Thermodynamics Formulas For JEE 2026
The Laws of Motion is one of the most important chapters in JEE Physics because it builds the foundation for many other topics in mechanics. In this chapter, students learn about Newton’s three laws of motion and how different forces act on objects in real situations. It also includes key concepts such as friction, momentum, impulse, and the use of free body diagrams to understand forces clearly. While preparing for JEE Mains, students should practice problems involving inclined planes, pulley systems, and objects under different constraints, as these are commonly asked in the exam. Once these concepts are understood well, solving mechanics problems becomes much easier.
In JEE Mains, the Laws of Motion chapter generally contributes around 2–4 questions every year, making it a highly scoring topic if prepared properly. Many mechanics problems depend on a clear understanding of force analysis and Newton’s laws, especially questions involving friction, circular motion forces, and connected bodies. Regular practice and strong conceptual clarity are the keys to performing well in this chapter. For quick revision before exams, students can refer to a well-structured JEE Mains Physics Formula PDF, which helps them revise important formulas quickly and improves accuracy while solving exam-level questions.
Thermal Equilibrium and Zeroth Law
Imagine you place a hot cup of coffee on a table. Over time, the coffee cools down and the table warms up slightly, until both reach the same temperature. At that point, heat stops flowing between them — they are in thermal equilibrium. This simple idea is the starting point of thermodynamics.
Thermal Equilibrium: Two objects are in thermal equilibrium when they are at the same temperature and no heat flows between them
Zeroth Law of Thermodynamics: If body A is in thermal equilibrium with body C, and body B is also in thermal equilibrium with body C, then A and B are in thermal equilibrium with each other
The Zeroth Law may sound obvious, but it is the foundation that allows us to define temperature and use thermometers. A thermometer (body C) in equilibrium with a patient (body A) reads the patient's temperature because of this law.
Heat, Work, and Internal Energy
Thermodynamics studies how heat (thermal energy transfer) and work (mechanical energy transfer) change the energy stored inside a system. Think of a gas inside a cylinder with a movable piston — you can heat the gas (add heat) or compress the piston (do work on it).
Internal Energy ($$U$$): The total energy stored within a system due to the random motion and interactions of its molecules. It depends only on the state (temperature, pressure, volume) of the system, not on how it got there.
Heat ($$Q$$): Energy transferred between a system and its surroundings because of a temperature difference. Heat flows from hot to cold.
Work ($$W$$): Energy transferred when a force acts through a displacement. For a gas, work is done when the gas expands or is compressed.
Work Done by a Gas
For a gas expanding or compressing against a piston:
$$$W = \int_{V_1}^{V_2} P \, dV$$$
where $$P$$ = pressure, $$V_1$$ = initial volume, $$V_2$$ = final volume.
- Gas expands ($$V_2 > V_1$$): $$W > 0$$ (work done by the gas)
- Gas compresses ($$V_2 < V_1$$): $$W < 0$$ (work done on the gas)
- On a PV diagram, $$W$$ = area under the curve between $$V_1$$ and $$V_2$$
Note: The sign convention used in JEE: $$Q > 0$$ means heat is added to the system; $$W > 0$$ means work is done by the system. Some textbooks use the opposite convention for $$W$$, so always check.
First Law of Thermodynamics
The First Law is simply the law of conservation of energy applied to thermal systems. Whatever heat you add to a system either increases its internal energy or is used to do work — energy cannot appear or disappear.
First Law of Thermodynamics
$$$Q = \Delta U + W$$$
where:
- $$Q$$ = heat added to the system
- $$\Delta U$$ = change in internal energy
- $$W$$ = work done by the system
Worked Example
A gas absorbs 500 J of heat and does 200 J of work on the surroundings. Find the change in internal energy.
Using the First Law: $$Q = \Delta U + W$$
$$500 = \Delta U + 200$$
$$\Delta U = 500 - 200 = 300$$ J
The internal energy of the gas increases by 300 J.
