Question

An ideal gas at initial temperature $T_0$ and initial volume $V_0$ is expanded adiabatically to a volume $2V_0$. The gas is then expanded isothermally to a volume $5V_0$ and there after compressed adiabatically so that the temperature of the gas becomes again $T_0$. If the final volume of the gas is $\alpha V_0$, then the value of constant $\alpha$ is

• A. 2.5
• B. 1.5
• C. 2.0
• D. 3.0

Answer 1

Answer: (A)

For adiabatic expansion,
$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}\Rightarrow \frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
Here, $T_1=T_0$
$\frac{T_2}{T_0}={\left(\frac{V_0}{2V_0}\right)}^{\gamma -1}$
$T_2={\left(\frac{1}{2}\right)}^{\gamma -1}{T}_0\left[\begin{array}{c}{V}_1=V_0\\ V_2=2V_0\end{array}\right]$
Again, gas is expanded isothermally, therefore
$p_1{V}_1^{\prime }=p_2{V}_2^{\prime }$
Here, $\left[\begin{array}{c}<br/><br/>V_1^{\prime }=V_2=2V_0\\ <br/><br/>V_2^{\prime }=5V_0<br/><br/>\end{array}\right]$
$\frac{p_1}{p_2}=\frac{V_2^{\prime }}{V_1^{\prime }}=\frac{5V_0}{2V_0}$
$\frac{p_1}{p_2}=\frac{V_2^{\prime }}{V_1^{\prime }}=\frac{5}{2}$
Finally gas compressed again adiabatically, therefore, by adiabatic relation,
$T{V}^{\gamma -1}=$ constant
$T_2{\left(V_2^{\prime }\right)}^{\gamma -1}=T_3{\left(V_3\right)}^{\gamma -1}\Rightarrow {\left(\frac{V_3}{V_2^{\prime }}\right)}^{\gamma -1}=\frac{T_2}{T_3}$
$\left[\therefore T_3=T_0\right]$
${\left(\frac{V_3}{V_2^{\prime }}\right)}^{\gamma -1}=\frac{{\left(\frac{1}{2}\right)}^{\gamma -1}{T}_0}{T_0}\Rightarrow V_3=\frac{V_2^{\prime }}{2}=\frac{5V_0}{2}=2.5V_0$
Hence, the value of constant is $2.5$.