Question

Two spherical vessels of equal volume are connected by a narrow tube. The apparatus contains an ideal gas at one atmosphere and $300K$. Now if one vessel is immersed in a bath of constant temperature $600K$ and the other in a bath of constant temperature $300K$, then the common pressure will be -

Answer 1

Answer: 1.33

$n_A=\frac{P_0{V}_0}{300R}$
$n_B=\frac{P_0{V}_0}{300R}\left(P_0=1atm\right)$

$n_A^{\prime }=\frac{P{V}_0}{600R},n_B^{\prime }=\frac{P{V}_0}{300R}$
As number of moles remain conserved
$n_A+n_B=n_A^{\prime }+n_B^{\prime }$
$\frac{P_0{V}_0}{300R}+\frac{P_0{V}_0}{300R}=\frac{P{V}_0}{600R}+\frac{P{V}_0}{300R}$
$\frac{P{V}_0}{R}\left(\frac{1}{600}+\frac{1}{300}\right)=\frac{2P_0{V}_0}{300R}$
$P\times \frac{3}{600}=\frac{2}{300}\times 1$
$P=\frac{2}{3}\times \frac{600}{300}=\frac{4}{3}atm$