Prove that the work done in an adiabatic process$i\to f\mathrm{is}{W}_s\equiv (p_f{V}_f-p_i{V}_i(1-\lambda )$

Given : Work is done in an Adiabatic process

Theory used :

An adiabatic process is defined as a process in which no heat transfer takes place. This does not mean that the temperature is constant, but rather that no heat is transferred into or out from the system.

The following will be delivered as part of the work:

$W={\int }_{V_1}^{V_2}pdV$

As a result, we must express the pressure in terms of volume in order to conduct the integration. We get $p_1{V_1}^{\lambda }=K=p_2{V_2}^{\lambda }$ from the adiabatic process, where$K$is a constant. We can see that the pressure can be expressed as

$p=\frac{K}{V^{\lambda }}$

where $K$is determined by the beginning or final pressure and volume conditions. As a result, our integral is

$W={\int }_{V_1}^{V_2}K{V}^{-\lambda }dV$

After performing the integral, we have :

role="math" localid="1649068687977" $$\begin{gathered} W=p_1{V_1}^{\lambda }{\left(\frac{V^{\lambda +1}}{1-\lambda }\right)}_{V_1}^{V_2} \\ =\frac{p_1}{1-\lambda }({V_2}^{-\lambda +1}{V_1}^{\lambda }-{V_1}^{-\lambda +1}{V_1}^{\lambda } \end{gathered}$$

Let us now express the first volume to the power of $\lambda$as

${V_1}^{\lambda }=\frac{p_2}{p_1}{V_2}^{\lambda }$

which allows us to reduce the expression even more to

role="math" localid="1649068830061" $$\begin{gathered} W=\frac{p_1}{1-\lambda }({V_2}^{-\lambda +1}\frac{p_2}{p_1}{V_2}^{\lambda }-V_1) \\ \Rightarrow \boxed{W=\frac{{\mathrm{p}}_2{\mathrm{V}}_2-{\mathrm{p}}_1{\mathrm{V}}_1}{1-\mathrm{\lambda }}} \end{gathered}$$