Question

A container has two chambers of volumes $V_1$ = 2 litres and $V_2$ = 3 litres separated by a partition made of a thermal insulator. The chambers contains $n_1$ = 5 and $n_2$ = 4 moles of ideal gas at pressures $P_1$ = 1 atm and $P_2$ = 2 atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of :

• A. 1.4 atm
• B. 1.8 atm
• C. 1.3 atm
• D. 1.6 atm

Answer 1

Answer: (D)

1.6 atm

Answer 2

The mixing process is assumed to be adiabatic and involve no work, leading to conservation of total internal energy. For ideal gases, internal energy $U=nC_vT$. Conservation of internal energy means $n_1C_vT_1 + n_2C_vT_2 = (n_1+n_2)C_vT_f$. Using the ideal gas law $PV=nRT$, we have $nT = PV/R$. Substituting this into the energy conservation equation gives $\frac{P_1V_1}{R}C_v + \frac{P_2V_2}{R}C_v = \frac{P_f(V_1+V_2)}{R}C_v$. This simplifies to $P_1V_1 + P_2V_2 = P_f(V_1+V_2)$. Solving for $P_f$: $P_f = \frac{P_1V_1 + P_2V_2}{V_1 + V_2}$. Plugging in the given values $V_1=2$ L, $P_1=1$ atm, $V_2=3$ L, $P_2=2$ atm: $P_f = \frac{(1)(2) + (2)(3)}{2+3} = \frac{2+6}{5} = \frac{8}{5} = 1.6$ atm.