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Kinetic Theory

JEE Main level, single correct, medium

3 of 168

Question 3 of 168

One gas of $n_1$ mole of molecules at temperature $T_1$ , volume $V_1$ , and pressure $P_1$ , and another gas of $n_2$ mole of molecules at temperature $T_2$ , volume $V_2$ , and pressure $P_2$ , are mixed resulting in pressure $P$ and volume $V$ of the mixture. The temperature of the mixture is _____.

Show answer and solution

Answer: (B) $T_1 T_2 PV/(T_2 P_1 V_1 + T_1 P_2 V_2)$

From the ideal gas equation, the number of moles of the first gas is

n_1 = \dfrac{P_1 V_1}{R T_1}

and the number of moles of the second gas is

n_2 = \dfrac{P_2 V_2}{R T_2}

.

For the mixture, the total number of moles is

n = n_1 + n_2

.

Applying the ideal gas equation for the mixture:

P V = n R T

T = \dfrac{P V}{n R}

Substituting the values of

n_1

and

n_2

T = \dfrac{P V}{\left( \dfrac{P_1 V_1}{R T_1} + \dfrac{P_2 V_2}{R T_2} \right) R}

T = \dfrac{P V}{\dfrac{P_1 V_1}{T_1} + \dfrac{P_2 V_2}{T_2}}

T = \dfrac{T_1 T_2 P V}{P_1 V_1 T_2 + P_2 V_2 T_1}

Answer:

T_1 T_2 PV/(T_2 P_1 V_1 + T_1 P_2 V_2)

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Kinetic Theory (Gas Laws), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.