JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Heat & Thermodynamics

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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✓ Correct
(C)24
(D)18

Answer: (B)

Solution

Writing

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

for each vessel and conserving the total number of moles,

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

. Substituting

P_{1} = 8

kPa,

T_{1} = 1000

P_{2} = 7

kPa,

T_{2} = 500

K and

T_{f} = 600

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

, so

\frac{30}{1000} = \frac{3 P _{f}}{600}

. This gives

P_{f} = \frac{30 \times 600}{1000 \times 3} = 6 kPa

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Heat & Thermodynamics

Solution

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