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:

Answer 1

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.

$\frac{2P_{i}V}{R{T}_1}=\frac{P_{f}V}{R{T}_1}+\frac{P_{f}V}{R{T}_2}$
Hence, $P_f=\frac{2P_i{T}_2}{T_1+T_2}$
Given,
$P_i=1atm$
$T_1=300K$
$T_2=450K$
$P_f=\frac{2\times 1\times 450}{300+450}=1.2atm$