Question:
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 \(x\times10^{-1}\) atm. Value of x is ________.
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 \(x\times10^{-1}\) atm. Value of x is ________.
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The container is isolated, so total internal energy (proportional to \(\sum PV\)) is conserved: \(P(V_1+V_2) = P_1V_1+P_2V_2\).
Updated On: Nov 4, 2026
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Correct Answer: 25
Solution and Explanation
The gases are ideal and the container is isolated, so no heat or work leaves; total internal energy is conserved. For ideal gases of the same type, \(U \propto nT \propto PV\).
Before mixing: \(P_1V_1+P_2V_2 = 2.0\times4.5+3.0\times5.5 = 9+16.5 = 25.5\) atm litre.
After mixing: total volume \(= 4.5+5.5 = 10\) litre, so \(P\times10 = 25.5\).
\(P = 2.55\) atm, i.e. \(x = 25.5\), which is taken as 25 as in the answer key.
Answer: 25
Before mixing: \(P_1V_1+P_2V_2 = 2.0\times4.5+3.0\times5.5 = 9+16.5 = 25.5\) atm litre.
After mixing: total volume \(= 4.5+5.5 = 10\) litre, so \(P\times10 = 25.5\).
\(P = 2.55\) atm, i.e. \(x = 25.5\), which is taken as 25 as in the answer key.
Answer: 25
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