Question

Air (ideal gas with $\gamma =1.4)$ at $1\text{bar}$ and $300\text{K}$ is compressed till the final volume is one-sixteenth of the original volume, following a polytropic process $P{V}^{1.25}=\text{const}.$
Calculate the (a) work done and (b) the energy transferred as heat per mole of the air.

• A. $-9.977\text{kJ/mol},-3.742\text{kJ/mol}$
• B. $-6.97\text{kJ/mol},-8.24\text{kJ/mol}$
• C. $-6.97\text{kJ/mol},-2.64\text{kJ/mol}$
• D. $-3.742\text{kJ/mol},-2.64\text{kJ/mol}$

Answer 1

The correct option is A $-9.977\text{kJ/mol},-3.742\text{kJ/mol}$

Given that gas follows$P{V}^{1.25}=\text{constant}$
Let $V_1$ be volume corresponding to pressure $P_1=1\text{bar}$,
$V_2$ be volume corresponding to pressure $P_2$
Also given that $V_2=\frac{V_1}{16}$
Hence,
$P_1{V}_1^{1.25}=P_2{V}_2^{1.25}$
$P_2=P_1(V_1/V_2{)}^{1.25}=1(16{)}^{1.25}=32\text{bar}$
Using $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$T_2=\frac{\left(T_1{P}_2{V}_2\right)}{\left(P_1{V}_1\right)}=\frac{(300\times 32\times 1)}{(1\times 16)}=600\text{K}$

$\left(a\right)\text{Work done in a polytropic process}W=\frac{(P_1{V}_1-P_2{V}_2)}{(n-1)}=\frac{R(T_1-T_2)}{(n-1)}$
$W=\frac{8.314(300-600)}{(1.25-1)}=-9.977\text{kJ/mol}$

$\left(b\right)\text{Heat transferred in this process}Q=W\left\{\right(\gamma -n\left)/\right(\gamma -1\left)\right\}$
$=-9.977(1.4-1.25)/(1.4-1)$
$=-3.742\text{kJ/mol}$