Question

A gas is present above and below the piston in a cylinder fitted with movable piston having a certain mass. On both sides of the piston there are equal number of moles of gas. The volume above is two times the volume below at a temperature of $300\text{K}$. At what temperature will the volume above be four times the volume below?

• A. $600\text{K}$
• B. $400\text{K}$
• C. $200\text{K}$
• D. $120\text{K}$

Answer 1

The correct option is D $120\text{K}$

Let the pressure exerted by the piston be $=P_0$
$P_A+P_0=P_B$ or $P_0=P_B-P_A$
From ideal gas equation we can write,
$P_0=\frac{nR\times 300}{V/3}-\frac{nR\times 300}{2V/3}$

Let the final temperature = $T$

$P_B^{\prime }=P_0+P_A^{\prime }$ or $P_0=P_B^{\prime }-P_A^{\prime }$
Now from ideal gas equation we have,

$P_0=\frac{nRT}{V/5}-\frac{nRT}{4V/5}$

Putting the value of $P_0$ from above we get,

$\therefore \frac{nRT}{V}\times 5\times \frac{3}{4}=\frac{nR\times 300}{V}\times 3\times \frac{1}{2}$

or $T\times \frac{15}{4}=300\times \frac{3}{2}$

or $T=\frac{4}{15}\times 300\times \frac{3}{2}=120\text{K}$.