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III. Long Answer Questions · Q16

Q.Derive the work done in an adiabatic process.

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Step 1. For a quasi-static adiabatic process, PVγ=constantPV^\gamma=\text{constant}, so P=constantVγP=\dfrac{\text{constant}}{V^\gamma}. The work done is W=∫ViVfP dV=constant∫ViVfV−γ dV=constant[V1−γ1−γ]ViVf.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 constant=PiViγ=PfVfγ\text{constant}=P_iV_i^\gamma=P_fV_f^\gamma) gives W=PiVi−PfVfγ−1.W=\dfrac{P_iV_i-P_fV_f}{\gamma-1}.

Step 3. Using the ideal gas law, PiVi=μRTiP_iV_i=\mu RT_i and PfVf=μRTfP_fV_f=\mu RT_f; substituting gives Wadia=μRγ−1(Ti−Tf).W_{adia}=\dfrac{\mu R}{\gamma-1}(T_i-T_f). …

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