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There are two vessels filled with an ideal gas where the volume of one is double the volume of the 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)

Asked in JEE Main 4th April 2nd Shift 2025 · Pistons and partitions

Answer: (2) 6

Step-by-step solution

Let the smaller vessel have volume V, so the larger has 2V; the total is 3V.

Moles before: n₁=(8(2V))/(1000R)=(0.016V)/R and n₂=(7V)/(500R)=(0.014V)/R.

Total n=(0.030V)/R, and it is conserved.

After: both vessels at 600 K share pressure P, so n=(P(3V))/(600R).

(0.030V)/R=(3PV)/(600R)⇒ P=(0.030×600)/3=6 kPa.

Why the other options are wrong

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