Question

A cylinder with piston contains $N_2$ and a small amount of liquid water at a temperature of ${25}^{o}C$ (aqueous tension of water = 23.8 mm). The total pressure inside the cylinder is 600 mm of Hg, the piston in the cylinder is pushed downwards keeping temperature constant to reduced the volume to half of its initial value. What is the pressure of this condition?

• A. 1176.2 mm of Hg
• B. 1152.4 mm of Hg
• C. 1226 mm of Hg
• D. 576.2 mm of Hg

Answer 1

The correct option is A 1176.2 mm of Hg

$P_1{V}_1=P_2{V}_2$
$P_2=\frac{P_1{V}_1}{V_2}$
= $\frac{600\times V}{\frac{V}{2}}$
$P_2=1200$
Final pressure = 1200 - 23.8
= 1176.2 mm Hg