Question

There are two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and volume are \(P_{A}\), \(T_{A}\) and V in the vessel A and \(P_{B}\), \(T_{B}\) and V in vessel B. The vessels are now connected through a small tube. In equilibrium, pressure is P and temperature is T, then \(\frac{P}{T}\) is {number of moles of gas in vessel A \(=\) number of moles of gas in vessel B}

• A. \(\frac{1}{2}\left(\frac{P_{A}}{T_{A}}+\frac{P_{B}}{T_{B}}\right)\)
• B. \(\frac{3}{2}\left(\frac{P_{A}}{T_{A}}+\frac{P_{B}}{T_{B}}\right)\)
• C. \(\frac{3}{4}\left(\frac{P_{A}}{T_{A}}+\frac{P_{B}}{T_{B}}\right)\)
• D. \(\frac{5}{4}\left(\frac{P_{A}}{T_{A}}+\frac{P_{B}}{T_{B}}\right)\)

Hint:

The total number of moles is conserved. Apply the ideal gas equation to each vessel before joining and to the combined volume after joining.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Moles before joining: \(n_{A}=\dfrac{P_{A}V}{RT_{A}}\) and \(n_{B}=\dfrac{P_{B}V}{RT_{B}}\). Walls are rigid and no gas escapes, so the total moles stay the same.

Step 2: Calculation
After joining the volume is \(2V\):
\(\dfrac{P(2V)}{RT}=\dfrac{P_{A}V}{RT_{A}}+\dfrac{P_{B}V}{RT_{B}}\)

Step 3: Result
Cancelling \(V/R\): \(\dfrac{P}{T}=\dfrac{1}{2}\left(\dfrac{P_{A}}{T_{A}}+\dfrac{P_{B}}{T_{B}}\right)\).

Final Answer:
P/T equals half the sum of \(P_{A}/T_{A}\) and \(P_{B}/T_{B}\). \[ \boxed{\text{(A) }\dfrac{1}{2}\left(\dfrac{P_{A}}{T_{A}}+\dfrac{P_{B}}{T_{B}}\right)} \]