0
$\begingroup$

I was reading the Wikipedia page about Helmholtz free energy and there is a sentence I don't get:

We see that the total amount of work that can be extracted in an isothermal process is limited by the free-energy decrease, and that increasing the free energy in a reversible process requires work to be done on the system. If no work is extracted from the system, then $$\Delta F \leq 0$$

and thus for a system kept at constant temperature and volume and not capable of performing electrical or other non-''PV'' work, the total free energy during a spontaneous change can only decrease.

This result seems to contradict the equation dF = −S dT − P dV, as keeping T and V constant seems to imply dF = 0, and hence F = constant. In reality there is no contradiction: In a simple one-component system, to which the validity of the equation dF = −S dT − P dV is restricted, no process can occur at constant T and V, since there is a unique P(T, V) relation, and thus T, V, and P are all fixed.

I really don't understand why there is no contradiction. For a one-component system in contact with heat bath, why there is no process that occurs at constant T and V? What is "unique P(T, V) relation"?

$\endgroup$

1 Answer 1

3
$\begingroup$

Because the question assumes that there are only two thermodynamic variables of that particular system setup, $T$ and $V$. If, for example, you have also allowed matter be exchanged then you would have three variables with the corresponding Gibbs relations:$$dE= TdS-pdV+\mu dn$$ where now $\mu$ is the chemical potential and $n$ denotes mass, or rather molar mass. The corresponding free energy is $$dF=-SdT-pdV+\mu dn.$$ In this form the functions are $$F=F(T,V,n)\\ p=p(T,V,n)\\ \mu=\mu(T,V,n)$$ and if you keep $V$ constant and $T$ constant by attaching it to a reservoir then the system moves spontaneously by absorbing or rejecting $dn$ so that $$dF=\mu dn \le 0.$$

Here is another example. Take two pieces of matter, $T_1, V_1, p_1$ and $T_2,V_2, p_2$, put them in a common container of fixed volume $V_1+V_2=V=const$ and let them connected to a single thermal reservoir at a fixed temperature $T$.

Assume that each piece of matter internally have such high thermal conductivity that during the process they are always homogeneous both in temperature and in pressure, of course they may differ from each other. Now we have not two, but four variables, with two thermal equations of state $p_1=p_1(T_1,V_1)$ and $p_2=p_2(T_2,V_2)$.

The free energies are additive when there is a common temperature, this is because both the internal energies and the entropies are unconditionally additive: $F_1+F_2 = E_1-T_1S_1 +E_2-T_2S_2=E-TS$ for $E=E_1+E_2$, $S=S_1+S_2$ and $T=T_1=T_2.$

The joint system is $$dF=dF_1+dF_2 = -S_1dT-p_1dV_1 -S_2dT-p_2dV_2.$$ Now take the constraint into account $dV=0$ $$dF=dF_1+dF_2 = -(S_1+S_2)dT+(p_2-p_1)dV_1.$$ Even if the individual temperatures are always at the reservoir temperature, so that $dT_1=dT_2=dT$ the volume $dV_1$ can still be variable and will change so that $(p_2-p_1)dV_1\le 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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