JEE Physics — Thermodynamics And Kinetic Theory Complete…

Chapter Overview & Weightage

Thermodynamics and Kinetic Theory is one of the most consistently tested chapters in JEE Main. Over the past few years, you can almost guarantee 2–3 questions from this chapter in every shift.

JEE Main Weightage: 8–10% This chapter contributes roughly 2–3 questions per paper. Combined with Waves and SHM, it forms the “scoring backbone” of JEE Physics for most students.

Year	JEE Main (Avg Questions)	JEE Advanced	Key Topics Tested
2024	3	2	Carnot efficiency, Cp/Cv, First Law
2023	2	2	KTG, Entropy, PV diagrams
2022	3	1	Adiabatic process, Mean free path
2021	2	2	Carnot, Specific heats, KTG
2020	3	2	Isothermal/adiabatic work, Degrees of freedom

The pattern is clear: PV diagrams, Carnot engine, and kinetic theory numericals appear almost every year. JEE Advanced adds entropy-based problems and multi-step thermodynamic cycles.

Key Concepts You Must Know

Ranked by how frequently they appear in JEE Main PYQs:

Tier 1 — High Frequency (appear almost every year)

First Law of Thermodynamics: $\Delta U = Q - W$ and sign conventions
Isothermal, adiabatic, isobaric, isochoric processes — work done expressions and PV relationships
Carnot efficiency: $\eta = 1 - \frac{T_2}{T_1}$ , and COP of refrigerator
Degrees of freedom and $C_p$ , $C_v$ , $\gamma$ for monoatomic, diatomic, polyatomic gases
Kinetic Theory: $\frac{1}{2}mv_{rms}^2 = \frac{3}{2}k_BT$ , pressure–KE relation

Tier 2 — Medium Frequency (2–3 times in last 5 years)

Mean free path and its dependence on temperature, pressure
PV diagram area = work done; identifying process from slope
Second Law and concept of entropy (qualitative for JEE Main, numerical for Advanced)
Heat engines: comparing efficiency of reversible vs irreversible engines
Equipartition theorem and internal energy of ideal gases

Tier 3 — Low but Non-Zero (watch for JEE Advanced)

Entropy change calculations: $\Delta S = \int \frac{dQ}{T}$
Multi-step cycles (not standard Carnot)
Real gas behaviour and van der Waals deviation

Important Formulas

\Delta U = Q - W

Sign convention (most common source of errors):

$Q > 0$ : heat absorbed by system
$W > 0$ : work done by system (expansion)

When to use: Any process where heat exchange or work done is given. Always establish sign convention before substituting.

W_{isothermal} = nRT \ln\frac{V_2}{V_1} = nRT \ln\frac{P_1}{P_2}

W_{adiabatic} = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}

W_{isobaric} = P\Delta V = nR\Delta T

W_{isochoric} = 0

When to use: Identify process type first. Adiabatic formula is directly from $Q=0$ applied to first law.

C_v = \frac{f}{2}R, \quad C_p = C_v + R = \frac{f+2}{2}R, \quad \gamma = \frac{C_p}{C_v} = \frac{f+2}{f}

Gas Type	$f$	$C_v$	$C_p$	$\gamma$
Monoatomic	3	$\frac{3}{2}R$	$\frac{5}{2}R$	1.67
Diatomic (rigid)	5	$\frac{5}{2}R$	$\frac{7}{2}R$	1.4
Polyatomic	6	$3R$	$4R$	1.33

When to use: Any question giving “monoatomic” or “diatomic” gas — immediately write down $\gamma$ and $C_v$ .

\eta_{Carnot} = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}

COP_{refrigerator} = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}

COP_{heat pump} = \frac{Q_1}{W} = \frac{T_1}{T_1 - T_2}

When to use: All temperatures must be in Kelvin. Note: $COP_{heat pump} = COP_{refrigerator} + 1$ .

