Question

One $\text{mole}$ of an ideal diatomic gas at $300\text{K}$ is compressed adiabatically to one-fourth of its volume. If $\gamma =1.5$, the work done on gas will be

• A. $1280\text{J}$
• B. $1610\text{J}$
• C. $2044\text{J}$
• D. $4986\text{J}$

Answer 1

The correct option is D $4986\text{J}$

Given,
$\gamma =1.5,V_1=V;V_2=\frac{V}{4};T_1=300\text{K}$
As, $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
We get,$T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=300(4{)}^{0.5}=600\text{K}$
for one mole of gas, the work done
$W=\left|\frac{R(T_1-T_2)}{\gamma -1}\right|=\left|\frac{8.31\times (300-600)}{0.5}\right|$
$=4986\text{J}$