We know that for an adiabatic process,$\mathit{p}{\mathit{V}}^{\mathbf{\gamma }}\mathbf{=}\mathit{a}\mathbf{}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}$.Evaluate “ constant” for an adiabatic process involving exactly$\mathbf{2}\mathbf{.}\mathbf{0}\mathit{m}\mathit{o}\mathbf{\text{l}}$of an ideal gas passing through the state having exactly$\mathit{p}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathit{a}\mathit{t}\mathit{m}$and $\mathit{T}\mathbf{=}\mathbf{300}\mathit{K}$. Assume a diatomic gas whose molecules rotate but do not oscillate

• No. of mole of gas are$n=2\text{\hspace{0.17em}mol}$.
• Pressure is$p=1\text{\hspace{0.17em}atm}$.
• Temperature is $T=300\text{\hspace{0.17em}K}$.

The expression for the$\gamma$ is given by,

$\gamma =\frac{C_P}{C_V}$ ...... (i)

Here$C_P$ is the specific heat capacity at constant pressure,$C_V$ is the specific heat capacity at constant volume.

For diatomic gas,

$C_P=\frac{7}{2}R$and

$C_V=\frac{5}{2}R$

Substitute $\frac{7}{2}R$for$C_P$and $\frac{5}{2}R$for $C_V$into the equation (i),

$\begin{array}{c}\gamma =\frac{C_p}{C_v}\\ =\frac{\left(\frac{7}{2}\right)R}{\left(\left(\frac{5}{2}\right)R\right)}\\ =\frac{7}{5}\end{array}$

$\Rightarrow \gamma =1.4$

We can find the volume of the gas

$1\text{\hspace{0.17em}atm}=1.01\times {10}^5\text{\hspace{0.17em}Pa}$

We can find the volume of gas using the gas law

$$\begin{gathered} pV=nRT \\ \Rightarrow V=\frac{nRT}{p} \end{gathered}$$

Substitute $8.3145J/mol\cdot K$ for R, $2mol$for n,$300K$ for T , $1.01\times {10}^5\text{\hspace{0.17em}Pa}$for p into the above equation,

$\begin{array}{c}V=\frac{(2\times 8.31\times 300)}{1.01\times {10}^5}\\ =0.049{\text{\hspace{0.17em}m}}^3\end{array}$

For an adiabatic process,

$p{V}^{\gamma }=a$

Substitute$1.01\times {10}^5\text{\hspace{0.17em}Pa}$ for p, $0.049{\text{\hspace{0.17em}m}}^3$for V and $1.4$for$\gamma$ into the above equation,

$\begin{array}{c}\text{a}=1.01\times {10}^5\times {0.049}^{1.4}\\ \approx 1.5\times {10}^3{\text{N.m}}^2\end{array}$

Therefore the value of a is $1.5\times {10}^3{\text{Nm}}^2$.