A van der Waals gas obeys the equation of sta | vedclass

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(C) For a quasistatic adiabatic process,the heat exchange $dQ = 0$.From the first law of thermodynamics,$dU = dQ - dW$,so $dW = -dU$.Given $U = CT - \

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

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(C) For a quasistatic adiabatic process,the heat exchange $dQ = 0$.From the first law of thermodynamics,$dU = dQ - dW$,so $dW = -dU$.Given $U = CT - \frac{n^2 a}{V}$,the differential change in internal energy is $dU = C dT + \frac{n^2 a}{V^2} dV$.Also,the work done is $dW = p dV$.Equating $dW = -dU$,we get $p dV = -(C dT + \frac{n^2 a}{V^2} dV)$.Substituting the van der Waals equation $p = \frac{nRT}{V-nb} - \frac{n^2 a}{V^2}$ into the equation above:$(\frac{nRT}{V-nb} - \frac{n^2 a}{V^2}) dV = -C dT - \frac{n^2 a}{V^2} dV$.The term $-\frac{n^2 a}{V^2} dV$ cancels out from both sides:$\frac{nRT}{V-nb} dV = -C dT$.Rearranging the variables,we get $\frac{dV}{V-nb} = -\frac{C}{nR} \frac{dT}{T}$.Integrating both sides: $\int \frac{dV}{V-nb} = -\frac{C}{nR} \int \frac{dT}{T}$.This yields $\ln(V-nb) = -\frac{C}{nR} \ln T + 	ext{constant}$.Rearranging gives $\ln(V-nb) + \ln(T^{C/nR}) = 	ext{constant}$.Thus,$T^{C/nR} \cdot (V-nb) = 	ext{constant}$.

$A$ van der Waals gas obeys the equation of state $\left(p+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$. Its internal energy is given by $U=C T-\frac{n^2 a}{V}$. The equation of a quasistatic adiabat for this gas is given by

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