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


1 Answer 1


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, $$ F(T,V,N)=-Nk_bT_0\ln{(V/V_0)}+\frac{Nk_bT_0}{(a+1)}\frac{V_0}{V}\left[1-\left(\frac{T}{T_0}\right)^{a+1}\right] $$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.