Question

One mole of Argon is heated using $P{V}^{3/2}$= const. Find the amount of heat obtained by the process when the temperature changes by $△T=-26k$

Answer 1

Given : Number of moles of the gas $n=1$

Change in temperature $\Delta T=-26K$

Argon is a monoatomic gas.

For monoatomic gas, $\gamma =\frac{5}{3}$

The process is given as $P{V}^{3/2}=constant$

Comparing with $P{V}^m=constant$, we get $m=\frac{3}{2}$

$\therefore$ Specific heat capacity of the process $C=\frac{R}{\gamma -1}+\frac{R}{1-m}$

$C=\frac{R}{\frac{5}{3}-1}+\frac{R}{1-\frac{3}{2}}=-\frac{R}{2}$

Amount of heat involved in the process $Q=nC\Delta T$

$\therefore$ $Q=1\times (\frac{-8.31}{2})\times (-26)$ $(R=8.31J{K}^{-1}mo{l}^{-1})$

$⟹Q=108$ J