The equation of state of n moles of a non-ide | vedclass

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(A) For an adiabatic process,the first law of thermodynamics states $dU = dQ + dW$. Since $dQ = 0$,we have $dU = dW = -p dV$.For a van der Waals gas,t

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$T(V-n b)^{R / C_V} = 	ext{constant}$

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(A) For an adiabatic process,the first law of thermodynamics states $dU = dQ + dW$. Since $dQ = 0$,we have $dU = dW = -p dV$.For a van der Waals gas,the internal energy $U$ depends on both $T$ and $V$. The differential change is $dU = C_V dT + \left(\frac{\partial U}{\partial V}ight)_T dV$.Using the thermodynamic relation $\left(\frac{\partial U}{\partial V}ight)_T = T \left(\frac{\partial p}{\partial T}ight)_V - p$,and from the equation of state $p = \frac{nRT}{V-nb} - \frac{n^2 a}{V^2}$,we find $\left(\frac{\partial p}{\partial T}ight)_V = \frac{nR}{V-nb}$.Thus,$\left(\frac{\partial U}{\partial V}ight)_T = T \left(\frac{nR}{V-nb}ight) - \left(\frac{nRT}{V-nb} - \frac{n^2 a}{V^2}ight) = \frac{n^2 a}{V^2}$.Substituting this into the energy equation: $C_V dT + \frac{n^2 a}{V^2} dV = -p dV$.$C_V dT + \frac{n^2 a}{V^2} dV = -\left(\frac{nRT}{V-nb} - \frac{n^2 a}{V^2}ight) dV$.$C_V dT = -\frac{nRT}{V-nb} dV$.Rearranging gives $\frac{dT}{T} = -\frac{nR}{C_V} \frac{dV}{V-nb}$.Integrating both sides: $\ln T = -\frac{nR}{C_V} \ln(V-nb) + 	ext{constant}$.$T(V-nb)^{nR/C_V} = 	ext{constant}$. Since $n$ is constant,this is equivalent to $T(V-nb)^{R/C_V} = 	ext{constant}$.

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