JEE Main202126 Feb 2021Morning ShiftPhysicsThermal Properties of MatterActual
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 × 10 -
Correct answer
0
Step-by-step solution
Let common equilibrium pressure of mixture is P atmp. then f 2 P 1   V 1 + f 2 P 2   V 2 = f 2 P V 1 + V 2 f 2 2 4 . 5 + f 2 3 5 . 5 = f 2 P 4 . 5 + 5 . 5 ⇒ P = 2 . 55 = x × 10 - 1 atmp So x = 25 . 5 ≈ 26 (Nearest integer)