Question

The change in the magnitude of the volume of an ideal gas when a small additional pressure $∆\mathrm{P}$ is applied at a constant temperature is the same as the change when the temperature is reduced by a small quantity $∆\mathrm{T}$at constant pressure. The initial temperature and pressure of the gas were $300\mathrm{K}$ and$2\mathrm{atm}$. respectively. If $\left|∆\mathrm{T}\right|=\mathrm{C}\left|∆\mathrm{P}\right|$then the value of C in ($\mathrm{K}{\mathrm{atm}}^{-1}$) is __________.

Answer 1

Step 1: Given data

The initial pressure of the gas$=2\mathrm{atm}$

Initial temperature$=300\mathrm{K}$

Step 2: Applying the formula

We know, $\mathrm{PV}=\mathrm{nRT}$, where P is pressure, V is volume, n is moles, R is gas constant, and T is temperature.

For constant temperature, $\mathrm{P}∆\mathrm{V}+\mathrm{V}∆\mathrm{P}=0$

and for constant pressure, $\mathrm{P}∆\mathrm{V}=\mathrm{nR}∆\mathrm{T}$

$\therefore ∆\mathrm{T}=\frac{\mathrm{P}∆\mathrm{V}}{\mathrm{nR}}$

$∆\mathrm{P}=-\frac{\mathrm{P}∆\mathrm{V}}{\mathrm{V}}$, $∆\mathrm{V}$ is the same in both the cases

$\frac{∆\mathrm{T}}{∆\mathrm{P}}=\frac{\mathrm{P}∆\mathrm{V}}{\mathrm{nR}}\frac{\mathrm{V}}{-\mathrm{P}∆\mathrm{V}}=\frac{-\mathrm{V}}{\mathrm{nR}}=-\mathrm{TP}$

As $\mathrm{PV}=\mathrm{nRT}$

$\frac{\mathrm{V}}{\mathrm{nR}}=\frac{\mathrm{T}}{\mathrm{P}}$

$\therefore \left|\frac{∆T}{∆P}\right|=-\frac{300}{2}\Rightarrow 150$

Therefore, the value of C is $150\mathrm{K}{\mathrm{atm}}^{-1}$