# Finding the Enthalpy of an Ideal Gas given internal energy

I am trying to find a formula for tying things up. Given an ideal gas like helium or nitrogen gas (diatomic) how can I find its enthalpy simply given internal energy?

I remember it was once taught in thermodynamics class but I cannot find the reference material anymore; I've tried searching on the net too.

I tried tying things up though:

PV = mRT
H = U + PV
H = U + mRT


Is this right? Am I missing anything? is there another more elegant way?

The internal energy of a system is directly proportional to its temperature. Formally, $$E_{sys}=\frac{3}{2}RT.$$ You could then note that $$PV=nRT=H_{sys}-E_{sys},$$ or $$H_{sys}=RT\bigg(\frac{3}{2}+n\bigg)$$ or, identically, $$H_{sys}=\frac{3}{2}RT +PV.$$ Your method should work, however, this is in my opinion a more "elegant" solution.
• You missed the $n$ in your $E_{sys}$ expression. Which will change your expression for $H_{sys}$ slightly. Commented Jul 5, 2015 at 21:12
Ideal gas equation of state - $$PV = Nk_bT$$ For ideal gas (can be calculated directly from entropy (Sakur-Tetrode) or via equipartition theorem) - $$E = \frac{3}{2}Nk_bT=\frac{3}{2}PV$$ thus - $$H = E+PV=\frac{3}{2}PV+PV=\frac{5}{2}PV=\frac{5}{2} \frac{2}{3}E=\frac{5}{3}E$$