1. An ideal gas initially at pressure${\mathbf{p}}_0$ undergoes free expansion until its volume is $\mathbf{3}\mathbf{.}\mathbf{00}$times its initial volume. What then is the ratio of its pressure to ${\mathbf{p}}_0$?
2. The gas is next slowly and adiabatically compressed back to its original volume. The pressure after compression is${(\mathbf{3}\mathbf{.}\mathbf{00})}^{1/3}{\mathbf{p}}_0$ . Is the gas monoatomic, diatomic or polyatomic?
3. What is the ratio of the average kinetic energy per molecule in this final state to that in initial state?

If a process is conducted rapidly,in an insulating chamber, then the process is known as adiabatic process. In the process, no exchange of heat can take place between the system and surrounding. Such a process must satisfy the relation-

${\mathbf{PV}}^{\gamma }\mathbf{=}\mathbf{constant}$

The average kinetic energy for each degree of freedom is equal to$\frac{1}{2}\mathbf{kT}$.

1. The volume of expanded gas is $V_1=3.00\times V_0$
2. The pressure after compression is$P_2={\left(3.00\right)}^{\frac{1}{3}}\times P_0$

The equation to be satisfied during the free expansion of an ideal gas is-

$P_i{V}_i=P_f{V}_f$

So, for the free expansion from the original state 0 to state 1, it becomes

$P_0{V}_0=P_1{V}_1$

For${\text{V}}_1=3.00\times {\text{V}}_0$; the equation becomes-

$\begin{array}{c}\frac{P_1}{P_0}=\frac{V_0}{V_1}\\ =\frac{V_0}{3.00\times V_0}\\ =\frac{1}{3}\end{array}$

The ratio of pressure in final state to initial state is$\frac{P_1}{P_0}=\frac{1}{3}$ .

For an adiabatic process, we have

$P{V}^{\gamma }=\text{constant}$

So, for this case, we can write

$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

For, $V_1=3.00\times V_0$ and $P_2={\left(3.00\right)}^{\frac{1}{3}}\times P_0$, equation becomes-

$$\begin{gathered} \left(\frac{1}{3}{P}_0\right)\times {(3.00V_0)}^{\gamma }=(\left(3.00\right){}^{\frac{1}{3}}{P}_0){\left(V_0\right)}^{\gamma } \\ \gamma -1=\frac{1}{3} \\ \gamma =\frac{4}{3} \end{gathered}$$

If f is the number of degree of freedom, then-

$\begin{array}{c}\gamma =1+\frac{2}{f}\\ =\frac{4}{3}\\ \begin{array}{c}\frac{4}{3}=1+\frac{2}{f}\\ f=6\end{array}\end{array}$

This shows that the gas is polyatomic.

We know that the average kinetic energy of an ideal gas is

$K_{avg}=\frac{3}{2}kT$

So,

$\begin{array}{c}\frac{K_{avg2}}{K_{avg1}}=\frac{T_2}{T_1}\\ =\frac{P_2}{P_1}\end{array}$

$$\begin{gathered} \frac{K_{avg2}}{K_{avg1}}=\frac{{\left(3.00\right)}^{\frac{1}{3}}{P}_0}{P_0}={3.00}^{\frac{1}{3}} \\ \frac{K_{avg2}}{K_{avg1}}=1.44 \end{gathered}$$

Ratio of the average kinetic energy per molecule is$\frac{K_{avg2}}{K_{avg1}}=1.44$