Question

In a close vessel $4$ L $C{H}_4$ gas is present at $60$ mm Hg pressure. If $1$ litre of water is filled into the container, then find the total pressure of gas mixture (Assuming vapour pressure of water at that temperature = $20$ mm Hg)

• A. $100$ mm Hg
• B. $106.7$ mm Hg
• C. $80$ mm Hg
• D. $40$ mm Hg

Answer 1

The correct option is A $100$ mm Hg

According to Boyle's law
$P_1{V}_1=P_2{V}_2$
$60\times 4=P_2\times 3$
$\Rightarrow P_2=80\text{mm Hg}$
Given, $P_{H_{2}O}=20\text{mm Hg}$
Therefore, total pressure, $P_T=(80+20)\text{mm Hg}=100\text{mm Hg}$