Two non-conducting containers having volume $V_{1}$ and $V_{2}$ contain mono-atomic and diatomic gases respectively. They are connected as shown in the figure. Pressure and temperature in the two containers are $P_{1}, T_{1}$ and $P_{2}, T_{2}$ respectively. Initially, the stopcock is closed, if the stopcock is opened, find the final pressure and temperature

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Solution

$n_{1} = \frac{P_{1} V_{1}}{R T_{1}}, n_{2} = \frac{P_{2} V_{2}}{R T_{2}}$
$n = n_{1} + n_{2}$ (number of moles are conserved)
Finally pressure in both parts & temperature of the both the gases will become equal.
$\frac{P (V_{1} + V_{2})}{R T} = \frac{P_{1} V_{1}}{R T_{1}} + \frac{P_{2} V_{2}}{R T_{2}}$
From energy conservation
$\frac{3}{2} n_{1} R T_{1} + \frac{5}{2} n_{2} R T_{2} = \frac{3}{2} n_{1} R T + \frac{5}{2} n_{2} R T$

$\Rightarrow T = \frac{(3 P_{1} V_{1} + 5 P_{2} V_{2}) T_{1} T_{2}}{3 P_{1} V_{1} T_{2} + 5 P_{2} V_{2} T_{1}} \Rightarrow P = (\frac{3 P_{1} V_{1} + 5 P_{2} V_{2}}{3 P_{1} V_{1} T_{2} + 5 P_{2} V_{2} T_{1}}) (\frac{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1}}{V_{1} + V_{2}})$

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