# Physical interpretation of the enthalpy of a free gas

The enthalpy $H$ of a thermodynamic system is defined as $$H = U + pV$$

where $U$ is the internal energy, $p$ is the pressure and $V$ is the volume of the system.

For the ideal gas, we have $$U = \frac{3}{2}NkT$$ which can be interpreted as $N$ particles with mean energy of $\frac{3}{2}kT$.

If we use the ideal gas law, we can calculate $$H = \frac{3}{2}NkT + pV = \frac{3}{2}NkT + NkT = \frac{5}{2}NkT$$

My question: What is the physical significance of this? Can it be interpreted on the level of the single particles as I have done for the internal energy above?

Thus, when you heat an ideal gas at constant pressure, more energy is required to obtain the same mean energy of $\frac{3}{2}kT$ per particle (relative to heating it at constant volume). In fact, the energy to expand the system boundary against the external atmosphere is higher by a factor of $\frac{5}{3}$, and an average energy of $\frac{2}{3}kT$ per ideal gas particle passes through momentum transfer into the environment, heating it (negligibly).