States of Matter - Result 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, $\quad \frac{2 p _i V}{R T _1}=\frac{p _f V}{R T _1}+\frac{p _f V}{R T _2} \Rightarrow p _f=2 p _i\left(\frac{T _2}{T _1+T _2}\right)$
