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Two identical vessels are connected by a tube with a valve letting gas pass from one vessel into the other if the pressure difference, ΔP >= 2.0 atm. Initially there was a vacuum in one vessel while the other contained ideal gas at 250 K and pressure 5.0 atm. Then both vessels were heated to 500 K. Up to what value will the pressure (in atm) in the first vessel (which had a vacuum initially) increase?

Answer:
9
Solution:

With no flow the filled vessel would reach 5.0 × 500/250 = 10.0 atm, far above the 2.0 atm threshold, so the valve opens and flow continues until Delta P = 2.0 atm.

Mole balance (identical volumes V): 5.0V/(250R) = (P₁ + P₂)V/(500R) ⇒ P₁ + P₂ = 20.0.

With P₂ - P₁ = 2.0: P₁ = 9.0 atm, P₂ = 11.0 atm.

Pressure in the first vessel rises to 9 atm.

Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Problems related with container/piston/SPM
⏱ 2mℹ️ Source: QB

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