Q.Derive the work done in an adiabatic process.

Step 1. For a quasi-static adiabatic process, $PV^\gamma=\text{constant}$, so $P=\dfrac{\text{constant}}{V^\gamma}$. The work done is $W=\int_{V_i}^{V_f}P\,dV=\text{constant}\int_{V_i}^{V_f}V^{-\gamma}\,dV=\text{constant}\left[\dfrac{V^{1-\gamma}}{1-\gamma}\right]_{V_i}^{V_f}.$

Step 2. Evaluating and simplifying (using $\text{constant}=P_iV_i^\gamma=P_fV_f^\gamma$) gives $W=\dfrac{P_iV_i-P_fV_f}{\gamma-1}.$

Step 3. Using the ideal gas law, $P_iV_i=\mu RT_i$ and $P_fV_f=\mu RT_f$; substituting gives $W_{adia}=\dfrac{\mu R}{\gamma-1}(T_i-T_f).$ …

Integrate $P=\text{const}/V^\gamma$ over $dV$, simplify using $P_iV_i^\gamma=P_fV_f^ …

• Sign confusion between $(T_i-T_f)$ and $(T_f-T_i)$ -- check against the physical expectation (gas cools on adiabatic expans …