Internal Energy of an Ideal Gas
For an ideal gas, the internal energy depends only on temperature, not on pressure or volume. This is a key simplification.
Internal Energy of an Ideal Gas
$$$U = \frac{f}{2} nRT$$$
$$$\Delta U = nC_v \Delta T = \frac{f}{2} nR\Delta T$$$
where $$f$$ = degrees of freedom, $$n$$ = number of moles, $$R$$ = gas constant ($$8.314 \text{ J mol}^{-1}\text{K}^{-1}$$), $$T$$ = temperature in kelvin, $$C_v$$ = molar specific heat at constant volume.
Thermodynamic Processes — Isothermal, Adiabatic, Isobaric, Isochoric
A thermodynamic process describes how a gas goes from one state to another. Different constraints (constant temperature, constant pressure, etc.) lead to different types of processes, each with its own formula for work, heat, and internal energy change.
Isothermal Process (Constant Temperature)
In an isothermal process, the temperature stays constant throughout. Since internal energy of an ideal gas depends only on temperature, $$\Delta U = 0$$, and all the heat added is converted into work.
Isothermal Process ($$T = \text{constant}$$)
- Equation: $$PV = \text{constant}$$ (Boyle's law)
- $$\Delta U = 0$$ (temperature doesn't change)
- $$Q = W$$ (all heat becomes work)
- Work done:
$$$W = nRT \ln\frac{V_2}{V_1} = nRT \ln\frac{P_1}{P_2}$$$
Worked Example
2 moles of an ideal gas expand isothermally at 300 K from volume $$V$$ to $$3V$$. Find the work done.
$$W = nRT \ln\frac{V_2}{V_1} = 2 \times 8.314 \times 300 \times \ln 3$$
$$= 4988.4 \times 1.099 = 5482$$ J $$\approx 5.48$$ kJ
Adiabatic Process (No Heat Exchange)
In an adiabatic process, the system is perfectly insulated — no heat enters or leaves ($$Q = 0$$). Any work done comes entirely from the system's internal energy. When a gas expands adiabatically, it cools down; when compressed, it heats up.
Adiabatic Index ($$\gamma$$): The ratio of specific heats: $$\gamma = C_p / C_v$$. For monoatomic gases $$\gamma = 5/3$$; for diatomic gases $$\gamma = 7/5$$.
Adiabatic Process ($$Q = 0$$)
- Equation: $$PV^{\gamma} = \text{constant}$$
- Also: $$TV^{\gamma-1} = \text{constant}$$ and $$T^{\gamma} P^{1-\gamma} = \text{constant}$$
- $$Q = 0$$, so $$\Delta U = -W$$
- Work done:
$$$W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}$$$
Worked Example
A monoatomic ideal gas ($$\gamma = 5/3$$) at 300 K is compressed adiabatically to half its volume. Find the final temperature.
Using $$TV^{\gamma - 1} = \text{constant}$$:
$$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$$
$$300 \times V^{2/3} = T_2 \times \left(\frac{V}{2}\right)^{2/3}$$
$$T_2 = 300 \times \left(\frac{V}{V/2}\right)^{2/3} = 300 \times 2^{2/3} = 300 \times 1.587 = 476.1$$ K
Isobaric Process (Constant Pressure)
In an isobaric process, the pressure stays constant while the gas expands or compresses. Heating a gas in a cylinder with a freely movable piston (open to atmosphere) is approximately isobaric.
Isobaric Process ($$P = \text{constant}$$)
- $$W = P(V_2 - V_1) = P\Delta V = nR\Delta T$$
- $$\Delta U = nC_v \Delta T$$
- $$Q = nC_p \Delta T$$
- Since $$C_p > C_v$$, more heat is needed than in a constant-volume process for the same $$\Delta T$$
Isochoric Process (Constant Volume)
In an isochoric process, the volume stays constant — the gas cannot expand or compress. Since there is no displacement, no work is done. All the heat added goes into increasing the internal energy (raising the temperature).
