# The change in the magnitude of the volume of an ideal gas when a small additional pressure

Question:

The change in the magnitude of the volume of an ideal gas when a small additional pressure $\Delta P$ is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $\Delta T$ at constant pressure. The initial temperature and pressure of the gas were $300 \mathrm{~K}$ and $2 \mathrm{~atm}$. respectively. If $|\Delta T|=C|\Delta P|$, then value of $C$ in (K/atm.) is_______

Solution:

(150)

In first case,

From ideal gas equation

$P V=n R T$

$P \Delta V+V \Delta P=0$        (As temperature is constant)

$\Delta V=-\frac{\Delta P}{P} V$         $\ldots$ (i)

In second case, using ideal gas equation again

$P \Delta V=-n R \Delta T$

$\Delta V=-\frac{n R \Delta T}{P}$             ...(ii)

Equating (i) and (ii), we get

$\frac{n R \Delta T}{P}=-\frac{\Delta P}{P} V \Rightarrow \Delta T=\Delta P \frac{V}{n R}$

Comparing the above equation with $|\Delta T|=C|\Delta P|$, we have

$C=\frac{V}{n R}=\frac{\Delta T}{\Delta P}=\frac{300 \mathrm{~K}}{2 \mathrm{~atm}}=150 \mathrm{~K} / \mathrm{atm}$