$$ Therefore pressure and density are connected by

$$P \;\propto\; \rho^{\gamma}.$$

Using this proportionality for the two states, we write

$$\frac{P_2}{P_1} \;=\;\Bigl(\frac{\rho_2}{\rho_1}\Bigr)^{\gamma}.$$

Now we substitute the given multiples:

$$\frac{P_2}{P_1} \;=\; n,$$

and

$$\frac{\rho_2}{\rho_1} \;=\; 32.$$

Hence

$$n \;=\; 32^{\gamma}.$$

Putting $$\gamma = \dfrac{7}{5}$$, we have

$$n \;=\; 32^{\,\tfrac{7}{5}}.$$

We rewrite 32 as a power of 2 for convenience:

$$32 \;=\; 2^{5}.$$

Therefore

$$n \;=\; (2^{5})^{\tfrac{7}{5}}.$$

Using the law of exponents $$(a^{b})^{c} = a^{\,bc},$$ we obtain

$$n \;=\; 2^{\,5 \times \tfrac{7}{5}} \;=\; 2^{7}.$$

Now $$2^{7} = 128.$$

So we have found

$$n \;=\; 128.$$

Hence, the correct answer is Option 3.

We are dealing with an ideal mono-atomic gas. For such a gas the internal energy is a pure function of temperature alone and is given by the well-known expression

$$U=\tfrac32\,nRT.$$

Consequently, the change in internal energy between any two states depends only on the initial and final temperatures:

$$\Delta U=\tfrac32\,nR\,(T_{\text{final}}-T_{\text{initial}}).$$

In the $$P\!-\!V$$ diagram all three processes start from the common state $$A(P_A,V_A,T_A)$$ and terminate at the three distinct points $$B,\;C,\;D$$. A careful inspection of the diagram shows that the points $$B,\;C,\;D$$ lie on the same isothermal curve that passes through none of the other points. Algebraically that statement reads

$$P_BV_B=P_CV_C=P_DV_D=k\quad(\text{constant}).$$

Using the ideal-gas equation $$PV=nRT$$ this instantly gives

$$T_B=T_C=T_D=T',$$

where $$T'$$ is some temperature (which may or may not equal $$T_A$$). Because all three final temperatures are equal, the change in internal energy from the common initial state $$A$$ to any of the three final states is identical:

$$E_{AB}=\Delta U_{AB}=\tfrac32\,nR\,(T'-T_A),$$

$$E_{AC}=\Delta U_{AC}=\tfrac32\,nR\,(T'-T_A),$$

$$E_{AD}=\Delta U_{AD}=\tfrac32\,nR\,(T'-T_A).$$

Hence

$$E_{AB}=E_{AC}=E_{AD}.$$

Next we examine the work done, remembering the definition

$$W=\int_{V_{\text{initial}}}^{V_{\text{final}}}P\,dV.$$

Sign conventions to keep in mind:

• When the gas expands, $$dV>0$$ so $$W>0$$(positive work done by the gas).
• When the gas undergoes an isochoric (constant-volume) change, $$dV=0$$ so $$W=0$$(no work).
• When the gas is compressed, $$dV<0$$ so $$W<0$$(negative work, work done on the gas).

Looking again at the geometry of the three paths:

• Along $$A\to B$$ the curve clearly moves to a larger volume: $$V_B>V_A$$. Therefore $$W_{AB}>0.$

• Along $$A\to C$$ the path is vertical, i.e. the volume is unchanged: $$V_C=V_A$$. Hence $$W_{AC}=0.$

• Along $$A\to D$$ the path proceeds to a smaller volume: $$V_D<V_A$$. Consequently $$W_{AD}<0.$

Collecting all the deductions we have

$$E_{AB}=E_{AC}=E_{AD},\qquad W_{AB}>0,\qquad W_{AC}=0,\qquad W_{AD}<0.$$

This set of equalities and inequalities is exactly the one listed in Option B of the question.

Hence, the correct answer is Option B.