Isochoric Process ($$V = \text{constant}$$)
- $$W = 0$$ (no change in volume)
- $$Q = \Delta U = nC_v \Delta T$$
- Pressure and temperature change: $$\dfrac{P}{T} = \text{constant}$$ (Gay-Lussac's law)
Summary of Thermodynamic Processes
Comparison Table of Thermodynamic Processes
| Process | Constant | $$W$$ | $$\Delta U$$ | $$Q$$ |
|---|---|---|---|---|
| Isothermal | $$T$$ | $$nRT\ln\frac{V_2}{V_1}$$ | 0 | $$W$$ |
| Adiabatic | $$Q=0$$ | $$\frac{nR(T_1-T_2)}{\gamma-1}$$ | $$-W$$ | 0 |
| Isobaric | $$P$$ | $$P\Delta V$$ | $$nC_v\Delta T$$ | $$nC_p\Delta T$$ |
| Isochoric | $$V$$ | 0 | $$nC_v\Delta T$$ | $$\Delta U$$ |
Tip: On a PV diagram, an adiabatic curve is steeper than an isothermal curve passing through the same point, because $$\gamma > 1$$. This is a common JEE conceptual question.
PV Diagrams and Work Done in Cyclic Processes
A PV diagram (Pressure vs. Volume graph) is an essential tool in thermodynamics. Each point on the diagram represents a state of the gas, and a curve represents a process. The area under the curve gives the work done.
Key Facts about PV Diagrams
- Work done = area under the curve on a PV diagram
- In a cyclic process (the gas returns to its initial state), $$\Delta U = 0$$ (since $$U$$ depends only on state)
- For a cycle: $$Q_{\text{net}} = W_{\text{net}}$$ = area enclosed by the cycle
- Clockwise cycle: net work is positive (heat engine)
- Anticlockwise cycle: net work is negative (refrigerator/heat pump)
Worked Example
A gas undergoes a cyclic process on a PV diagram forming a rectangle with corners at $$(V_1, P_1)$$, $$(V_2, P_1)$$, $$(V_2, P_2)$$, $$(V_1, P_2)$$ where $$P_2 > P_1$$. Find the net work done if $$P_1 = 1$$ atm, $$P_2 = 3$$ atm, $$V_1 = 2$$ L, $$V_2 = 5$$ L.
Net work = area of rectangle = $$(P_2 - P_1)(V_2 - V_1)$$
$$= (3 - 1) \times 1.013 \times 10^5 \times (5 - 2) \times 10^{-3}$$
$$= 2 \times 1.013 \times 10^5 \times 3 \times 10^{-3} = 607.8$$ J
Second Law of Thermodynamics
The First Law tells us that energy is conserved, but it doesn't tell us which processes are possible. For example, heat always flows from hot to cold — never the reverse on its own. The Second Law captures this directionality.
Kelvin–Planck Statement: It is impossible to build a heat engine that converts all the heat absorbed from a source into work, with no other effect. (Some heat must always be rejected.)
Clausius Statement: It is impossible for heat to flow from a colder body to a hotter body without external work being done.
Carnot Engine
The Carnot engine is a theoretical "perfect" heat engine that operates between two temperatures. No real engine can be more efficient than a Carnot engine operating between the same temperatures.
Carnot Engine
A Carnot cycle consists of: isothermal expansion $$\rightarrow$$ adiabatic expansion $$\rightarrow$$ isothermal compression $$\rightarrow$$ adiabatic compression.
