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 :

2 P_{i} (\frac{T_{2}}{T_{1} + T_{2}})

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2 P_{i} (\frac{T_{1} T_{2}}{T_{1} + T_{2}})

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P_{i} (\frac{T_{1} T_{2}}{T_{1} + T_{2}})

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2 P_{i} (\frac{T_{1}}{T_{1} + T_{2}})

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Solution

The correct option is A $2 P_{i} (\frac{T_{2}}{T_{1} + T_{2}})$
Since the above system is a closed one, the total number of moles of the ideal gas will be equal before and after the temperature increase.
Hence in the given case.
Initial moles and final moles are rqual $(n_{T})_{i} = (n_{T})_{f}$
Applying $n = \frac{P V}{R T}$
$\frac{P_{i} V}{R T_{1}} + \frac{P_{i} V}{R T_{1}} = \frac{P_{f} V}{R T_{1}} + \frac{P_{f} V}{R T_{2}}$

$2 \frac{P_{i}}{T_{1}} = \frac{P_{f}}{T_{1}} + \frac{P_{f}}{T_{2}} \Rightarrow P_{f} = \frac{2 P_{i} T_{2}}{T_{1} + T_{2}}$

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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: