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 $300\text{K}$ and $2\text{atm}$, respectively. If$|\Delta T|=C|\Delta P|$, value of $C$ in $\left(\text{K/atm}\right)$ is

Answer 1

In first case,
From ideal gas equation
$PV=nRT$

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

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

In second case, using ideal gas equation again

$P\Delta V=-nR\Delta T$

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

Equating (i) and (ii), we get

$\frac{nR\Delta T}{P}=\frac{\Delta P}{P}V\Rightarrow \Delta T=\Delta P\frac{V}{nR}$

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

$C=\frac{V}{nR}=\frac{\Delta T}{\Delta P}=\frac{300}{2}=150\text{K/atm}$.