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 contains $3.0$ moles of gas at pressure $2.0$ atm and second chamber contains $4.0$ 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 $x$ (nearest integer) is ______
Numerical value type. Enter your answer.
Answer: 26
The container is isolated, so the total internal energy is unchanged. For ideal gases with the same $C_V$, $U = \dfrac{f}{2}PV$, hence $\sum P_iV_i$ stays the same:
$$P(V_1 + V_2) = P_1V_1 + P_2V_2$$
$$P = \frac{2.0\times4.5 + 3.0\times5.5}{10} = \frac{25.5}{10} = 2.55\ \text{atm} = 25.5\times10^{-1}\ \text{atm}$$
To the nearest integer, $x = 26$.
Solution by Sreeraj P, M.Sc Physics