Question

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

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

Answer 1

The correct option is B 817 mm of Hg

$P_{total}=P_{gas}+V.P=830\text{mm of Hg}$
$\therefore P_{gas}=830-V.P=830-30=800$
Since, ${}^{\prime }{V}^{\prime }$ and ${}^{\prime }{n}^{\prime }$ are constant for this gas, $P\propto T$
$\therefore$ $P$ also reduces by $1\%$
$P_{gas}=\frac{99}{100}\times 800=792\text{mm}$
$V.P$ @ lower $T$ is $25\text{mm}$
$P_{total}=792+25=817\text{mm}$