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Pressure

Two closed bu...

Question

Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure 1 atm and temperature 300 K 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 450 K. The final pressure is:

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Solution

Let,
Initial pressure = $P_{i}$
Initial Volume = V
Initial temperature = $T_{1}$

We know total moles of the two bulbs is constant. Hence

Total moles before heating = total moles after heating.
Since the volume is constant, pressure will change upon heating.

$\frac{2 P_{i} V}{R T_{1}} = \frac{P_{f} V}{R T_{1}} + \frac{P_{f} V}{R T_{2}}$
Hence, $P_{f} = \frac{2 P_{i} T_{2}}{T_{1} + T_{2}}$
Given,
$P_{i} = 1 a t m$
$T_{1} = 300 K$
$T_{2} = 450 K$
$P_{f} = \frac{2 \times 1 \times 450}{300 + 450} = 1.2 a t m$

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