Verified Solution • 2025–26 Session moderate

QJEEM21Ph026b: JEE Main 2021 Physics - Class 11 JEEM21Ph026b Solution: A container is divided into two chambers by a partition. The…

A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x ×\times 10 −- 1 atm. Value of x is ________.

Step-by-Step Verified Solution

By energy conservation

32n1RT1+32n2RT2=32(n1+n2)RT{3 \over 2}{n_1}R{T_1} + {3 \over 2}{n_2}R{T_2} = {3 \over 2}({n_1} + {n_2})RT

Using PV = nRT

P1V1 + P2V2 = P(V1 + V2)

P=P1V1+P2V2V1+V2=2×4.5+3×5.54.5+5.5P = {{{P_1}{V_1} + {P_2}{V_2}} \over {{V_1} + {V_2}}} = {{2 \times 4.5 + 3 \times 5.5} \over {4.5 + 5.5}}

P=9+16.510=25.510P = {{9 + 16.5} \over {10}} = {{25.5} \over {10}}

≈25×10−1\approx 25 \times {10^{ - 1}} atm

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring