Question

Two glass bulbs of internal volume $0.5$ and $0.2L$ respectively are connected by a narrow tube of negligible volume. The pressure of air in the vessel is $75cm$ at ${17}^{\circ }C$. The smaller bulb is immersed in melting ice and the larger bulb in boiling water. Calculate final pressure in the bulbs neglecting the expansion of glass.

Answer 1

At first at ${17}^{0}C$ or $290K$,

the moles in 1st bulb, $n_1=\frac{75\times 0.5}{R\times 290}$

the moles in 2nd bulb, $n_2=\frac{75\times 0.2}{R\times 290}$

after immersing smaller bulb (2) in ice water and larger bulb (1) in boiling water, same gas will be transferred between the bulbs, until pressure of both bulb become equal.

In final conditions,

$n_1^1=\frac{P\times 0.5}{373R}$ and $n_2^1=\frac{P\times 0.2}{273R}$

now, $n_1+n_2=n_1^1+n_2^1$

$\Rightarrow \frac{75\times 0.5}{R\times 290}+\frac{75\times 0.2}{R\times 290}=\frac{P\times 0.5}{373R}+\frac{P\times 0.2}{273R}$

$\Rightarrow 0.18=P(1.34+0.73)\times {10}^{-3}$

$\Rightarrow P=86.95$ cm of $Hg$