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}.
- A) left [ left ({ large{P_B over P_A}} { large{T_A over T_B }} + β right ) / left (β + { large{1 over alpha}} { large{T_A over T_B}} right ) right ] P_A
- B) left [ left (1+β{ large{P_B over P_A}} { large{T_A over T_B}} right ) / left (1- { large{β over alpha}} { large{T_A over T_B}} right ) right ]P_A
- C) left [ left (1+β{ large{P_B over P_A}} { large{T_A over T_B}} right ) / left (1+ { large{β over alpha}} { large{T_A over T_B}} right ) right ]P_A
- D) left [ left (1+β{ large{P_B over P_A}} { large{T_A over T_B}} right ) / left (alpha+β { large{T_A over T_B}} right ) right ] P_A
- E) left [ left (β{ large{P_B over P_A}} { large{T_A over T_B}}-1 right ) / left (alpha+β { large{T_A over T_B}} right ) right ] P_A
Resposta:
The correct answer is C) $left[left(1+frac{beta}{alpha}right) left(frac{P_B}{P_A}right) left(frac{T_A}{T_B}right)right]frac{1}{left(1+frac{beta}{alpha}right) left(frac{T_A}{T_B}right)}P_A$.
To understand why this is the correct answer, - let's break down the problem step by step. We have two containers A and B, each with a volume of V_A and V_B, respectively, connected by a tube with a valve that regulates the flow of an ideal gas. Initially, the gas in container A is at temperature T_A under pressure P_A, and the gas in container B is at temperature T_B under pressure P_B.
When the valve is opened, the pressures in both containers will equalize, and the temperature will remain constant. We need to find the final pressure P_{Af} in terms of the initial pressures and temperatures.
Using the ideal gas law, we can write the initial conditions as:
P_A V_A = n_A R T_A ... (1)
P_B V_B = n_B R T_B ... (2)
where n_A and n_B are the number of moles of gas in containers A and B, respectively, and R is the gas constant.
Since the volumes are constant, we can write:
V_B = β V_A ... (3)
Substituting (3) into (2), we get:
P_B β V_A = n_B R T_B ... (4)
Now, when the valve is opened, the total number of moles of gas becomes n_A + n_B, and the total volume is V_A + V_B = V_A (1 + β). The final pressure P_{Af} is given by:
P_{Af} (V_A + V_B) = (n_A + n_B) R T_A
Substituting (1) and (4) into the above equation, we get:
P_{Af} V_A (1 + β) = P_A V_A + P_B β V_A
Simplifying the equation, we get:
P_{Af} = $left[left(1+frac{beta}{alpha}right) left(frac{P_B}{P_A}right) left(frac{T_A}{T_B}right)right]frac{1}{left(1+frac{beta}{alpha}right) left(frac{T_A}{T_B}right)}P_A$
which is the correct answer.
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