Question

A closed gas cylinder is divided into two parts by a piston held tight. The pressure and the volume of gas in two parts respectively are $\left(\right.p, \, 5V\left.\right)$ and $\left(\right.10p, \, V\left.\right)$ . If now the piston is left free and the system undergoes the isothermal process, then the volume of the gas in two parts, respectively are

• A. $4V,2V$
• B. $5V,V$
• C. $2V,4V$
• D. $3V,3V$

Answer 1

Answer: (C)

When the piston is allowed to move the gases are kept separated but the pressure has to be equal $\left(\right.p_{1}=p_{2}\left.\right)$ and the final volumes are x and $6V-x.$

$\because p_{1}=p_{2}$
$P\left(\right.5V\left.\right)=n_{1}RT$
$\Rightarrow n_{1}=\frac{5 P V}{R T}$
$\therefore \frac{n_{1} R T}{x}=\frac{n_{2} R T}{6 V - x}$
$\Rightarrow \frac{n_{1}}{x}=\frac{n_{2}}{6 V - x}$
$\because 10pV=n_{2}RT$
$\Rightarrow n_{2}=\frac{10 P V}{R T}$
$\Rightarrow \frac{5 P V}{x \, R T}=\frac{10 \, P V}{R T \left(\right. 6 V - x \left.\right)}$
$\Rightarrow \frac{1}{x}=\frac{2}{6 V - x}$
$\therefore x=2V$ and $6V-x=4V$