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

Answer 1

$PV = nRT$
$P \Delta V + V \Delta P =0 \,\,\,\,\,$ (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 )
$\frac{\Delta T }{\Delta P }=\frac{ P \Delta V }{ nR } \frac{ V }{- P \Delta V }=\frac{- V }{ nR }=-\frac{ T }{ P }$
$( PV = nRT )$
$\left(\frac{ V }{ nR }=\frac{ T }{ P }\right)$
$\left|\frac{\Delta T }{\Delta P }\right|=\left|\frac{-300}{2}\right|=150\, a$