Question

One mole of a monoatomic gas is heated using $P{V}^{\frac{3}{2}}=\text{const}$. Find the amount of heat obtained in the process, when the temperature changes by $\Delta T=-26\text{K}$

• A. $92\text{J}$
• B. $108\text{J}$
• C. $127\text{J}$
• D. $256\text{J}$

Answer 1

The correct option is B $108\text{J}$

Given:
Number of moles of the gas $n=1$
Change in temperature $\Delta T=-26\text{K}$
For monoatomic gas, $\gamma =\frac{5}{3}$
The process is given as $P{V}^{\frac{3}{2}}=\text{constant}$
Comparing with $P{V}^x=\text{constant}$, we get $x=\frac{3}{2}$
$\therefore$ Specific heat capacity of the process $C=\frac{R}{\gamma -1}+\frac{R}{1-x}$
$C=\frac{R}{\frac{5}{3}-1}+\frac{R}{1-\frac{3}{2}}=-\frac{R}{2}$
Amount of heat involved in the process $Q=nC\Delta T$
$\therefore Q=1\times \left(\frac{-8.31}{2}\right)\times (-26)(\because R=8.31{\text{J K}}^{-1}{\text{mol}}^{-1})$
$\Rightarrow Q=108\text{J}$