Two closed bulbs of equal volume (V)containing an ideal gas initially at pressure p_iand temperature T₁ are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulb is then raised to T₂. The final pressure p_f is

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$2 p_{i} (\frac{T_{1}}{T_{1} + T_{2}})$

$2 p_{i} (\frac{T_{2}}{T_{1} + T_{2}})$

$p_{i} (\frac{T_{1} T_{2}}{T_{1} + T_{2}})$

$2 p_{i} (\frac{T_{1} T_{2}}{T_{1} + T_{2}})$

answer is B.

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Detailed Solution

It is clear that moles of gas and volume remains constant during operation.
Initial moles = final moles
$\begin{array}{l} \frac{P_{i} \times V}{{RT}_{1}} + \frac{P_{i} \times V}{{RT}_{1}} = \frac{P_{f} \times V}{{RT}_{2}} + \frac{P_{f} \times V}{{RT}_{1}} \\ \frac{P_{i}}{T_{1}} + \frac{P_{i}}{T_{1}} = \frac{P_{f}}{T_{2}} + \frac{P_{f}}{T_{1}} \\ 2 \frac{P_{i}}{T_{1}} = P_{f} (\frac{1}{T_{2}} + \frac{1}{T_{1}}) \\ 2 \frac{P_{i}}{T_{1}} = P_{f} (\frac{T_{1} + T_{2}}{T_{1} T_{2}}) \\ P_{f} = 2 P_{i} \times (\frac{T_{2}}{T_{1} + T_{2}}) \end{array}$