We have a Carnot engine whose efficiency when it works as a heat engine is given as $$\dfrac{1}{10}$$. First, recall the definition of efficiency for a Carnot heat engine:

$$\eta \;=\;\dfrac{W}{Q_h}\;=\;1-\dfrac{T_c}{T_h}$$

Here $$\eta$$ is the efficiency, $$W$$ is the work obtained from the engine, $$Q_h$$ is the heat absorbed from the high-temperature reservoir, $$T_h$$ is the absolute temperature of the hot reservoir, and $$T_c$$ is the absolute temperature of the cold reservoir.

Substituting the given efficiency, we write

$$\eta \;=\;\dfrac{1}{10}\;=\;1-\dfrac{T_c}{T_h}.$$

Now we isolate the temperature ratio. Moving the fraction to the right-hand side and the constant term to the left, we obtain

$$\dfrac{T_c}{T_h}\;=\;1-\dfrac{1}{10}.$$

The right side simplifies as

$$1-\dfrac{1}{10}\;=\;\dfrac{10}{10}-\dfrac{1}{10}\;=\;\dfrac{9}{10}.$$

So,

$$\dfrac{T_c}{T_h}\;=\;\dfrac{9}{10}=0.9.$$

Next, we change the mode of operation: the same Carnot device is now used as a refrigerator. For a Carnot refrigerator, the coefficient of performance (COP) is defined by the formula

$$\text{COP}_{\text{ref}}\;=\;\dfrac{Q_c}{W}\;=\;\dfrac{T_c}{T_h-T_c},$$

where $$Q_c$$ is the heat drawn from the cold (low-temperature) reservoir and $$W$$ is the work input to run the refrigerator.

We already know $$\dfrac{T_c}{T_h}=0.9$$, so let us express $$T_c$$ and $$T_h-T_c$$ in terms of $$T_h$$:

$$T_c = 0.9\,T_h,$$

$$T_h - T_c = T_h - 0.9\,T_h = 0.1\,T_h.$$

Substituting these values into the COP expression, we get

$$\text{COP}_{\text{ref}} = \dfrac{0.9\,T_h}{0.1\,T_h}.$$

The common factor $$T_h$$ cancels out, leaving

$$\text{COP}_{\text{ref}} = \dfrac{0.9}{0.1}=9.$$

This means that for every 1 joule of work supplied to the refrigerator, it absorbs 9 joules of heat from the low-temperature reservoir.

Now, the problem states that the work done on the refrigerator is 10 J. Using the definition of COP once again,

$$\text{COP}_{\text{ref}} = \dfrac{Q_c}{W}.$$

Substituting $$\text{COP}_{\text{ref}} = 9$$ and $$W = 10\;\text{J}$$, we find

$$9 = \dfrac{Q_c}{10\;\text{J}}.$$

Multiplying both sides by 10 J, we get

$$Q_c = 9 \times 10\;\text{J} = 90\;\text{J}.$$

So, the refrigerator absorbs 90 J of heat from the low-temperature reservoir.

Hence, the correct answer is Option D.

First we recall the Carnot‐engine efficiency formula, valid for all temperatures measured on the absolute (kelvin) scale:

$$\text{Efficiency} = 1 - \dfrac{T_\text{cold}}{T_\text{hot}}.$$

The first engine works between the hot reservoir at temperature $$T_1$$ and the intermediate reservoir at temperature $$T$$. Its efficiency is therefore

$$\eta_1 = 1 - \dfrac{T}{T_1}.$$

Let us denote by $$Q_1$$ the heat absorbed from the hot reservoir $$T_1$$. The work produced by this first engine is

$$W_1 = \eta_1 Q_1 = \left(1 - \dfrac{T}{T_1}\right)Q_1.$$

The heat rejected by the first engine becomes the input heat for the second engine. Using the definition $$Q_\text{rejected} = Q_\text{absorbed} - W$$ we write

$$Q_{\text{rejected},1} = Q_1 - W_1.$$

Substituting the value of $$W_1$$ we get

$$Q_{\text{rejected},1} = Q_1 - \left(1 - \dfrac{T}{T_1}\right)Q_1 = Q_1 \left[1 - 1 + \dfrac{T}{T_1}\right] = Q_1 \dfrac{T}{T_1}.$$

