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 $|\Delta T| = C|\Delta P|$ then value of $C$ in $(\text{K/atm})$ is ______.
Numerical value type. Enter your answer.
Answer: 150
Isothermal ($PV$ constant): $|\Delta V| = V\dfrac{\Delta P}{P}$. Isobaric ($V/T$ constant): $|\Delta V| = V\dfrac{\Delta T}{T}$.
$$\frac{\Delta P}{P} = \frac{\Delta T}{T} \Rightarrow \Delta T = \frac TP\Delta P = \frac{300}{2}\Delta P$$
So $C = 150$ K/atm.
Solution by Sreeraj P, M.Sc Physics