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

v_{avg} = \sqrt{\frac{8RT}{\pi M}}, \quad v_{mp} = \sqrt{\frac{2RT}{M}}

v_{mp} : v_{avg} : v_{rms} = 1 : 1.13 : 1.22

When to use: The ratio is a direct 1-mark question in JEE Main. The $v_{rms}$ formula appears in energy calculations too.

P = \frac{1}{3}\rho v_{rms}^2 = \frac{1}{3}\frac{mN}{V}v_{rms}^2

KE_{per molecule} = \frac{f}{2}k_BT, \quad KE_{total} = \frac{f}{2}nRT

When to use: When pressure is related to molecular motion, or when internal energy of $n$ moles is asked.

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 (Shift 1)

Q: One mole of an ideal monoatomic gas undergoes an adiabatic expansion. If the temperature falls from 400 K to 300 K, the work done by the gas is:

Solution:

For an adiabatic process, $Q = 0$ , so from the First Law:

W = -\Delta U = -(nC_v\Delta T)

For monoatomic gas, $C_v = \frac{3}{2}R$ :

W = -n \cdot \frac{3}{2}R \cdot (T_2 - T_1) = -1 \cdot \frac{3}{2} \times 8.314 \times (300 - 400)

W = -\frac{3}{2} \times 8.314 \times (-100) = 1247 \text{ J} \approx 1250 \text{ J}

Many students write $W = nC_v(T_1 - T_2)$ without the negative sign reasoning, and sometimes get the sign backwards. Remember: gas expands → temperature falls → gas does positive work. Check your answer’s sign against physical intuition.

PYQ 2 — JEE Main 2023

Q: A Carnot engine works between 227°C and 27°C. If the engine absorbs 6000 J of heat from the hot reservoir per cycle, the work done per cycle is:

Solution:

Convert to Kelvin first (a step many rush through): $T_1 = 500$ K, $T_2 = 300$ K.

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

Work done per cycle:

W = \eta \times Q_1 = 0.4 \times 6000 = 2400 \text{ J}

Heat rejected to cold reservoir: $Q_2 = Q_1 - W = 6000 - 2400 = 3600$ J.

In Carnot problems, always find $\eta$ first, then $W$ , then $Q_2 = Q_1 - W$ . This sequence never fails. If a question asks for COP of the same engine run as refrigerator, use $COP = T_2/(T_1 - T_2) = 300/200 = 1.5$ .

PYQ 3 — JEE Main 2022

Q: The ratio of average kinetic energy of $H_2$ and $O_2$ at the same temperature is:

(A) 1 : 16 (B) 1 : 1 (C) 4 : 1 (D) 16 : 1

Solution:

This is a direct application of the equipartition theorem. Average kinetic energy per molecule is:

KE = \frac{f}{2}k_BT

Both $H_2$ and $O_2$ are diatomic (rigid), so $f = 5$ for both.

At the same temperature, average KE per molecule is identical for both gases.

Answer: (B) 1 : 1

Trap: students confuse average KE per molecule with total internal energy of a gas sample. If the question asked “1 mole of each”, it’s still 1:1 because $U = \frac{5}{2}RT$ for both. The mass does NOT appear in the equipartition result. KE depends only on temperature and degrees of freedom.

Difficulty Distribution

For JEE Main, the chapter breaks down like this:

Difficulty	% of Questions	What it Tests
Easy	40%	Direct formula application (Carnot $\eta$ , $C_v$ values, KTG ratios)
Medium	45%	Multi-step problems (First Law + process identification, PV diagrams)
Hard	15%	Combined concepts (mixing gases + temperature change, multi-step cycles)

For JEE Advanced, the distribution shifts: roughly 30% easy, 40% medium, 30% hard. Advanced loves entropy questions and multi-process cycles where you have to track $\Delta U$ across each step.

