91Ó°ÊÓ

• Initial pressure is$P_i=1.20\text{ a³Ù³¾}$
• Initial volume is$V_i=0.200{\text{″¾}}^3$

• Final pressure is$P_f=2.40\text{ a³Ù³¾}$

In case of adiabatic process,

$P{V}^{\mathrm{γ}}=Constant$

Here P is the pressure, V is the volume and$\mathrm{γ}$ is the ratio specific heat capacity at constant pressure to the specific heat capacity at constant volume.

$\mathrm{γ}=\frac{C_P}{C_V}$

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,

$W=\frac{P_i{V}_i−P_f{V}_f}{\mathrm{γ}−1}$

But first, we have to findusing the relation:

$P_i{V}_i^{\mathrm{γ}}=P_f{V}_f^{\mathrm{γ}}$

Substitute $1.20\text{ }atm$for$P_i$ , $0.200\text{ }{m}^3$for$V_i$ , $2.40\text{ }atm$ for $P_f$and $1.4$for$\mathrm{γ}$ into the above equation,

$\begin{array}{c}1.20×{0.200}^{1.4}=2.40×V_f^{1.4}\\ V_f=0.1219{\text{″¾}}^3\end{array}$

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

$W=\frac{P_i{V}_i−P_f{V}_f}{\mathrm{γ}−1}$

Substitute$1.20×1.0135×{10}^5\text{ }Pa$ for $P_i$,$0.200\text{ }{m}^3$ for$V_i$ , $2.40×1.0135×{10}^5\text{ }Pa$ for $P_f$, $0.1219{\text{″¾}}^3$for$V_f$ ,$1.4$for$\mathrm{γ}$ into the above equation,

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

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