How does enthalpy vary with pressure? Doesn't it increase at higher pressure?
I think the confusion is in considering the variation of
I calculated the
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 (
#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
After some derivation, we would have:
#color(blue)(((delH)/(delP))_T = T((delV)/(delT))_P - V)#
This says that the variation in
- 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:
#((delV)/(delT))_P = (del)/(delT)[(nRT)/P]_P = (nR)/P#
This was one possibility for what you meant, but it's likely not what you were thinking.