Question

A jar contains a gas and a few drops of water at $T\text{K}$. 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 the jar.

• A. 817
• B. 971
• C. 771
• D. 671

Answer 1

The correct option is A 817

Here $P_{1\text{(gas)}}=830\text{mm}Hg-30\text{mm}Hg$
$P_{1_{\text{(gas)}}}=800\text{mm}Hg$
$P_2=?$
$T_1=TK,T_2=0.99TK$ as temperature is reduced by 1%
Then from $P\propto T$ (at const. $V$)
$\frac{P_1}{P_2}=\frac{T_1}{T_2}\Rightarrow \frac{800}{P_2}=\frac{T}{0.99T}$
$\Rightarrow P_2=792\text{mm}$ of $Hg$ (of gas)
Net new pressure = 792 + 25
$=817\text{mm}$ of $Hg$