Q181 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
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 $\dfrac{P}{T}$ is {number of moles of gas in vessel A $=$ number of moles of gas in vessel B}
- Correct answerA$\dfrac{1}{2}\left(\dfrac{P_A}{T_A}+\dfrac{P_B}{T_B}\right)$
- B$\dfrac{3}{2}\left(\dfrac{P_A}{T_A}+\dfrac{P_B}{T_B}\right)$
- C$\dfrac{3}{4}\left(\dfrac{P_A}{T_A}+\dfrac{P_B}{T_B}\right)$
- D$\dfrac{5}{4}\left(\dfrac{P_A}{T_A}+\dfrac{P_B}{T_B}\right)$
Solution
Using the ideal gas equation, the initial number of moles in vessel A is:
$n_A = \dfrac{P_A V}{R T_A}$
The initial number of moles in vessel B is:
$n_B = \dfrac{P_B V}{R T_B}$
When the vessels are connected, the total volume becomes $V_{total} = V + V = 2V$. Let the equilibrium pressure be $P$ and the equilibrium temperature be $T$. The total number of moles in the system remains conserved:
$n_{total} = n_A + n_B$
$\dfrac{P(2V)}{RT} = \dfrac{P_A V}{RT_A} + \dfrac{P_B V}{RT_B}$
Dividing the entire equation by $\dfrac{V}{R}$:
$\dfrac{2P}{T} = \dfrac{P_A}{T_A} + \dfrac{P_B}{T_B}$
$\dfrac{P}{T} = \dfrac{1}{2}\left(\dfrac{P_A}{T_A} + \dfrac{P_B}{T_B}\right)$
Answer: $\dfrac{1}{2}\left(\dfrac{P_A}{T_A}+\dfrac{P_B}{T_B}\right)$
$n_A = \dfrac{P_A V}{R T_A}$
The initial number of moles in vessel B is:
$n_B = \dfrac{P_B V}{R T_B}$
When the vessels are connected, the total volume becomes $V_{total} = V + V = 2V$. Let the equilibrium pressure be $P$ and the equilibrium temperature be $T$. The total number of moles in the system remains conserved:
$n_{total} = n_A + n_B$
$\dfrac{P(2V)}{RT} = \dfrac{P_A V}{RT_A} + \dfrac{P_B V}{RT_B}$
Dividing the entire equation by $\dfrac{V}{R}$:
$\dfrac{2P}{T} = \dfrac{P_A}{T_A} + \dfrac{P_B}{T_B}$
$\dfrac{P}{T} = \dfrac{1}{2}\left(\dfrac{P_A}{T_A} + \dfrac{P_B}{T_B}\right)$
Answer: $\dfrac{1}{2}\left(\dfrac{P_A}{T_A}+\dfrac{P_B}{T_B}\right)$
Hence, the correct answer is option: 'A'.