Dois recipientes A e B de respectivos volumes V_A e V_B=beta V_A, constantes, contêm um gás ideal e são conectados por um tubo fino com válvula que regula a passagem do gás, conforme a figura. Inicialmente o gás em A está na temperatura T_A sob pressão P_A e em B, na temperatura T_B sob pressão P_B. A válvula é então aberta até que as pressões finais P_{Af} e P_{Bf} alcancem a proporção P_{Af}/P_{Bf} = α, mantendo as temperaturas nos seus valores iniciais. Assinale a opção com a expressão de P_{Af}.

The correct alternative is letter C) $left[left(1+betafrac{P_B}{P_A}right)left(frac{T_A}{T_B}right) / left(1 + frac{beta}{alpha}frac{T_A}{T_B}right)right] P_A$.

To explain, (and prove) this answer, when the valve is opened, the pressure in both containers A and B will equalize, and the volume of the gas will be distributed between the two containers. Since the volume of container A is $V_A$ and the volume of container B is $V_B = beta V_A$, the total volume is $V_A + V_B = V_A + beta V_A = V_A(1 + beta)$. The number of moles of gas in container A is $n_A = frac{P_AV_A}{RT_A}$, and the number of moles of gas in container B is $n_B = frac{P_BV_B}{RT_B} = frac{P_Bbeta V_A}{RT_B}$. After the valve is opened, the total number of moles of gas is $n_A + n_B = frac{P_AV_A}{RT_A} + frac{P_Bbeta V_A}{RT_B}$. Since the temperature remains constant, the total number of moles of gas is also constant, so $n_A + n_B = frac{P_fV_A(1+beta)}{RT_A}$, where $P_f$ is the final pressure. Equating the two expressions, we get $frac{P_AV_A}{RT_A} + frac{P_Bbeta V_A}{RT_B} = frac{P_fV_A(1+beta)}{RT_A}$. Solving for $P_f$, we get $P_f = frac{P_A + P_Bbetafrac{T_A}{T_B}}{1+beta}$. Finally, since $P_f = P_{Af} = alpha P_{Bf}$, we have $P_{Af} = frac{P_A + P_Bbetafrac{T_A}{T_B}}{1+beta} = left[left(1+betafrac{P_B}{P_A}right)left(frac{T_A}{T_B}right) / left(1 + frac{beta}{alpha}frac{T_A}{T_B}right)right] P_A$. Therefore, the correct alternative is letter C).