Chapter 11: Thermodynamics

Given: rate of water flow $$= 3.0$$ litres per minute, temperature rises from $$27\,{}^\circ\mathrm{C}$$ to $$77\,{}^\circ\mathrm{C}$$, heat of combustion of the fuel $$= 4.0 \times 10^4 \, \mathrm{J/g}$$.

Step 1 — Mass of water heated per minute. Taking the density of water as $$1 \, \mathrm{kg/L}$$, the mass of water flowing through the geyser each minute is $$m = 3.0 \, \mathrm{kg}$$.

Step 2 — Heat required per minute. The rise in temperature is $$\Delta T = 77 - 27 = 50\,{}^\circ\mathrm{C} = 50 \, \mathrm{K}$$. With specific heat of water $$c = 4.2 \times 10^3 \, \mathrm{J\,kg^{-1}\,K^{-1}}$$, the heat absorbed by the water each minute is

$$Q = mc\,\Delta T = 3.0 \times (4.2 \times 10^3) \times 50 = 6.3 \times 10^5 \, \mathrm{J}$$

Step 3 — Mass of fuel burnt per minute. All this heat is supplied by burning the fuel. Since burning $$1 \, \mathrm{g}$$ of fuel releases $$4.0 \times 10^4 \, \mathrm{J}$$, the mass of fuel needed per minute is

$$m_{\text{fuel}} = \dfrac{Q}{\text{heat of combustion}} = \dfrac{6.3 \times 10^5}{4.0 \times 10^4} = 15.75 \, \mathrm{g}$$

Hence the geyser consumes fuel at the rate of about $$15.75 \, \mathrm{g}$$ per minute.

Given: mass of nitrogen $$m = 2.0 \times 10^{-2} \, \mathrm{kg} = 20 \, \mathrm{g}$$, molecular mass $$M = 28 \, \mathrm{g\,mol^{-1}}$$, rise in temperature $$\Delta T = 45\,{}^\circ\mathrm{C} = 45 \, \mathrm{K}$$, $$R = 8.3 \, \mathrm{J\,mol^{-1}\,K^{-1}}$$, and the heating is at constant pressure.

Step 1 — Number of moles.

$$n = \dfrac{m}{M} = \dfrac{20}{28} = 0.714 \, \mathrm{mol}$$

Step 2 — Molar specific heat at constant pressure. Nitrogen ($$\mathrm{N_2}$$) is a diatomic gas. For a diatomic gas the molar specific heat at constant pressure is

$$C_p = \dfrac{7}{2}R = \dfrac{7}{2}\times 8.3 = 29.05 \, \mathrm{J\,mol^{-1}\,K^{-1}}$$

Step 3 — Heat supplied at constant pressure. At constant pressure the heat required is

$$Q = nC_p\,\Delta T = 0.714 \times 29.05 \times 45 \approx 933.4 \, \mathrm{J}$$

Hence about $$933 \, \mathrm{J}$$ of heat must be supplied to the nitrogen.

Identifying the process. The walls of the cylinder are heat insulators and the piston is also insulated (by the sand on it). So no heat can enter or leave the gas: $$Q = 0$$. The compression is therefore an adiabatic process.

Step 1 — Adiabatic relation. For an adiabatic process, $$PV^{\gamma} = \text{constant}$$. Comparing the initial state $$(P_1, V_1)$$ with the final state $$(P_2, V_2)$$,

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

Step 2 — Ratio of pressures.

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

The gas is compressed to half its volume, so $$V_2 = \dfrac{V_1}{2}$$, which gives $$\dfrac{V_1}{V_2} = 2$$.

Step 3 — Value of $$\gamma$$. Hydrogen ($$\mathrm{H_2}$$) is a diatomic gas, for which

$$\gamma = \dfrac{C_p}{C_v} = \dfrac{7}{5} = 1.4$$

Step 4 — Factor by which the pressure increases.

$$\dfrac{P_2}{P_1} = (2)^{1.4} \approx 2.64$$

Hence the pressure of the gas increases by a factor of about $$2.64$$. (Note that the number of moles of gas does not affect this ratio.)

First law of thermodynamics: $$\Delta U = Q - W$$, where $$Q$$ is the heat absorbed by the system, $$W$$ is the work done by the system, and $$\Delta U$$ is the change in internal energy.

