Question

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at $0^0$ C at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at ${62}^0$ C. What is the new value of the pressure inside the bulbs ? The volume of the connecting tube is negligible.

Answer 1

When the bulbs are maintained at two different temperatures.

The total heat gained by 'B' is the heat lost by 'A'. Let the final temperature be X

So, $m_{1}SDt=m^{2}SDt$

$\Rightarrow n_{1}X=62n^2-n^{2}X$

So, X = ${31}^0$ C = 304K

[Since, Initial temperature = $0^0$ C]

P = 76 cm of Hg, $V_1=V_{2}Hence{n}_1=n_2]$

for a single ball

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$\Rightarrow \frac{76\times V}{273}=\frac{P_2\times V}{304}$

$\Rightarrow P^2=\frac{304\times 76}{273}$

= 84.630 = ${84}^0$ C