Question

Two closed bulbs of equal volume $(V)$ containing an ideal gas initially at pressure $p_i$ and temperature $T_1$ are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to $T_2$. The final pressure $p_f$ is :

• A. $p_{i} \left( \frac{T_{1}T_{2}}{T_{1} +T_{2}}\right) $
• B. $2 p_{i} \left( \frac{T_{1}}{T_{1} +T_{2}}\right) $
• C. $2 p_{i} \left( \frac{ T_{2}}{T_{1} +T_{2}}\right) $
• D. $2 p_{i} \left( \frac{T_{1}T_{2}}{T_{1} +T_{2}}\right) $

Answer 1

The Correct Option is C

To determine the final pressure $p_f$ in the closed bulbs, we can use the ideal gas law and principles of equilibrium. Let's break our explanation into logical steps:

1. Initially, each bulb contains an ideal gas at pressure $p_i$ and temperature $T_1$, with a volume $V$ for each bulb.
2. According to the Ideal Gas Law, for the first bulb: PV = nRT_1, where n is the number of moles and R is the gas constant.
3. After connecting the bulbs, the temperature of the second bulb is raised to $T_2$.
4. In thermal equilibrium, the pressure in both bulbs becomes uniform. The total number of moles is conserved, so we write: n_1 = \frac{p_i V}{RT_1} and n_2 = \frac{p_i V}{RT_1} initially in both bulbs.
5. After heating, using conservation of moles and equal final pressure $p_f$: \[ n_1 + n_2 = \frac{p_f V}{RT_1} + \frac{p_f V}{RT_2} \] \[ 2\frac{p_i V}{RT_1} = \frac{p_f V}{RT_1} + \frac{p_f V}{RT_2} \]
6. Solving for $p_f$ gives: \[ 2p_i = p_f \left(\frac{1}{T_1} + \frac{1}{T_2}\right) \] \[ p_f = 2p_i \left(\frac{T_1 T_2}{T_1 + T_2}\right) \]
7. The final pressure $p_f$ by calculation is: $2 p_{i} \left( \frac{ T_{2}}{T_{1} +T_{2}}\right) $, making option (c) correct.

Conclusion: The correct option is $2 p_{i} \left( \frac{ T_{2}}{T_{1} +T_{2}}\right) $, since it aligns with the derived equation and reflects the physical conditions after heating one of the bulbs.