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?


The enthalpy includes the energy required to move the atmosphere out of the way of the system. That action requires the particles to do work on the exterior atmosphere. which you can idealize as each particle bouncing off of and transferring momentum to an invisible boundary around the system until the pressure is equalized.

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).

Is this the type of interpretation you're looking for?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.