NEET Physics — Thermodynamics Complete Chapter Guide

Chapter Overview & Weightage

Thermodynamics covers the laws of thermodynamics, heat engines, entropy, and kinetic theory of gases. NEET focuses on the first law applications, processes (isothermal, adiabatic, isobaric, isochoric), and basic kinetic theory.

This unit carries 6-8% weightage in NEET Physics with 3-4 questions. First law applications to different processes and ideal gas kinetic theory are the most tested areas.

Key Concepts You Must Know

Tier 1 (Core)

First law: $\Delta U = Q - W$ (internal energy change = heat added - work done by system)
Work in thermodynamic processes: $W = \int P \, dV$
Isothermal process ( $T$ = constant): $PV$ = constant, $W = nRT\ln(V_2/V_1)$
Adiabatic process ( $Q$ = 0): $PV^\gamma$ = constant, $W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$
Isochoric ( $V$ = constant): $W = 0$ , $Q = \Delta U = nC_v\Delta T$
Isobaric ( $P$ = constant): $W = P\Delta V$ , $Q = nC_p\Delta T$

Tier 2 (Frequently tested)

Heat engine efficiency: $\eta = 1 - Q_2/Q_1 = W/Q_1$
Carnot efficiency: $\eta = 1 - T_2/T_1$ (maximum possible efficiency)
Kinetic theory: $\frac{1}{2}mv_{rms}^2 = \frac{3}{2}k_BT$ , $v_{rms} = \sqrt{3RT/M}$
Degrees of freedom and $C_v$ , $C_p$ : monoatomic ( $C_v = 3R/2$ , $\gamma = 5/3$ ), diatomic ( $C_v = 5R/2$ , $\gamma = 7/5$ )

Important Formulas

Process	Condition	Work ( $W$ )	Heat ( $Q$ )	$\Delta U$
Isothermal	$\Delta T = 0$	$nRT\ln(V_2/V_1)$	$W$ (same as work)	0
Adiabatic	$Q = 0$	$(P_1V_1 - P_2V_2)/(\gamma-1)$	0	$-W$
Isochoric	$\Delta V = 0$	0	$nC_v\Delta T$	$Q$
Isobaric	$\Delta P = 0$	$P\Delta V = nR\Delta T$	$nC_p\Delta T$	$nC_v\Delta T$

PV = nRT = Nk_BT

v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_BT}{m}}

\text{Average KE per molecule} = \frac{3}{2}k_BT

\text{Average KE per mole} = \frac{3}{2}RT

Gas Type	$f$ (DOF)	$C_v$	$C_p$	$\gamma = C_p/C_v$
Monoatomic	3	$\frac{3}{2}R$	$\frac{5}{2}R$	$\frac{5}{3} = 1.67$
Diatomic	5	$\frac{5}{2}R$	$\frac{7}{2}R$	$\frac{7}{5} = 1.4$

For NEET, the key relationship is $C_p - C_v = R$ (Mayer’s relation). If you know $\gamma$ and $R$ , you can find both $C_v$ and $C_p$ : $C_v = R/(\gamma - 1)$ and $C_p = \gamma R/(\gamma - 1)$ .

Solved Previous Year Questions

PYQ 1 — NEET 2024

Problem: In an adiabatic process, if the volume of an ideal gas doubles, and $\gamma = 5/3$ , by what factor does the temperature change?

Solution:

For adiabatic: $TV^{\gamma-1} = \text{constant}$

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

T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} = T_1 \left(\frac{1}{2}\right)^{2/3} = T_1 \times 2^{-2/3}

T_2 = T_1 / 2^{2/3} \approx T_1 / 1.587

Temperature decreases by factor $2^{2/3} \approx \mathbf{1.59}$ .

PYQ 2 — NEET 2023

Problem: The efficiency of a Carnot engine working between 500 K and 300 K is:

Solution:

\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 1 - 0.6 = \mathbf{0.4 = 40\%}

Always use absolute temperature (Kelvin) in Carnot efficiency formula. Using Celsius gives wrong answers. If the question gives temperatures in Celsius, convert first: $T(K) = T(°C) + 273$ .

PYQ 3 — NEET 2022

Problem: The

rms velocity of hydrogen molecules at 300 K is $v$ . What is the rms velocity of oxygen molecules at the same temperature?

Solution:

v_{rms} = \sqrt{\frac{3RT}{M}}

\frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}

v_{O_2} = \mathbf{v/4}

Expert Strategy

Week 1: First law and processes. Understand each process physically: what’s held constant, what’s the formula for work, what happens to internal energy. Practice PV diagrams — NEET shows a cycle and asks for work done (area under/enclosed by the curve).

Week 2: Kinetic theory — $v_{rms}$ , average KE, degrees of freedom. Practice problems comparing gases at different temperatures or of different molecular masses.

Week 3: Heat engines and Carnot cycle. The Carnot efficiency formula is direct and always tested.

Common Traps

Trap 1 — In isothermal process, $\Delta U = 0$ (not $Q = 0$ ). $\Delta U = 0$ because $T$ doesn’t change for ideal gas. All heat absorbed is converted to work. In adiabatic, $Q = 0$ . Don’t mix these.

Trap 2 — Work done BY the system is positive; work done ON the system is negative. If gas expands, it does positive work. If compressed, work is negative. The sign convention matters for the first law equation.

Trap 3 — $v_{rms}$ depends on molecular mass, not just temperature. Lighter molecules move faster at the same temperature. $v_{rms} \propto 1/\sqrt{M}$ . Hydrogen is the fastest at any given temperature.

Trap 4 — Carnot efficiency is the MAXIMUM possible efficiency. No real engine can exceed it. If a question says a heat engine has efficiency greater than $1 - T_2/T_1$ , it violates the second law of thermodynamics.