# Adiabatic compression of liquid water and vapor in equilibrium [closed]

I'm trying to solve Exercise 6.16 in Garrod's Statistical Mechanics and Thermodynamics.

Here is what I've noticed so far. Since

$$V=n_lV_l+n_gV_g$$

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

$$H=n[C_{p,l}(T-T_r)+V_l(p-p_r)]+n_gL(T)$$.

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?

• 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 $V_l=n_lv_l$ – 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 $$dn_g$$.
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, $$dh_l=C_{pl}dT$$.
• 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. – Chet Miller Dec 6 '19 at 12:48