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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 = Pi
Initial Volume = V
Initial temperature = T1

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.

2PiVRT1=PfVRT1+PfVRT2
Hence, Pf=2PiT2T1+T2
Given,
Pi=1 atm
T1=300 K
T2=450 K
Pf=2×1×450300+450=1.2 atm

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