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 $x \times 10^{-1} \mathrm{~atm}$. Value of $x$ is $\_\_\_\_$ .
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Solution
Let common equilibrium pressure of mixture is P atmp. then
$$
\begin{aligned}
& U 1+U 2=U_{\text {mixutre }} \\
& \frac{f}{2} P_1 V_1+\frac{f}{2} P_2 V_2=\frac{f}{2} P\left(V_1+V_2\right) \\
& \frac{f}{2}(2)(4.5)+\frac{f}{2}(3)(5.5)=\frac{f}{2} P(4.5+5.5) \\
& \Rightarrow P=2.55=x \times 10^{-1} \mathrm{atmp}
\end{aligned}
$$
So $\mathrm{x}=25.5 \approx 26$ (Nearest integer)
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