This rejected heat is the heat absorbed by the second engine, so we set

$$Q_2 = Q_{\text{rejected},1} = Q_1 \dfrac{T}{T_1}.$$

The second engine operates between the source at $$T$$ and the sink at $$T_2$$. Its efficiency is

$$\eta_2 = 1 - \dfrac{T_2}{T}.$$

Hence the work done by the second engine is

$$W_2 = \eta_2 Q_2 = \left(1 - \dfrac{T_2}{T}\right) \left(Q_1 \dfrac{T}{T_1}\right).$$

We are told that both engines perform equal amounts of work, so we impose the condition

$$W_1 = W_2.$$

Substituting the expressions for $$W_1$$ and $$W_2$$, we have

$$\left(1 - \dfrac{T}{T_1}\right)Q_1 = \left(1 - \dfrac{T_2}{T}\right) \left(Q_1 \dfrac{T}{T_1}\right).$$

Dividing both sides by $$Q_1$$ (which is non-zero) gives

$$1 - \dfrac{T}{T_1} = \left(1 - \dfrac{T_2}{T}\right)\dfrac{T}{T_1}.$$

Now we multiply every term by $$T_1$$ to clear the denominators:

$$T_1 - T = \left(1 - \dfrac{T_2}{T}\right)T.$$

Expanding the right-hand side:

$$T_1 - T = T - \dfrac{T_2 T}{T}.$$

Simplifying the last term $$\dfrac{T_2 T}{T} = T_2$$, we obtain

$$T_1 - T = T - T_2.$$

Bringing like terms together:

$$T_1 + T_2 = 2T.$$

Finally we solve for the intermediate temperature $$T$$:

$$T = \dfrac{T_1 + T_2}{2}.$$

Thus the required temperature is simply the arithmetic mean of $$T_1$$ and $$T_2$$.

Hence, the correct answer is Option B.

For an adiabatic process of an ideal gas we first recall the fundamental relation

$$P\,V^{\gamma}= \text{constant},$$

where $$\gamma=\dfrac{C_P}{C_V}$$ is the ratio of molar heat capacities at constant pressure and at constant volume.

Because the gas is diatomic and rigid at room temperature, it possesses three translational and two rotational degrees of freedom, making a total of $$f=5$$ degrees of freedom. The vibrational modes are not excited at room temperature, so we neglect them.

Using the formula $$C_V=\dfrac{f}{2}R,$$ we obtain

$$C_V=\dfrac{5}{2}R.$$

Next we employ the relation $$C_P=C_V+R.$$ Substituting the value of $$C_V$$ gives

$$C_P=\dfrac{5}{2}R+R=\dfrac{7}{2}R.$$

Now we find $$\gamma$$:

$$\gamma=\dfrac{C_P}{C_V}=\dfrac{\dfrac{7}{2}R}{\dfrac{5}{2}R}=\dfrac{7}{5}=1.4.$$

The ideal-gas equation is $$P V = n R T.$$ Solving it for pressure, we have

$$P=\dfrac{nRT}{V}.$$

We substitute this expression for $$P$$ into the adiabatic condition $$P\,V^{\gamma}= \text{constant}$$:

$$\left(\dfrac{nRT}{V}\right) V^{\gamma}= \text{constant}.$$ $$nRT\,V^{\gamma-1}= \text{constant}.$$

The factors $$n$$ and $$R$$ are fixed for a given sample, so they can be absorbed into the constant. Thus we may write

$$T\,V^{\gamma-1}= \text{constant}.$$

Comparing this with the form given in the problem statement, namely $$T V^{x}= \text{constant},$$ we immediately identify

$$x=\gamma-1.$$

Substituting the value $$\gamma=\dfrac{7}{5},$$ we obtain

$$x=\dfrac{7}{5}-1=\dfrac{2}{5}.$$

Hence, the correct answer is Option B.

Given:

$$ΔU_{ca}=−180J,\ Q_{ab}=250J,\ Q_{bc}=60J$$

Step 1: Internal energy change along ac

$$ΔU_{ac}=−ΔU_{ca}=−(−180)=180J$$

Step 2: Total heat absorbed along abc

$$Q=Q_{ab}+Q_{bc}=250+60=310J$$

Step 3: Apply First Law of Thermodynamics

Q=ΔU+W

310=180+W

⇒W=130 J

Based on the First Law of Thermodynamics and the properties of state functions, the solution is as follows:

1. Internal Energy Change ($$\Delta U$$):

Internal energy is a state function, meaning it only depends on the initial and final states of the system, not the path taken. Since both processes A and B start and end at the same points on the P-V diagram:

$$\Delta U_A = \Delta U_B$$

2. Work Done (W):

Work done in a thermodynamic process is equal to the area under the curve on a P-V diagram.