Efficiency:
$$$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{Q_C}{Q_H} = \frac{W}{Q_H}$$$
where:
- $$T_H$$ = temperature of hot source (in kelvin)
- $$T_C$$ = temperature of cold sink (in kelvin)
- $$Q_H$$ = heat absorbed from source
- $$Q_C$$ = heat rejected to sink
- $$W = Q_H - Q_C$$ = net work done
Worked Example
A Carnot engine operates between 600 K and 300 K. It absorbs 1000 J of heat per cycle. Find the efficiency and work done.
$$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.5 = 50\%$$
$$W = \eta \times Q_H = 0.5 \times 1000 = 500$$ J
$$Q_C = Q_H - W = 1000 - 500 = 500$$ J
Refrigerator (Heat Pump)
A refrigerator works in the reverse direction of a heat engine — it uses work to transfer heat from a cold body to a hot body (the opposite of natural heat flow).
Coefficient of Performance (COP)
$$$\text{COP} = \frac{Q_C}{W} = \frac{Q_C}{Q_H - Q_C} = \frac{T_C}{T_H - T_C}$$$
where $$Q_C$$ = heat extracted from the cold body, $$W$$ = work input.
Tip: Carnot efficiency is always less than 100% (since $$T_C > 0$$ K in practice). JEE problems often give temperatures in $$^\circ$$C — always convert to kelvin first using $$T(\text{K}) = T(^\circ\text{C}) + 273$$.
Entropy Formulas
Entropy is a measure of the disorder or randomness of a system. The Second Law can be restated as: the total entropy of an isolated system never decreases. In any natural (irreversible) process, entropy always increases.
Entropy ($$S$$): A thermodynamic quantity that measures the degree of disorder. Change in entropy is defined as the heat exchanged reversibly divided by the temperature.
Entropy Change
$$$\Delta S = \int \frac{dQ_{\text{rev}}}{T}$$$
For a process at constant temperature:
$$$\Delta S = \frac{Q_{\text{rev}}}{T}$$$
For an ideal gas going from state 1 to state 2:
$$$\Delta S = nC_v \ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1}$$$
Note: For a reversible process, $$\Delta S_{\text{universe}} = 0$$. For an irreversible (natural) process, $$\Delta S_{\text{universe}} > 0$$. Entropy of the universe never decreases.
Kinetic Theory of Gases Formulas
So far, we treated a gas as a continuous substance. Kinetic theory takes a microscopic view — it explains the behaviour of a gas by considering the random motion of billions of tiny molecules. Remarkably, macroscopic quantities like pressure and temperature emerge from this molecular chaos.
Ideal Gas Equation
An ideal gas is a theoretical gas where molecules are point particles with no intermolecular forces (except during collisions). Real gases behave approximately as ideal gases at low pressures and high temperatures.
Ideal Gas Equation
$$$PV = nRT = Nk_BT$$$
where:
- $$P$$ = pressure, $$V$$ = volume, $$T$$ = temperature (in kelvin)
- $$n$$ = number of moles, $$R = 8.314 \text{ J mol}^{-1}\text{K}^{-1}$$ (universal gas constant)
- $$N$$ = number of molecules, $$k_B = 1.38 \times 10^{-23} \text{ J K}^{-1}$$ (Boltzmann constant)
- $$R = N_A k_B$$ where $$N_A = 6.022 \times 10^{23}$$ (Avogadro's number)
Pressure from Kinetic Theory
When gas molecules collide with the walls of a container, they exert a force. The cumulative effect of billions of such collisions per second is what we measure as pressure. Kinetic theory derives this pressure from the molecular speeds.
Pressure of an Ideal Gas
$$$P = \frac{1}{3}\frac{mN}{V}v_{\text{rms}}^2 = \frac{1}{3}\rho \, v_{\text{rms}}^2$$$
where $$m$$ = mass of one molecule, $$N$$ = total number of molecules, $$\rho$$ = density of the gas, $$v_{\text{rms}}$$ = root mean square speed.
Kinetic Energy and Temperature
One of the most beautiful results of kinetic theory: the average kinetic energy of a gas molecule depends only on temperature, not on the type of gas. A hydrogen molecule and an oxygen molecule at the same temperature have the same average kinetic energy.
