# How does enthalpy vary with pressure? Doesn't it increase at higher pressure?

Feb 23, 2017

I think the confusion is in considering the variation of $\Delta H$ with pressure, vs. the variation of $H$ with pressure.

Variation in $\boldsymbol{\Delta H}$ with pressure at a constant temperature

I calculated the $\Delta {H}_{\text{vap}}$ from here at various pressures in PSI:
http://www.tlv.com/global/TI/calculator/steam-table-pressure.html

Here is a graph I constructed in Excel plotting data acquired from the above calculations for the variation in the enthalpy of vaporization of water at 10 - 2000 PSI: Enthalpy of vaporization ($\Delta {H}_{v a p}$) is the heat required at a given temperature and pressure to turn a liquid into a gas. If $\Delta {H}_{v a p}$ decreases, then vaporization is easier. That should make sense at higher pressure because of Henry's Law:

$s = {k}_{H} P$

The solubility of a gas increases with a higher vapor pressure of gas above the solution.

If you increase the pressure of the atmosphere, the gas becomes more soluble in solution (because it's pushed into the solution), so its vapor pressure above the solution increases.

Since higher vapor pressure means easier boiling, it also means easier vaporization and thus lower $\Delta {H}_{v a p}$. So, $\Delta {H}_{v a p}$ decreases at higher pressure for a constant temperature.

Variation in $\boldsymbol{H}$ with pressure at a constant temperature

After some derivation, we would have:

$\textcolor{b l u e}{{\left(\frac{\partial H}{\partial P}\right)}_{T} = T {\left(\frac{\partial V}{\partial T}\right)}_{P} - V}$

This says that the variation in $H$ with pressure at constant temperature is dependent on:

• The variation in volume due to temperature changes at constant pressure
• The volume of the system

Let's say we have an ideal gas. Then:

${\left(\frac{\partial V}{\partial T}\right)}_{P} = \frac{\partial}{\partial T} {\left[\frac{n R T}{P}\right]}_{P} = \frac{n R}{P}$

Since $\frac{n R}{P}$ is positive, this partial derivative is always positive. In units of $\text{K}$ like temperature is in many equations, we thus have that $T {\left(\frac{\partial V}{\partial T}\right)}_{P}$ is always positive.

But $V = \frac{n R T}{P}$ for an ideal gas, so for an ideal gas, this quantity is zero. For real gases, the variation is small, but may be negative or positive.

This was one possibility for what you meant, but it's likely not what you were thinking.