Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : If the average kinetic energy of H_{2} and O_{2} molecules, kept in two different sized containers are same, then their temperatures will be same. Reason R : The r.m.s speed of H_{2} and O_{2} molecules are same at same temperature. Choose the correct answer from the options given below

A is true but \mathbf{R} is false

Both \mathbf{A} and \mathbf{R} are true and \mathbf{R} is the correct explanation of \mathbf{A}

Both \mathbf{A} and \mathbf{R} are true but \mathbf{R} is NOT the correct explanation of \mathbf{A}

A is false but \mathbf{R} is true

JEE Main · Physics · Chapter 10

Heat and Thermodynamics JEE Main PYQs (2020–2026)

353 questions from Heat and Thermodynamics have appeared in JEE Main between 2020 and 2026, including 244 in the last five papers (2022–2026). The most-asked topic is First Law of Thermodynamics and Thermodynamic Processes (116 questions). Practise them all with instant answers and step-by-step solutions.

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Solved Heat and Thermodynamics PYQs

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JEE Main 2026EasyDegrees of Freedom, Specific Heats of Gases, and Gas Mixtures
Q1
If 2 mole of an ideal monoatomic gas at temperature $T$ , is mixed with 6 mole of another ideal monoatomic gas at temperature $2 T$ then the temperature of mixture is:
- A $\frac{5}{2} T$
- B $\frac{5}{4} T$
- C $\frac{7}{2} T$
- D $\frac{7}{4} T$
Show answer & solution
Correct answer: (D)
The internal energy $(U)$ of $n$ moles of an ideal gas at temperature $T$ is given by $= n C_{v} T$ , where $C_{v}$ is the molar specific heat at constant volume.

For the mixture, total initial internal energy $U_{initial} = U_{1} + U_{2}$

Total final internal energy $U_{final} = (n_{1} + n_{2}) C_{v, mix} T_{mix}$

Since both gases are monoatomic, their molar specific heats are the same,

$C_{v 1} = C_{v 2} = C_{v, mi x} = \frac{3}{2} R$

Therefore, using conservation of energy (total initial internal energy is equal to final internal energy).

$n_{1} C_{v} T_{1} + n_{2} C_{v} T_{2} = (n_{1} + n_{2}) C_{v} T_{mi x}$

$\Rightarrow T_{mi x} = \frac{n _{1} T _{1} + n _{2} T _{2}}{( n _{1} + n _{2} )}$

Moles of first gas is $n_{1} = 2$

Temperature of first gas is $T_{1} = T$

Moles of second gas $n_{2} = 6$

Temperature of second gas $T_{2} = 2 T$

On substituting the values into expression for the temperature of mixture,

$T_{mi x} = \frac{( 2 \times T ) + ( 6 \times 2 T )}{2 + 6}$

$\Rightarrow T_{mi x} = \frac{2 T + 12 T}{8}$

$\Rightarrow T_{mi x} = \frac{14 T}{8}$

$\Rightarrow T_{mi x} = \frac{7}{4} T$

Therefore, the correct option is (D).
JEE Main 2026MediumIdeal Gas Laws and Real Gases
Q2
Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure 90 kPa and temperature 400 K . Keeping the temperature of one vessel constant at 400 K the second vessel temperature is raised to 500 K . The final pressure in the vessels is $____$ kPa .
- A100
- B120
- C90
- D105
Show answer & solution
Correct answer: (A)
Using Ideal Gas Law:

PV = nRT

Where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is absolute temperature. Since the system is closed, the total number of moles ( $n_{total}$ ) remains constant before and after heating.

Initial State:

Both vessels (each of volume V ) are at $P_{0} = 90 k P a$ and $T_{0} = 400 K$ .

$n_{total} = \frac{P _{0} V}{RT _{0}} + \frac{P _{0} V}{RT _{0}} = \frac{2 P _{0} V}{RT _{0}} \dots (i)$

Final State: Pressure becomes $P_{f}$ in both vessels. Vessel 1 stays at $T_{1} = 400 K$ , and vessel 2 is at $T_{2} = 500 K$ .

$n_{total} = \frac{P _{f} V}{RT _{1}} + \frac{P _{f} V}{RT _{2}} = \frac{P _{f} V}{R} (\frac{1}{T _{1}} + \frac{1}{T _{2}}) \dots (ii)$

From both the equations,

$\frac{2 P _{0} V}{RT _{0}} = \frac{P _{f} V}{R} (\frac{1}{T _{1}} + \frac{1}{T _{2}})$

$\Rightarrow \frac{2 P _{0}}{T _{0}} = P_{f} (\frac{1}{T _{1}} + \frac{1}{T _{2}})$

Substituting the values ( $P_{0} = 90, T_{0} = 400, T_{1} = 400, T_{2} = 500$ ):

$2 \times \frac{90}{400} = P_{f} (\frac{1}{400} + \frac{1}{500})$

$\Rightarrow \frac{180}{400} = P_{f} (\frac{5 + 4}{2000})$

$\Rightarrow 0.45 = P_{f} (\frac{9}{2000})$

$\Rightarrow P_{f} = \frac{0.45 \times 2000}{9} = \frac{900}{9} = 100 k P a$

Therefore, the final pressure of the vessel is 100 kPa . Thus, the correct option is (A).
JEE Main 2026EasyKinetic Theory of Gases: Molecular Speeds and Mean Free Path
Q3
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A : If the average kinetic energy of $H_{2}$ and $O_{2}$ molecules, kept in two different sized containers are same, then their temperatures will be same.

Reason R : The r.m.s speed of $H_{2}$ and $O_{2}$ molecules are same at same temperature.

Choose the correct answer from the options given below
- ABoth $A$ and $R$ are true and $R$ is the correct explanation of $A$
- BBoth $A$ and $R$ are true but $R$ is NOT the correct explanation of $A$
- CA is true but $R$ is false
- DA is false but $R$ is true
Show answer & solution
Correct answer: (C)
According to the kinetic theory of gases, the average translational kinetic energy ( $K_{avg}$ ) of a gas molecule is directly proportional to its absolute temperature (T).

$K_{a vg} = \frac{3}{2} k_{B} T$

Where $k_{B}$ is the Boltzmann constant.

So, the average kinetic energy depends only on the temperature and not on the mass of the molecule or the size/volume of the container.

Since $K_{a vg}$ is the same for both $H_{2}$ and $O_{2}$ , their temperatures (T) must be identical.

Therefore, Assertion A is true.

The root mean square (r. m. s.) speed ( $v_{rms}$ ) of a gas molecule is given by the formula:

$v_{r m s} = \frac{3 RT}{M}$

Where R is the universal gas constant, T is the absolute temperature, and M is the molar mass of the gas.

At the same temperature (T), $v_{r m s}$ is inversely proportional to the square root of the molar mass (M).

Since the molar mass of $H_{2} (2 g / m o l)$ is different from $O_{2} (32 g / m o l)$ , their r.m.s. speeds will be different even at the same temperature.

Therefore, Reason R is false.

Hence, Assertion A is true, but Reason R is false.

So, the correct option is (C).