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A container has two chambers of volumes $V_1 = 2\text{ litres}$ and $V_2 = 3\text{ litres}$ separated by a partition made of a thermal insulator. The chambers contains $n_1 = 5$ and $n_2 = 4\text{ moles}$ of ideal gas at pressures $p_1 = 1\text{ atm}$ and $p_2 = 2\text{ atm}$, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:
A
1.3 atm
B
1.6 atm
C
1.4 atm
D
1.8 atm
Explanation
Total internal energy is conserved, so $P_f V_{total} = P_1 V_1 + P_2 V_2 \implies P_f = \frac{1(2) + 2(3)}{2 + 3} = \frac{8}{5} = 1.6\text{ atm}$.
Detailed Solution
For ideal gases insulated from the surroundings, internal energy is conserved upon free mixing: $U = U_1 + U_2$. Since internal energy of an ideal gas is $U = \frac{f}{2} n R T = \frac{f}{2} P V$, we have: $\frac{f}{2} P_f (V_1 + V_2) = \frac{f}{2} P_1 V_1 + \frac{f}{2} P_2 V_2 \implies P_f = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2}$. Substituting the values: $P_f = \frac{(1\text{ atm} \times 2\text{ L}) + (2\text{ atm} \times 3\text{ L})}{2\text{ L} + 3\text{ L}} = \frac{2 + 6}{5} = \frac{8}{5} = 1.6\text{ atm}$.