The heat capacity of one mole of an ideal gas | vedclass

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Question

(A) For a reversible adiabatic process,the first law of thermodynamics is $dQ = dU + dW = 0$,so $dU = -dW$.Given $dU = C_V dT$ and $dW = P dV = \frac{

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$TV^{3/2} e^{aRT} = 	ext{constant}$

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(A) For a reversible adiabatic process,the first law of thermodynamics is $dQ = dU + dW = 0$,so $dU = -dW$.Given $dU = C_V dT$ and $dW = P dV = \frac{RT}{V} dV$ (for one mole of ideal gas).Thus,$C_V dT = -\frac{RT}{V} dV$.Substituting $C_V = \frac{3R(1 + aRT)}{2}$:$\frac{3R(1 + aRT)}{2} dT = -\frac{RT}{V} dV$.Rearranging the terms:$\frac{3(1 + aRT)}{2T} dT = -\frac{R}{V} dV$.$\frac{3}{2} (\frac{1}{T} + aR) dT = -\frac{R}{V} dV$.Integrating both sides:$\int \frac{3}{2} (\frac{1}{T} + aR) dT = -\int \frac{R}{V} dV$.$\frac{3}{2} (\ln T + aRT) = -R \ln V + 	ext{constant}$.$\ln T^{3/2} + \frac{3}{2} aRT = -\ln V^R + 	ext{constant}$.$\ln (T^{3/2} V^R) + \frac{3}{2} aRT = 	ext{constant}$.Since $R$ is a constant,we can write $T^{3/2} V e^{aRT} = 	ext{constant}$ (adjusting for the power of $R$ in the exponent).Comparing with the given options,the correct form is $TV^{3/2} e^{aRT} = 	ext{constant}$.

The heat capacity of one mole of an ideal gas is found to be $C_V = \frac{3R(1 + aRT)}{2}$,where $a$ is a constant. The equation obeyed by this gas during a reversible adiabatic expansion is

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