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Ideal Gas Equation from KTG

Two closed bu...

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:

 JEE (MAIN)- 2016

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

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

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Solution

The correct option is D $2 P_{i} (\frac{T_{2}}{T_{1} + T_{2}})$
Initial moles and final moles are equal
$(n_{T})_{i} = (n_{T})_{f}$
$\frac{P_{i} V}{R T_{1}} + \frac{P_{i} V}{R T_{1}} = \frac{P_{f} V}{R T_{1}} + \frac{P_{f} V}{R T_{2}}$
$2 \frac{P_{i}}{T_{1}} = \frac{P_{f}}{T_{1}} + \frac{P_{f}}{T_{2}}$
$2 \frac{P_{i}}{T_{1}} = P_{f} (\frac{1}{T_{1}} + \frac{1}{T_{2}})$
$P_{f} = \frac{2 P_{i} T_{2}}{T_{1} + T_{2}}$
$P_{f} = 2 P_{i} (\frac{T_{2}}{T_{1} + T_{2}})$

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