An ideal gas, at initial temperature ${\mathrm{T}}_1$and initial volume$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{m}}^3\mathbf{}$ , is expanded adiabatically to a volume of$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{m}}^3$ , then expanded isothermally to a volume of$\mathbf{10}\mathbf{}{\mathbf{m}}^3$ , and then compressed adiabatically back to${\mathbf{T}}_1$ .What is its final volume?

If during a thermodynamic process, the heat exchange between the system and surroundings is zero, then the process is called an Adiabatic process. Practically, such processes are conducted very rapidly, so that no exchange of heat takes place between the system and its surrounding.For the adiabatic process,the following relation must be satisfied:

${\mathbf{T}}_1{\mathbf{V}}_1^{\gamma \mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{T}}_2{\mathbf{V}}_2^{\gamma \mathbf{-}\mathbf{1}}$

Here, $\gamma$ is the ratio of molar-specific heat at constant pressure $\left({\mathbf{C}}_P\right)$to that at constant volume $\left({\mathbf{C}}_V\right)$.

1. Initial temperature =$T_1$
2. Initial volume,$V_1=2.0{\text{\hspace{0.17em}m}}^3$
3. Final volume in adiabatic process,$V_2=4{\text{\hspace{0.17em}m}}^3$
4. Volume at the end of isothermal process, $V_3=10{\text{\hspace{0.17em}m}}^3$

For the adiabatic expansion, we have-

$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

If we solve it for ${\text{T}}_2$we get,

$T_2=\frac{T_1{V}_1^{\gamma -1}}{V_2^{\gamma -1}}$

The second process is isothermal, so the temperature remains the same.

The next process is an adiabatic compression. For this process,the starting volume is${\text{V}}_3=10{\text{\hspace{0.17em}m}}^3$,and it ends at an unknown volume.

$T_1{V}_x^{\gamma -1}=T_2{V}_3^{\gamma -1}$

Substituting value of $T_2$

$T_1{V}_x^{\gamma -1}=\frac{T_1\left(V_1^{\gamma -1}{V}_3^{\gamma -1}\right)}{V_2^{\gamma -1}}$

Simplifying further,

$$\begin{gathered} V_x=\frac{V_1{V}_3}{V_2} \\ {V}_x=2{\text{\hspace{0.17em}m}}^3\times \frac{{\displaystyle 10{\text{\hspace{0.17em}m}}^3}}{{\displaystyle 4{\text{\hspace{0.17em}m}}^3}} \\ {V}_x=5.0{\text{\hspace{0.17em}m}}^3 \end{gathered}$$

The volume at the end of the adiabatic compression is $5.0{\text{\hspace{0.17em}m}}^{\text{3}}$.