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I have a N particle system for which I know the entropy as a function of temperature T and the quasistatic work as a function of V. From this I should compute the

a)Helmholtz free energy b)and then out of this the pressure c)And last the work done under any temperature

The work done under quasistatic expansion from $V_{0}$ to $V$ ($V_{0}<V$) at fixed temperature $T_{0}$: $$ \Delta W = Nk_{b}T_{0}ln\left( \frac{V}{V_{0}} \right) $$ And the entropy is given by:

$$ S=Nk_{b}\frac{V_{0}}{V}\left ( \frac{T}{T_{0}} \right )^{a} $$

with $a=const$,$V_{0}=const$ and $T_{0}=const$ for the entropy equation.

To start with I would use $$ S=-\frac{\partial F}{\partial T} $$

by integration I obtain:

$$ F(T,V,N)=-\frac{Nk_{b}V_{0}}{(a+1)VT_{0}^{a}}T^{a+1}+f(V) $$

Since I know the work done I can just insert the given work function for f(V) $$ F(T,V,N)=-\frac{Nk_{b}V_{0}}{(a+1)VT_{0}^{a}}T^{a+1}+Nk_{b}T_{0}ln\left( \frac{V}{V_{0}} \right) $$

The pressure is given by the derivative of F with respect to V

$$ P(V,T,N)=-\frac{\partial F}{\partial V}=-\frac{Nk_{b}V_{0}}{(a+1)V^{2}T_{0}^{a}}-\frac{Nk_{b}T_{0}}{V} $$

This would result in a negative pressure what does not make any sense but i can't find my error

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1 Answer 1

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The equation $$ F(T,V,N)=-\frac{Nk_{b}V_{0}}{(a+1)VT_{0}^{a}}T^{a+1}+f(V) $$is correct. The pressure is given by: $$P(T,V,N)=-\frac{\partial F}{\partial V}=-\frac{Nk_{b}V_{0}}{(a+1)V^2T_{0}^{a}}T^{a+1}-\frac{df}{dV}$$From the quasistatic work equation at constant temperature $T_0$, we know that: $$P(T_0,V,N)=\frac{Nk_bT_0}{V}$$Therefore,$$P(T_0,V,N)=\frac{Nk_bT_0}{V}=-\frac{Nk_{b}V_{0}T_0}{(a+1)V^2}-\frac{df}{dV}$$ Just integrate this ODE to get f(V)

ADDENDUM From the previous equation, it follows that $$\frac{df}{dV}=-\frac{Nk_bT_0}{V}-\frac{Nk_{b}V_{0}T_0}{(a+1)V^2}$$ If I substitute this into the equation for the pressure, I obtain: $$P(T,V,N)=\frac{Nk_{b}V_{0}T_0}{(a+1)V^2}\left[1-\left(\frac{T}{T_0}\right)^{a+1}\right]+\frac{Nk_bT_0}{V}$$ If I integrate the differential equation for f, I obtain: $$f=-Nk_bT_0\ln{(V/V_0)}+\frac{Nk_bT_0}{(a+1)}\frac{V_0}{V}$$ So, $$ F(T,V,N)=-Nk_bT_0\ln{(V/V_0)}+\frac{Nk_bT_0}{(a+1)}\frac{V_0}{V}\left[1-\left(\frac{T}{T_0}\right)^{a+1}\right] $$

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  • $\begingroup$ Thank you!! But the sign of the first term for the pressure should be a minus. If I make the derivative of $V^{-1}$ I obtain $-V^{-2}$ therefore the sign of the first term of the pressure would be minus the same as I have written above. Or am I going crazy? ;-) $\endgroup$
    – zodiac
    Oct 1, 2018 at 19:06
  • $\begingroup$ And one has to determine the f(V) for the Helmholtz free energy since this is asked. If I determine the pressure like you suggest I would run into the same problem again like it should be solved in this exercise but with an undetermined $\hat{f}(T)$ $\endgroup$
    – zodiac
    Oct 1, 2018 at 19:14
  • $\begingroup$ Please recheck your algebra. Also, notice that my final two equations involve the pressure at $T_0$, not at arbitrary T. And also notice that my final equation is a function only of V, so there is no f(T) required. I don't feel that you read over what I did carefully enough. If you would like me to complete the solution, I will do that. But I'm sure, once you have better digested what I have done, you can easily do that yourself. $\endgroup$ Oct 1, 2018 at 19:26
  • $\begingroup$ I am very sorry I don't mean to offend you. But I have to do a derivative of the following form $\frac{\partial- KV^{-1}}{\partial V}=KV^{-2}$ and then I have to take a minus again because I have to take the negative derivative. Like you wrote in your answer. This gives me in total -KV^{-2}. I checked this now on wolfram alpha what gives me the same. I don't get it I am sorry. $\endgroup$
    – zodiac
    Oct 1, 2018 at 19:46
  • $\begingroup$ Oops. You are absolutely right. My error. Thanks. I have made the correction. But, the rest of what I did is correct. $\endgroup$ Oct 1, 2018 at 19:55

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