States of Matter 1 Question 10
10. 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
(2016 Main)
(a) $2 p_{i}\left(\frac{T_{1}}{T_{1}+T_{2}}\right)$
(b) $2 p_{i}\left(\frac{T_{2}}{T_{1}+T_{2}}\right)$
(c) $2 p_{i}\left(\frac{T_{1} T_{2}}{T_{1}+T_{2}}\right)$
(d) $p_{i}\left(\frac{T_{1} T_{2}}{T_{1}+T_{2}}\right)$
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Solution:
- Initially,
Number of moles of gases in each container $=\frac{p_{i} V}{R T_{1}}$
Total number of moles of gases in both containers $=2 \frac{p_{i} V}{R T_{1}}$
After mixing, number of moles in left chamber $=\frac{p_{f} V}{R T_{1}}$
Number of moles in right chamber $=\frac{p_{f} V}{R T_{2}}$
Total number of moles $=\frac{p_{f} V}{R T_{1}}+\frac{p_{f} V}{R T_{2}}=\frac{p_{f} V}{R}\left(\frac{1}{T_{1}}+\frac{1}{T_{2}}\right)$
As total number of moles remains constant.
Hence, $\quad \frac{2 p_{i} V}{R T_{1}}=\frac{p_{f} V}{R T_{1}}+\frac{p_{f} V}{R T_{2}} \Rightarrow p_{f}=2 p_{i}\left(\frac{T_{2}}{T_{1}+T_{2}}\right)$
