Question

A jar contains a gas and a few drops of water. The pressure in the jar is $830\text{mm of Hg}$. The temperature of the jar is reduced by $1\%$. The vapour pressure of water at two temperatures are $30$ and $25\text{mm of Hg}$. Calculate the new pressure in jar.

• A. $792\text{mm of Hg}$
• B. $817\text{mm of Hg}$
• C. $800\text{mm of Hg}$
• D. $840\text{mm of Hg}$

Answer 1

The correct option is B $817\text{mm of Hg}$

The pressure inside the jar is due to the pressure of dry gas and pressure of moisture.
$P_{\text{gas}}=P_{\text{dry gas}}+P_{\text{moisture}}\Rightarrow P_{\text{dry}}=830-30=800\text{mm of Hg}$
Now according to the question, $T_2=0.99T_1$
At constant volume, $\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow P_2=\frac{800\times 0.99T_1}{T_1}=792\text{mm of Hg}$
$\therefore$ The new pressure in the jar after reduction in temperature is
$P_{\text{gas}}=P_{\text{dry}}+P_{\text{moisture}}=792+25=817\text{mm of Hg}$