The “easy” questions here are genuinely easy only if you’ve memorised the $\gamma$ values and degree-of-freedom table. In JEE Main, 1 mark is 1 mark — don’t lose direct-formula questions to memorisation gaps.

Expert Strategy

Week 1: Build the table. Make a 2×5 table in your notes: rows are the 5 process types (isothermal, adiabatic, isobaric, isochoric, free expansion), columns are $Q$ , $W$ , $\Delta U$ , $\Delta T$ , and “shape on PV diagram”. Fill every cell. This single table solves ~40% of JEE Main questions from this chapter.

Week 2: PV diagrams are non-negotiable. Practice identifying the process from the slope. On a PV diagram, for the same state change: $|slope_{adiabatic}| > |slope_{isothermal}|$ because $\gamma > 1$ . JEE Main has given this comparison at least twice in recent years.

Week 3: KTG in one sitting. Kinetic theory is actually a small sub-topic — degrees of freedom, the three speed formulas, pressure derivation, mean free path. One focused 3-hour session with 20 PYQs covers it completely.

Topper approach: Solve the Carnot/heat engine question first in the exam. It’s almost always 1–2 steps, high accuracy, and gets you into a confident rhythm. Save PV diagram questions with area calculations for later.

For JEE Advanced specifically: Practice entropy calculations: $\Delta S = nC_v \ln(T_2/T_1) + nR\ln(V_2/V_1)$ for a general ideal gas process. Advanced has given 1–2 questions requiring this in the last three years. It looks scary but the calculation is mechanical once you’ve done 5 practice problems.

Mixing problems — a separate skill: When two gases at different temperatures mix in an insulated container, use energy conservation: $n_1 C_{v1}(T_f - T_1) + n_2 C_{v2}(T_f - T_2) = 0$ . This appears in JEE Advanced almost every alternate year.

Common Traps

Trap 1: Celsius instead of Kelvin in Carnot. The efficiency formula $\eta = 1 - T_2/T_1$ requires absolute temperatures. A question giving “127°C and 27°C” expects you to convert to 400 K and 300 K. Using 127 and 27 gives $\eta = 0.787$ , which is physically wrong (it exceeds the true value). Every JEE Main paper seems to have at least one temperature-unit trap.

Trap 2: Work done ON vs BY the gas. JEE questions sometimes ask “work done on the gas” after you’ve calculated work done by the gas. They’re equal in magnitude but opposite in sign. Read the question’s last line carefully.

Trap 3: Free expansion has $W = 0$ , $Q = 0$ , so $\Delta U = 0$ . For an ideal gas undergoing free expansion into vacuum, temperature does NOT change. Many students apply the adiabatic formula here and get wrong answers. Free expansion is not an adiabatic process in the usual sense — it’s irreversible and no work is done against any external pressure.

Trap 4: Diatomic $\gamma$ at high temperature. At high temperatures, vibrational modes activate and diatomic $\gamma$ drops below 1.4. JEE Advanced has tested this. At “normal” temperatures (room temperature), use $f = 5$ and $\gamma = 1.4$ . If a question mentions “high temperature” explicitly, $f = 7$ and $\gamma = 9/7 \approx 1.28$ .

Trap 5: “Efficiency of a heat engine cannot exceed Carnot efficiency” — direction matters. This is the Second Law. An irreversible engine operating between the same reservoirs always has $\eta < \eta_{Carnot}$ . If a numerical answer gives efficiency greater than Carnot, you’ve made an error somewhere — this is a built-in self-check.

Last-minute revision checklist:

$\gamma$ values for mono/di/polyatomic — memorised cold?
PV diagram shapes for all 4 processes — can you draw them from memory?
Carnot sign: temperatures in Kelvin, $\eta$ between 0 and 1
Free expansion: $\Delta T = 0$ for ideal gas
$v_{mp} : v_{avg} : v_{rms} = 1 : 1.13 : 1.22$

If you can answer these five without notes, you’re ready.