# Vaporization at constant volume

my question is the following: how much energy does it take to evaporate a liquid at constant volume?

For constant pressure, it’s very easy: Q = ΔHvap (Q in J/mol). However, the enthalpy of vaporization con only be used at constant pressure and isothermal evaporation, so what about isochoric evaporation, where both pressure (corresponding to vapor pressure) and temperature increase along the liquid-gas phase boundary during the heating process? Yet I still must rely on ΔHvap, because it's the only tabulated value available most of times.

More in detail, let us imagine that we have a sealed container with liquid in it and that we supply heat until all liquid has become vapor. For simplicity, let us forget that the liquid initially occupied part of the available volume.

I have thought of two methods but I am not sure whether they make sense:

One approach is to perform two transformations to go from the initial state to the final one. First, isothermal evaporation at constant volume: Q = ΔU = ΔHvap - pΔV. Second, temperature increase: Q = ΔU = cvΔT.

A second approach involves three steps: First, evaporation at constant temperature and pressure: Q = ΔHvap. Second, isobaric temperature increase: Q = cpΔT. Third, isothermal compression: Q = 0? The work is given back to the external environment in form of heat.

Any ideas or clarifications? Thank you!

• The answer is that you would use $\Delta U$ instead of \Delta H\$ since it has the equivalent relationship to heat at constant volume as enthalpy does under conditions of constant pressure. Oct 27, 2023 at 10:45
• You can't neglect the condition that water only takes up part of the volume. Do you know how to apply the first law of thermodynamics to this system to determine the amount of liquid and vapor in the system at any equilibrium state, and the amount of heat that must be added to transition from one saturated thermodynamic mixture at one state to a saturated thermodynamic mixture at a higher temperature? Oct 27, 2023 at 10:57
• Key clarification: I must rely on the enthalpy of vaporization! Because it's the only tabulated value available. Good luck finding dUvap...
– Peye
Oct 27, 2023 at 11:56
• @Chez Miller: how would you do that then? My problem is exactly that I cannot solve it myself...
– Peye
Oct 27, 2023 at 12:01
• @Chez Miller. Moving along the equilibrium line and determining p and T at any point of the process is no problem: p(T)V = nRT, where p(T) is the vapor pressure at the temperature T. Also, I'm going all the way to the point where there is no liquid left, so the problem is even simpler from this point of view because the moles you had at the beginning in the liquid phase are later all in the vapor phase. From the first law of thermodynamics: dQ = dU since the volume does not change. However, finding an expression for dU using accessible thermodynamic values such as dHvap is my issue.
– Peye
Oct 27, 2023 at 12:16

Let m be the mass of water in the container, V be the volume of the container, and x be the fraction of the water that is vapor. Then for the volume to match at temperature T: $$m(1-x)v_L+mxv_V=V$$ or $$x=\frac{(V/m)-v_L}{v_V-v_L}$$where $$v_L$$ and $$v_V$$ are the specific volume of saturated liquid water and saturated water vapor, respectively, at temperature T. This equation gives the mass fraction of saturated vapor. The data on specific volume of the saturated liquid and vapor can be found in the steam tables.
At temperature T, the internal energy of the container contents can now be determined: $$U(T)=m(1-x)u_L+mxu_V$$where $$u_L$$ and $$u_V$$ are the specific internal energies of the saturated liquid and saturated vapor, respectively. The specific internal energies of the saturated liquid and the saturated vapor can be found in the steam tables.
To get the amount of heat Q required to raise the temperature of the constant volume system from $$T_1$$ to $$T_2$$ and evaporate the corresponding amount of liquid, we simply write $$Q=U(T_2)-U(T_1)$$