JEE Main 2025 Apr 4 Shift 2, Physics Q8: Kinetic Theory of an Ideal Gases

JEE Main2025Apr 4, Shift 2Physics

There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at 7 kPa at 500 K. If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 K, at steady state the pressure in the vessels will be (in kPa).

A
4.4
B
6
C
24
D
18

Solution

Let the small vessel have volume $V$ and the large vessel $2 V$ . The connecting tube equalises pressure at steady state; let the common pressure be $P_{f}$ and the common temperature $T_{f} = 600$ K. The total volume is $3 V$ .

Conservation of moles ( $n = P V / R T$ ):

$\frac{P _{1} ( 2 V )}{T _{1}} + \frac{P _{2} V}{T _{2}} = \frac{P _{f} ( 3 V )}{T _{f}}$

$\frac{8 \times 2}{1000} + \frac{7 \times 1}{500} = \frac{3 P _{f}}{600}$

$\frac{16}{1000} + \frac{14}{1000} = \frac{30}{1000} = \frac{3 P _{f}}{600}$ .

Solving: $P_{f} = \frac{30 \times 600}{1000 \times 3} = 6$ kPa.

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JEE Main 2025 Apr 4 Shift 2, Physics Q8: Kinetic Theory of an Ideal Gases

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