• Process A stays "above" process B.
• Therefore, the area under curve A is greater than the area under curve B.$$W_A > W_B$$

3. Heat Absorbed ($$\Delta Q$$):

According to the First Law of Thermodynamics:

$$\Delta Q = \Delta U + W$$

Applying this to both processes:

• $$\Delta Q_A = \Delta U_A + W_A$$
• $$\Delta Q_B = \Delta U_B + W_B$$

Since $$\Delta U$$ is the same for both but $$W_A > W_B$$, it follows that:

$$\boxed{\Delta Q_A > \Delta Q_B \quad \text{and} \quad \Delta U_A = \Delta U_B}$$

We have one mole of an ideal mono-atomic gas. Throughout the process the variables obey $$VT = K$$ where $$K$$ is a positive constant. Because the gas is ideal, the equation of state is also valid at every instant:

$$PV = RT \qquad\text{(for }n = 1\text{ mol).}$$

The temperature is raised by $$\Delta T,$$ so if we call the initial temperature $$T_i,$$ the final temperature is $$T_f = T_i + \Delta T.$$

First we express the volume in terms of temperature using the given relation:

$$VT = K \;\;\Longrightarrow\;\; V = \dfrac{K}{T}.$$

Next we write the pressure in terms of temperature alone. Substituting this $$V$$ into the ideal-gas equation gives

$$P = \dfrac{RT}{V} = \dfrac{RT}{K/T} = \dfrac{RT^{2}}{K}.$$

To find the work done by the gas we use the formula

$$W = \int_{V_i}^{V_f} P\,dV.$$

Because it is more convenient to integrate over temperature, we change variables. Differentiate the relation $$V = K/T$$:

$$dV = -\,\dfrac{K}{T^{2}}\;dT.$$

Now substitute both $$P$$ and $$dV$$ into the integral:

$$\displaystyle W = \int_{T_i}^{T_f} \Bigl(\dfrac{RT^{2}}{K}\Bigr)\,\Bigl(-\,\dfrac{K}{T^{2}}\;dT\Bigr) = \int_{T_i}^{T_f} \bigl(-R\bigr)\,dT = -R\,(T_f - T_i).$$

Since $$T_f - T_i = \Delta T,$$ the work done by the gas is

$$W = -\,R\Delta T.$$

The negative sign means the gas has been compressed (its volume decreased as temperature rose), so work is done on the gas; nevertheless, we keep the sign in further calculations exactly as obtained.

For a mono-atomic ideal gas the molar heat capacity at constant volume is

$$C_V = \dfrac{3}{2}R.$$

The change in internal energy is therefore, by the formula $$\Delta U = nC_V\Delta T,$$

$$\Delta U = (1)\Bigl(\dfrac{3}{2}R\Bigr)\Delta T = \dfrac{3}{2}R\Delta T.$$

Finally, we apply the first law of thermodynamics, $$Q = \Delta U + W.$$ Substituting the values just found gives

$$Q = \dfrac{3}{2}R\Delta T + \bigl(-R\Delta T\bigr) = \Bigl(\dfrac{3}{2} - 1\Bigr)R\Delta T = \dfrac{1}{2}R\Delta T.$$

This $$Q$$ is positive, so that amount of heat is indeed absorbed by the gas.

Hence, the correct answer is Option A.

We are given that a Carnot engine has an efficiency of $$\frac{1}{6}$$, and when the temperature of the sink is reduced by $$62^\circ C$$, its efficiency is doubled.

Step 1: Set up equations using Carnot efficiency

The efficiency of a Carnot engine is given by:

$$\eta = 1 - \frac{T_2}{T_1}$$

where $$T_1$$ is the source temperature and $$T_2$$ is the sink temperature (both in Kelvin).

