Let common equilibrium pressure of mixture is P atmp. then
2fP1V1+2fP2V2=2fP(V1+V2)
2f(2)(4.5)+2f(3)(5.5)=2fP(4.5+5.5)
⇒P=2.55=x×10−1 atmp
So x=25.5≈26 (Nearest integer)
JEE Main 2021 — Physics Thermodynamics
A container is divided into two chambers by a partition. The volume of first chamber is 4.5litre and second chamber is 5.5litre. The first chamber contain 3.0 moles of gas at pressure 2.0atm and second chamber contain 4.0 moles of gas at pressure 3.0atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x×10−1atm. Value of x (nearest integer) is
Held on 26 Feb 2021 · Verified 6 Jul 2026.
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