Kinetic Energy of Gas Molecules
Average kinetic energy per molecule:
$$$\frac{1}{2}mv_{\text{rms}}^2 = \frac{3}{2}k_B T$$$
Average kinetic energy per mole:
$$$KE = \frac{3}{2}RT$$$
Total kinetic energy of $$n$$ moles:
$$$KE_{\text{total}} = \frac{3}{2}nRT$$$
Worked Example
Find the average kinetic energy of an oxygen molecule at 300 K.
$$KE = \frac{3}{2}k_BT = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300$$
$$= 6.21 \times 10^{-21}$$ J
Molecular Speeds
Not all molecules in a gas move at the same speed. Some are fast, some are slow. The distribution of speeds follows the Maxwell–Boltzmann distribution. Three important characteristic speeds are defined.
Three Molecular Speeds
For a gas with molar mass $$M$$ (in kg/mol) at temperature $$T$$:
- RMS speed (root mean square): $$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_BT}{m}}$$$
- Average speed: $$$v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8k_BT}{\pi m}}$$$
- Most probable speed: $$$v_{\text{mp}} = \sqrt{\frac{2RT}{M}} = \sqrt{\frac{2k_BT}{m}}$$$
Ratio: $$v_{\text{mp}} : v_{\text{avg}} : v_{\text{rms}} = 1 : 1.128 : 1.225$$
Or equivalently: $$\sqrt{2} : \sqrt{8/\pi} : \sqrt{3}$$
Worked Example
Find the RMS speed of nitrogen ($$N_2$$, $$M = 28 \times 10^{-3}$$ kg/mol) at 300 K.
$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{28 \times 10^{-3}}}$$
$$= \sqrt{\frac{7482.6}{0.028}} = \sqrt{267236} = 517$$ m/s
Tip: Remember the speed ordering: $$v_{\text{mp}} < v_{\text{avg}} < v_{\text{rms}}$$. A quick mnemonic: MAP (Most probable, Average, RMS) — in alphabetical order of increasing value.
Degrees of Freedom
A molecule can store energy in different ways: moving through space (translational), spinning (rotational), and vibrating (vibrational). Each independent mode of motion is a degree of freedom.
Degrees of Freedom ($$f$$): The number of independent ways a molecule can store energy
Degrees of Freedom for Different Molecules
| Type | Example | $$f$$ | $$\gamma = C_p/C_v$$ |
|---|---|---|---|
| Monoatomic | He, Ne, Ar | 3 (translation only) | 5/3 = 1.67 |
| Diatomic (rigid) | $$\text{H}_2, \text{N}_2, \text{O}_2$$ | 5 (3 trans + 2 rot) | 7/5 = 1.4 |
| Diatomic (with vib.) | at high $$T$$ | 7 (3 trans + 2 rot + 2 vib) | 9/7 = 1.29 |
| Linear triatomic | $$\text{CO}_2$$ | 7 | 9/7 |
| Non-linear triatomic | $$\text{H}_2\text{O}$$ | 6 (3 trans + 3 rot) | 8/6 = 1.33 |
Law of Equipartition of Energy
Energy distributes itself equally among all available degrees of freedom. Each degree of freedom gets exactly $$\frac{1}{2}k_BT$$ of energy per molecule.
Equipartition of Energy
Energy per molecule $$= \dfrac{f}{2}k_BT$$
Energy per mole $$= \dfrac{f}{2}RT$$
This gives us the specific heats:
- $$C_v = \dfrac{f}{2}R$$
- $$C_p = C_v + R = \dfrac{f+2}{2}R$$ (Mayer's relation)
- $$\gamma = \dfrac{C_p}{C_v} = \dfrac{f+2}{f}$$
Mayer's Relation
For any ideal gas:
$$$C_p - C_v = R$$$
This is independent of the type of gas and is one of the most fundamental relations in thermodynamics.
Worked Example
Find $$C_v$$, $$C_p$$, and $$\gamma$$ for a diatomic gas (rigid molecules).
For rigid diatomic: $$f = 5$$
$$C_v = \frac{f}{2}R = \frac{5}{2} \times 8.314 = 20.79$$ J mol$$^{-1}$$K$$^{-1}$$
$$C_p = C_v + R = 20.79 + 8.314 = 29.10$$ J mol$$^{-1}$$K$$^{-1}$$
$$\gamma = \frac{C_p}{C_v} = \frac{29.10}{20.79} = 1.4$$
Specific Heat (Cp and Cv) Formulas
A gas can be heated in many ways, but two are especially important: heating at constant volume (no expansion, all heat raises temperature) and heating at constant pressure (gas expands, some heat goes into doing work).
Specific Heats and Gas Mixtures
- At constant volume: $$Q = nC_v\Delta T$$ (all heat $$\rightarrow$$ internal energy)
- At constant pressure: $$Q = nC_p\Delta T$$ (heat $$\rightarrow$$ internal energy + work)
- $$C_p > C_v$$ always, because extra heat is needed for the expansion work
For a mixture of gases:
$$$C_v^{\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}, \quad \gamma_{\text{mix}} = \frac{C_p^{\text{mix}}}{C_v^{\text{mix}}}$$$
Worked Example
1 mole of monoatomic gas is mixed with 1 mole of diatomic gas. Find $$\gamma$$ of the mixture.
$$C_{v1} = \frac{3}{2}R$$ (monoatomic), $$C_{v2} = \frac{5}{2}R$$ (diatomic)
$$C_v^{\text{mix}} = \frac{1 \times \frac{3}{2}R + 1 \times \frac{5}{2}R}{1 + 1} = \frac{4R}{2} = 2R$$
$$C_p^{\text{mix}} = C_v^{\text{mix}} + R = 3R$$
$$\gamma_{\text{mix}} = \frac{3R}{2R} = 1.5$$
Tip: JEE often tests the relation between $$\gamma$$ and degrees of freedom. Remember: $$\gamma = 1 + \frac{2}{f}$$. For monoatomic ($$f=3$$): $$\gamma = 5/3$$. For diatomic ($$f=5$$): $$\gamma = 7/5$$. As $$f$$ increases, $$\gamma$$ decreases.
Key Formulas at a Glance
Quick Reference
- First Law: $$Q = \Delta U + W$$
- Ideal gas: $$PV = nRT$$
- Internal energy: $$\Delta U = nC_v\Delta T$$ (always, for any process of an ideal gas)
- Isothermal work: $$W = nRT\ln(V_2/V_1)$$
- Adiabatic work: $$W = \frac{nR(T_1 - T_2)}{\gamma - 1}$$
- Carnot efficiency: $$\eta = 1 - T_C/T_H$$
- RMS speed: $$v_{\text{rms}} = \sqrt{3RT/M}$$
- Average KE per molecule: $$\frac{3}{2}k_BT$$
- Mayer's relation: $$C_p - C_v = R$$
Thermodynamics Formulas For JEE 2026: Conclusion
Thermodynamics is one of the most important chapters in JEE Physics because it connects concepts from heat, energy, and molecular motion. Students must clearly understand the laws of thermodynamics, thermodynamic processes, entropy, and kinetic theory of gases to solve both conceptual and numerical problems in the exam. Learning the Thermodynamics Formula For JEE 2026 helps students quickly apply equations and analyze different physical situations involving gases and energy transfer.
For effective exam preparation, regular practice and quick revision of formulas are essential. Students should revise important equations such as the ideal gas law, Carnot efficiency, RMS speed formulas, and entropy relations. Using a well-organized JEE Mains Physics Formula PDF can help students revise these formulas quickly before the exam and improve accuracy while solving JEE-level problems.