Step 2: First condition

$$\frac{1}{6} = 1 - \frac{T_2}{T_1}$$

$$\frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6}$$

$$T_2 = \frac{5}{6}T_1$$ $$-(1)$$

Step 3: Second condition

When the sink temperature is reduced by $$62^\circ C$$ (i.e., by 62 K), the new sink temperature is $$T_2 - 62$$ and the efficiency doubles to $$\frac{1}{3}$$:

$$\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$$

$$\frac{T_2 - 62}{T_1} = \frac{2}{3}$$

$$T_2 - 62 = \frac{2}{3}T_1$$ $$-(2)$$

Step 4: Solve the equations

Substituting $$(1)$$ into $$(2)$$:

$$\frac{5}{6}T_1 - 62 = \frac{2}{3}T_1$$

$$\frac{5}{6}T_1 - \frac{2}{3}T_1 = 62$$

$$\frac{5T_1 - 4T_1}{6} = 62$$

$$\frac{T_1}{6} = 62$$

$$T_1 = 372 \text{ K}$$

From $$(1)$$:

$$T_2 = \frac{5}{6} \times 372 = 310 \text{ K}$$

Step 5: Convert to Celsius

$$T_1 = 372 - 273 = 99^\circ C$$ (source temperature)

$$T_2 = 310 - 273 = 37^\circ C$$ (sink temperature)

The temperatures of the source and sink are $$99^\circ C$$ and $$37^\circ C$$, respectively.

The correct answer is Option C.

Based on the principles of thermodynamics and the slopes of the curves on a $$P-V$$ diagram, the processes are assigned as follows:

• Isobaric (Process $$a$$): Pressure remains constant, resulting in a horizontal line.
• Isochoric (Process $$d$$): Volume remains constant, resulting in a vertical line.
• Isothermal (Process $$b$$): Temperature remains constant ($$P \propto \frac{1}{V}$$).
• Adiabatic (Process $$c$$): No heat exchange occurs. The adiabatic curve is steeper than the isothermal curve because the adiabatic index $$\gamma > 1$$.

Correct Assignment:

• Isochoric $$\rightarrow$$ Process d
• Isobaric $$\rightarrow$$ Process a
• Isothermal $$\rightarrow$$ Process b
• Adiabatic $$\rightarrow$$ Process c

Final Order:

$$\boxed{d, a, b, c}$$

We are given three Carnot engines working one after the other (in series). The temperatures of the four reservoirs, arranged from hottest to coldest, are $$T_1, T_2, T_3,$$ and $$T_4$$ with $$T_1 > T_2 > T_3 > T_4.$$

For any Carnot engine operating between a hot reservoir at temperature $$T_h$$ and a cold reservoir at $$T_c,$$ the thermal efficiency is, by definition,

$$\eta = 1-\dfrac{T_c}{T_h}.$$

Because the engines are in series, the first engine works between $$T_1$$ (hot) and $$T_2$$ (cold), the second between $$T_2$$ and $$T_3,$$ and the third between $$T_3$$ and $$T_4.$$ Their efficiencies are therefore

$$\eta_1 = 1-\dfrac{T_2}{T_1}, \qquad \eta_2 = 1-\dfrac{T_3}{T_2}, \qquad \eta_3 = 1-\dfrac{T_4}{T_3}.$$

The statement “the three engines are equally efficient’’ means

$$\eta_1 = \eta_2 = \eta_3.$$

We first equate $$\eta_1$$ and $$\eta_2.$$ Starting with the equality

$$1-\dfrac{T_2}{T_1} = 1-\dfrac{T_3}{T_2},$$

we cancel the 1’s on both sides, giving

$$-\dfrac{T_2}{T_1} = -\dfrac{T_3}{T_2}.$$

Multiplying through by $$-1$$ to remove the negative sign, we have

$$\dfrac{T_2}{T_1} = \dfrac{T_3}{T_2}.$$

Cross-multiplication now yields

$$T_2^2 = T_1\,T_3,$$

and solving for $$T_3$$ gives

$$T_3 = \dfrac{T_2^2}{T_1}. \quad -(1)$$

Next we equate $$\eta_2$$ and $$\eta_3.$$ Setting

$$1-\dfrac{T_3}{T_2} = 1-\dfrac{T_4}{T_3},$$

we again cancel the 1’s, obtaining

$$-\dfrac{T_3}{T_2} = -\dfrac{T_4}{T_3}.$$

After multiplying by $$-1,$$ we have