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 $300K$ and $2atm$ respectively. If $|\Delta T|=C|\Delta P|$ then value of $C$ in $(K/atm)$ is____.

Answer 1

Answer: 150

$PV=nRT$
$P\Delta V+V\Delta P=0$ (for constant temp.)
$P\Delta V=nR\Delta T$ (for constant pressure)
$\Delta T=\frac{P\Delta V}{nR}$
$\Delta P=-\frac{P\Delta V}{V}(\Delta V$ is same in both cases )
$\frac{\Delta T}{\Delta P}=\frac{P\Delta V}{nR}\frac{V}{-P\Delta V}=\frac{-V}{nR}=-\frac{T}{P}$
$(PV=nRT)$
$\left(\frac{V}{nR}=\frac{T}{P}\right)$
$\left|\frac{\Delta T}{\Delta P}\right|=\left|\frac{-300}{2}\right|=150a$