An ideal diatomic gas, with rotation but no oscillation, undergoes an adiabatic compression. Its initial pressure and volume are$\mathbf{1}\mathbf{.}\mathbf{20}\mathit{a}\mathit{t}\mathit{m}$and$\mathbf{0}\mathbf{.}\mathbf{200}{\mathit{m}}^{\mathbf{3}}$. Its final pressure is$\mathbf{2}\mathbf{.}\mathbf{40}\mathit{a}\mathit{t}\mathit{m}$. How much work is done by the gas?

• Initial pressure is$P_i=1.20\text{\hspace{0.17em}atm}$
• Initial volume is$V_i=0.200{\text{\hspace{0.17em}m}}^3$

• Final pressure is$P_f=2.40\text{\hspace{0.17em}atm}$

In case of adiabatic process,

$P{V}^{\gamma }=Constant$

Here P is the pressure, V is the volume and$\gamma$ is the ratio specific heat capacity at constant pressure to the specific heat capacity at constant volume.

$\gamma =\frac{C_P}{C_V}$

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

Work done in case of adiabatic process is given by,

$W=\frac{P_i{V}_i-P_f{V}_f}{\gamma -1}$

But first, we have to findusing the relation:

$P_i{V}_i^{\gamma }=P_f{V}_f^{\gamma }$

Substitute $1.20atm$for$P_i$ , $0.200m^3$for$V_i$ , $2.40atm$ for $P_f$and $1.4$for$\gamma$ into the above equation,

$\begin{array}{c}1.20\times {0.200}^{1.4}=2.40\times V_f^{1.4}\\ V_f=0.1219{\text{\hspace{0.17em}m}}^3\end{array}$

The expression for work done in case of adiabatic process is given by,

$W=\frac{P_i{V}_i-P_f{V}_f}{\gamma -1}$

Substitute$1.20\times 1.0135\times {10}^{5}Pa$ for $P_i$,$0.200m^3$ for$V_i$ , $2.40\times 1.0135\times {10}^{5}Pa$ for $P_f$, $0.1219{\text{\hspace{0.17em}m}}^3$for$V_f$ ,$1.4$for$\gamma$ into the above equation,

$\begin{array}{c}W=\frac{(1.2\times 1.0135\times {10}^5\times 0.2)-(2.40\times 1.0135\times {10}^5\times 0.1219)}{1.4-1}\\ =-13317.39\text{J}\\ =-1.33\times {10}^4\text{J}\end{array}$

Therefore work done by the gas is $-1.33\times {10}^4\text{J}$.