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I'm getting frustrated finding my mistake, because I get the exact same answer as my professor, except for a constant. The problem is the following:

For a system, we know that $c_P = c + \frac{2aP}{RT^2}$ and $Pv = RT + > P(b-\frac{a}{RT})$, and that it undergoes a Joule-Kelvin process where the temperature halves. Find the change in entropy.

In order to find the change in entropy, which can be calculated as an integral in terms of $c_P(T,P)$ and $\alpha(T,P)$, I need to know the change in pressure, which I could find out by establishing that $\Delta h = 0$. So, since I need to know $h(P,T)$ I can write its differential: $$dh = \Big{(}\frac{\partial h}{\partial T}\Big{)}_P dT + \Big{(}\frac{\partial h}{\partial P}\Big{)}_T dP$$ And, knowing that $dh = Tds + vdP$, it follows: $$dh = \Bigg{(} T\Big{(}\frac{\partial s}{\partial P}\Big{)}_T + v\Bigg{)}dP + T\Big{(}\frac{\partial s}{\partial T}\Big{)}_P dT$$

From Gibbs' potential we can find an adequate Maxwell relation: $$\Big{(}\frac{\partial s}{\partial P}\Big{)}_T = -\Big{(}\frac{\partial v}{\partial T}\Big{)}_P$$ And therefore: $$dh = \Bigg{(} -T\Big{(}\frac{\partial v}{\partial T}\Big{)}_P + v\Bigg{)}dP + T\Big{(}\frac{\partial s}{\partial T}\Big{)}_P dT$$ $$dh = (v-Tv\alpha)dP + c_P dT$$

I have proceeded directly to calculate the integral: $$h=\int dh = \int \Big{(} c + \frac{2aP}{RT^2} \Big{)} dT + \int\Big{(} \frac{RT}{P}+b-\frac{a}{RT} - T(\frac{R}{P} + \frac{a}{RT^2}) \Big{)}dP$$

And my result is: $$h = cT +(b-\frac{4a}{RT})P + C$$ While my professor's is: $$h = cT +(b-\frac{2a}{RT})P + C$$

Where did I go wrong?

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  • $\begingroup$ $\int\frac{2aP}{RT^2} dT=\frac{2aP}{R}\int dTT^{-2}=-\frac{2aP}{RT}$ where $n=-2+1=-1$ using the power rule for integration. $\endgroup$ – Cinaed Simson Jun 29 at 3:07
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You integrated the PDE incorrectly. Integrating first with respect to T, we get: $$h=cT-\frac{2aP}{RT}+f(P)$$So, $$\frac{\partial h}{\partial P}=-\frac{2a}{RT}+\frac{df}{dP}=\frac{RT}{P}+b-\frac{a}{RT} - T(\frac{R}{P} + \frac{a}{RT^2})=b-\frac{2a}{RT}$$So, $$\frac{df}{dP}=b$$ So, $$f=bP+C$$

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  • $\begingroup$ Thanks a lot! I couldn't realize that last night. $\endgroup$ – Manuel Jun 29 at 14:32

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