Process 1 — adiabatic path $$A \to B$$. Work done on the system is $$22.3 \, \mathrm{J}$$, so the work done by the system is

$$W = -22.3 \, \mathrm{J}$$

For an adiabatic process no heat is exchanged, so $$Q = 0$$. Applying the first law,

$$\Delta U = Q - W = 0 - (-22.3) = +22.3 \, \mathrm{J}$$

Key idea — internal energy is a state function. $$\Delta U$$ depends only on the initial state $$A$$ and the final state $$B$$, not on the path joining them. Hence for any process taking the gas from $$A$$ to $$B$$,

$$\Delta U = 22.3 \, \mathrm{J}$$

Process 2 — path $$A \to B$$ with heat absorbed. The heat absorbed is $$Q = 9.35 \, \mathrm{cal}$$. Converting to joules,

$$Q = 9.35 \times 4.19 = 39.18 \, \mathrm{J}$$

Applying the first law to this process and solving for the work done by the system,

$$W = Q - \Delta U = 39.18 - 22.3 = 16.88 \, \mathrm{J}$$

Hence the net work done by the system in the second process is about $$16.9 \, \mathrm{J}$$.

Given: the heater supplies heat to the system at the rate $$\dfrac{\Delta Q}{\Delta t} = 100 \, \mathrm{W} = 100 \, \mathrm{J/s}$$, and the system performs work at the rate $$\dfrac{\Delta W}{\Delta t} = 75 \, \mathrm{J/s}$$.

First law of thermodynamics: the heat supplied to a system is shared between the increase in its internal energy and the work done by it:

$$\Delta Q = \Delta U + \Delta W$$

Rearranging for the change in internal energy,

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

Dividing throughout by the time interval $$\Delta t$$ gives the rate of increase of internal energy:

$$\dfrac{\Delta U}{\Delta t} = \dfrac{\Delta Q}{\Delta t} - \dfrac{\Delta W}{\Delta t} = 100 - 75 = 25 \, \mathrm{J/s}$$

Hence the internal energy of the system is increasing at the rate of $$25 \, \mathrm{W}$$.

The work done by a gas in a process equals the area under the path on the $$P$$-$$V$$ diagram. It is positive when the volume increases (gas does work) and negative when the volume decreases (work is done on the gas).

The three states are: $$D\,(V_D = 2.0 \, \mathrm{m^3},\ P_D = 600 \, \mathrm{N/m^2})$$, $$E\,(V_E = 5.0 \, \mathrm{m^3},\ P_E = 300 \, \mathrm{N/m^2})$$ and $$F\,(V_F = 2.0 \, \mathrm{m^3},\ P_F = 300 \, \mathrm{N/m^2})$$.

Process $$D \to E$$ (straight-line process). The volume increases from $$2.0 \, \mathrm{m^3}$$ to $$5.0 \, \mathrm{m^3}$$ while the pressure falls linearly from $$600 \, \mathrm{N/m^2}$$ to $$300 \, \mathrm{N/m^2}$$. The region under the straight line $$DE$$ is a trapezium, whose area is

$$W_{DE} = \dfrac{1}{2}\,(P_D + P_E)\,(V_E - V_D)$$

$$W_{DE} = \dfrac{1}{2}\,(600 + 300)\,(5.0 - 2.0) = \dfrac{1}{2}\times 900 \times 3.0 = 1350 \, \mathrm{J}$$

The volume increases, so this work is done by the gas: $$W_{DE} = +1350 \, \mathrm{J}$$.

Process $$E \to F$$ (isobaric). The pressure is constant at $$P = 300 \, \mathrm{N/m^2}$$ while the volume decreases from $$5.0 \, \mathrm{m^3}$$ to $$2.0 \, \mathrm{m^3}$$. The work done is

$$W_{EF} = P\,(V_F - V_E) = 300 \times (2.0 - 5.0) = 300 \times (-3.0) = -900 \, \mathrm{J}$$

The volume decreases, so this work is done on the gas: $$W_{EF} = -900 \, \mathrm{J}$$.

Total work done by the gas.

$$W = W_{DE} + W_{EF} = 1350 + (-900) = 450 \, \mathrm{J}$$

Hence the total work done by the gas in going from $$D$$ to $$E$$ to $$F$$ is $$450 \, \mathrm{J}$$.