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).

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q45 – JEE-MAIN 2025

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Home Physics ThermodynamicsQuestion #532

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

1000 K

while the smaller vessel contains the gas at

7 kPa

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).

Clue 1 of 3: Conservation of Moles

When two vessels containing gas are connected, the total number of moles of gas remains constant.

🥷

Ninja Strategy•Order of Magnitude Check

The final pressure should be in the same order of magnitude as the initial pressures (7-8 kPa), and likely an average or slightly adjusted value, making very high options like 18 or 24 kPa improbable.

🧠 Full Solution Path

Step 1: Calculate initial moles in each vessel✦ Active

Let the volume of the smaller vessel be $V$ . Then the volume of the larger vessel is $2 V$ . Using the ideal gas law $n = \frac{P V}{R T}$ :

n_{L} = \frac{P_{L} V_{L}}{R T_{L}} = \frac{8 kPa \times 2 V}{R \times 1000 K} = \frac{16 V}{1000 R}

n_{S} = \frac{P_{S} V_{S}}{R T_{S}} = \frac{7 kPa \times V}{R \times 500 K} = \frac{14 V}{1000 R}

n_{t o t a l} = n_{L} + n_{S} = \frac{16 V}{1000 R} + \frac{14 V}{1000 R} = \frac{30 V}{1000 R} = \frac{3 V}{100 R}

Step 2: Determine final conditions and apply Ideal Gas Law○ Expand

When the vessels are connected, the gas occupies the total volume, and the total number of moles remains constant. The final temperature is given.

V_{t o t a l} = V_{L} + V_{S} = 2 V + V = 3 V

T_{f} = 600 K

P_{f} V_{t o t a l} = n_{t o t a l} R T_{f}

Step 3: Calculate the final pressure○ Expand

Substitute the total moles, total volume, and final temperature into the ideal gas law equation to find the final pressure $P_{f}$ .

P_{f} (3 V) = (\frac{3 V}{100 R}) R (600 K)

P_{f} (3 V) = \frac{3 V \times 600}{100}

P_{f} (3 V) = 3 V \times 6

P_{f} = 6 kPa

💡 Teacher's Secret Hint

Ensure units are consistent throughout the calculation.

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Physics Question 45 – JEE-MAIN 2025

Clue 1 of 3: Conservation of Moles

🧠 Full Solution Path