Question

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

• A. $792mm$ of $Hg$
• B. $817mm$ of $Hg$
• C. $800mm$ of $Hg$
• D. $840mm$ of $Hg$

Answer 1

Answer: (B)

$817mm$ of $Hg$

Pressure in the jar $=830$ mm of Hg.

Pressure of moisture $=30$ mm of Hg.

Pressure of dry gas $=830-30=800$ mm of Hg.

Since the volume is constant and the temperature is changing, using the ideal gas equation,

$\frac{T_1}{P_1}=\frac{T_2}{P_2}\Rightarrow \frac{T_1}{800}=\frac{0.99T_1}{P_2}\Rightarrow P_2=0.99\times 800=792$ mm of Hg.

Therefore, the total pressure $=792+25=817$ mm of Hg.

So, the correct option is $B$