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The differential relations for Helmholtz free energy are

$$S = \left.-\left( \frac{\partial F}{\partial T} \right) \right|_{V,N}$$ $$P = \left.-\left( \frac{\partial F}{\partial V} \right) \right|_{T,N}$$ $$\mu_i = \left.\left( \frac{\partial F}{\partial N_i} \right) \right|_{T,V,N_{j\ne i}}$$

These are valid for a single-component system. I am wondering how differential relations change for multi-component systems.

Consider a system with fixed volume $V$ and particle count $N$ in thermal equilibrium with a reservoir of temperature $T$. It is divided into two subsystems 1 and 2, each having its own thermodynamic quantities $U_i, S_i, P_i, V_i, \mu_i, N_i$. The total Helmholtz energy is

$$F_{1,2}=F_1+F_2=U_1+U_2-T(S_1+S_2)$$

and its differential is

$$dF_{1,2}=-(S_1+S_2)dT-P_1dV_1-P_2dV_2+\mu_1dN_1+\mu_2dN_2$$

I am not sure what are differential relations for this system. It seems like there is no way to find $S_1$ and $S_2$ individually. I can only calculate the total entropy:

$$S_1+S_2=-\left(\frac{\partial F_{1,2}}{\partial T}\right)_{V_1,V_2,N_1,N_2}$$

Not sure if they are correct but I can calculate the other quantities without issues:

$$P_1 = \left.-\left( \frac{\partial F_{1,2}}{\partial V_1} \right) \right|_{T,V_2,N_1,N_2}$$ $$P_2 = \left.-\left( \frac{\partial F_{1,2}}{\partial V_2} \right) \right|_{T,V_1,N_1,N_2}$$ $$\mu_1 = \left.\left( \frac{\partial F_{1,2}}{\partial N_1} \right) \right|_{T,V_1,V_2,N_2}$$ $$\mu_2 = \left.\left( \frac{\partial F_{1,2}}{\partial N_2} \right) \right|_{T,V_1,V_2,N_1}$$

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  • $\begingroup$ If there is a physical separator like a piston separating the space into two parts, then you should be considering each single part separately. i.e. you can do just the system separately from the reservoir. Of course, you can also do both together, but it is not by the method you are going for. If there is no physical separator and both are allowed to mix, then the pressure and volume terms are one and the same, not separable into two. $\endgroup$
    – naturallyInconsistent
    Apr 23 at 2:56
  • $\begingroup$ I think you are confusing a two component system with a two compartment system. $\endgroup$
    – Chet Miller
    Apr 23 at 10:48
  • $\begingroup$ @ChetMiller What is the difference between these two? $\endgroup$
    – John Davies
    Apr 23 at 14:12
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    $\begingroup$ In thermodynamics, a two component system is a mixture or solution of two chemical species. A two. compartment system refers to two separate containers. $\endgroup$
    – Chet Miller
    Apr 23 at 16:19

1 Answer 1

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To find the individual entropies you need the caloric equation of state for the components $S_k=S_k(T_k, V_k, N_k)$, and the thermal equations of state $p_k=p_k(T_k, V_k, N_k)$ and $\mu_k=\mu_k(T_k, V_k, N_k)$ for $k=1,2$.

You know that in equilibrium $$T_1=T_2=T \\ p_1=p_2\\ \mu_1=\mu_2 \\ V_1+V_2=V_0 \\ N_1+N_2=N_0$$ This implies that for fixed $T,V_0,N_0$ you must find the solutions of the following pair of equations for $V_1$ and $N_1$ $$p_1(T,V_1,N_1)=p_2(T,V_0-V_1,N_0-N_1)\\ \mu_1(T,V_1,N_1)=\mu_2(T,V_0-V_1,N_0-N_1)$$ after which they can be substituted into the caloric equations $S_1(T,V_1, N_1)$ and $S_2(T,V_0-V_1, N_0-N_1)$ to get you the individual entropies.

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