I'm trying to solve Exercise 6.16

Garrods Statmech Exercise 6.16 in Garrod's Statistical Mechanics and Thermodynamics.

Here is what I've noticed so far. Since


where $n_{(\cdot)}, V_{(\cdot)}$ are molar numbers and molar volumes of liquid and gas, respectively,

I can write

$dV=V_ldn_l+V_gdn_g+n_gdV_g=(V_g-V_l)dn_g+n_gdV_g\approx V_gdn_g+n_gdV_g$.

I'm curious about if I can express $dn_g$ and $dV_g$ in terms of $n_g, n_l, V_g, V_l, p$ and $L(T)$, the last of which is the latent heat of the phase transition $l\rightarrow g$.

First, $dV_g=d(\frac{RT}{p})=\frac{R}{p}dT-\frac{RT}{p^2}dp$, so that expressing $dV_g$ reduces to that of $dT$.

Q1. At this point can I use the Clausius–Clapeyron relation $\frac{dp}{dT}=\frac{L}{T(V_g-V_l)}$? If so, how is it justified?

On the other hand,

$H_l=n_l[C_{p,l}(T-T_r)+V_l(p-p_r)]\;$ and $\;H_g=n_g[C_{p,l}(T-T_r)+V_l(p-p_r)]+n_gL(T)$

where the subscript $r$ stands for a fixed reference state and $C_{p,l}$ is the molar heat capacity of liquid water at a fixed pressure. Now the total enthalpy is given by


So in principle, I can equate the differential of this expression with $Vdp$ and see what it says about $dn_g$. But the calculation will be complicated, and I don't see how to reduce $dC_{p,l}$ in terms of other quantities.

Q2. What can I expect about $dn_g$? Are there better methods than my attempt?

  • $\begingroup$ You used the symbols $V_l$ and $V_g$ both for the molar volumes and for the actual volumes. You should use lower case for the molar volumes. So $V_g=n_gv_g$ and $V_l=n_lv_l$ $\endgroup$ – Chet Miller Dec 5 '19 at 21:54

In my judgment, you have the right idea, but I would be using lower case for molar volumes and enthalpies. For the initial condition, I would introduce subscript zeros to represent the undisturbed quantities: $$V_{go}=n_{g0}v_g(T_0, p_0)$$ $$V_{lo}=n_{l0}v_l(T_0, p_0)$$$$V_0=V_{go}+V_{l0}=n_{g0}v_g(T_0, p_0)+n_{l0}v_l(T_0, p_0)$$Also, you are correct in writing: $$dV=v_{go}dn_{g}+n_{go}dv_g$$where $$dv_g=\frac{\partial v_g}{\partial T}dT+\frac{\partial v_g}{\partial p}dp$$ You decided that it would be OK to use the ideal gas law at this point. So all that would be needed now would be to determine $dn_g$.

It was correct to use the Clausius-Clapeyron equation to get the change in temperature with pressure, since you will be moving differentially along the vapor-liquid equilibrium line. This eliminates dT from the equations. But, don't forget that vl is negligible compared to vg, and that vg can be determined from the ideal gas law.

And you were correct to use the enthalpy change to get $dn_g$. But, I would have set it up a little differently: $$dH=V_0dP=n_{l0}dh_l+n_{g0}dh_g+(h_{g0}-h_{l0})dn_g$$ with $$dh_g=\frac{\partial h_g}{\partial T}dT+\frac{\partial h_g}{\partial p}dp$$and $$dh_l=\frac{\partial h_l}{\partial T}dT+\frac{\partial h_l}{\partial p}dp$$In the limit of an ideal gas, $dh_g=C_{pg}dT$ and, neglecting the pressure dependence for the liquid, $dh_l=C_{pl}dT$.

This is all basically your attempt. In my judgment, there is not a better way of approaching this.

  • $\begingroup$ Thank you. Both your careful notation and elaboration on my attempt clear things up. Could you explain more on why the system will move along the equilibrium line? Is it because escaping from the line requires change in entropy? $\endgroup$ – Sea watcher Dec 6 '19 at 5:54
  • $\begingroup$ No. In the final state, you are going to still have liquid water and water vapor at equilibrium, although at a slightly different temperature and pressure. So you have moved along the equilibrium line. Incidentally, a slightly more direct route to the solution would be to just set $dS=n_l ds_l+n_gds_g+(s_g-s_l)dn_g$ equal to zero, with $ds=\frac{C_p}{T}dT-\left(\frac{\partial v}{\partial T}\right)_pdp$ for either the liquid or the gas. $\endgroup$ – Chet Miller Dec 6 '19 at 12:48
  • $\begingroup$ Thank you very much! I've learned a lot from your comments. $\endgroup$ – Sea watcher Dec 6 '19 at 13:31

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