Question

Two $\text{moles}$ of an ideal monoatomic gas at ${27}^{\circ }\text{C}$ occupies a volume of $V$. If the gas is expanded adiabatically to the volume $2^{3/2}V$, then the work done by the gas will be $(\gamma =\frac{5}{3},R=8.31\text{J/mol K})$

• A. $3739.5\text{J}$
• B. $2627.23\text{J}$
• C. $2500\text{J}$
• D. $-2500\text{J}$

Answer 1

The correct option is A $3739.5\text{J}$

Given, gas is monoatomic $\therefore \gamma =\frac{5}{3}$
Initial Temperature $\left(T_i\right)=300\text{K}$
Initial volume $\left(V_i\right)=V$
Final volume $\left(V_f\right)=2^{3/2}V$
Work done by the gas $\left(W\right)=\frac{nR(T_i-T_f)}{\gamma -1}$
$=\frac{nR{T}_i}{\gamma -1}[1-\frac{T_f}{T_i}]$
As, $T_i{V}_i^{\gamma -1}=T_f{V}_f^{\gamma -1}$
$\therefore W=\frac{nR{T}_i}{\gamma -1}[1-{\left(\frac{V_i}{V_f}\right)}^{\gamma -1}]$
$=\frac{2\times 8.31\times 300}{(\frac{5}{3}-1)}[1-{\left(\frac{1}{2^{3/2}}\right)}^{\frac{2}{3}}]$
$=3739.5\text{J}$