Question

I have been struggling to understand how the molecules behave during the Joule–Thomson effect. I would love to get some help on this concept. Here is what I got so far.

Under adiabatic and isenthalpic conditions, when a gas expand from high pressure to low pressure, depending on the initial pressure, initial temperature, and the molecular properties, the gas can either experience cooling or heating.

Specifically, if the gas was initially in the state where attractive intermolecular forces were dominant (typically at regions of moderate temperature and low pressure), as it expand, the kinetic energy of molecules are used to overcome the attraction reach greater separation. Also, the internal energy is overall decreased as the gas need to do work to expand. Thus the temperature of the gas decreases.

If repulsive forces were dominant (at regions of very high temperature or high pressure), as the gas expand, the repulsion kind of pushes(?) the molecules to reach greater distance. In other words, the potential energy is converted to kinetic energy here. And here although internal energy may be decreased to provide work for expansion, the molecule would have gained more kinetic energy from the repulsive forces such that the temperature of the molecules increases.

My question is, how does the system being isenthalpic play a role in this effect? I get that we need the adiabatic condition so that only the internal energy is being used to provide the work required for expansion. But what about enthalpy? How is isenthalpic + adiabatic expansion different from just adiabatic expansion?

Additionally, I would appreciate if someone could also explain how pV from the enthalpy formula H=U+pV, changes during this expansion.

Answer 1

The Joule–Thomson effect is an isenthalpic adiabatic process where a gas expands through a valve or porous plug without heat exchange. The enthalpy of the gas remains constant, as changes in internal energy are balanced by changes in the pV term, leading to cooling or heating depending on intermolecular forces.

Step-by-step explanation:

The Joule–Thomson effect describes the temperature change of a real gas or liquid when it is forced through a valve or porous plug while keeping it insulated from heat exchange (adiabatic). This process is isenthalpic, meaning that the enthalpy of the system (H = U + pV) remains constant.

Since the system is adiabatic, there is no heat exchange (Q=0), and thus the first law of thermodynamics simplifies to ∆U = -W, where W is the work done by the gas. The work done by the expanding gas can either increase its potential energy due to the intermolecular forces overcoming or reduce its kinetic energy, resulting in cooling or heating.

During the Joule–Thomson expansion, the volume (V) increase accumulated by the gas as pressure (p) drops is the critical factor for understanding enthalpy. Since the process is isenthalpic, any internal energy change (∆U) is offset by a change in the pV term, ensuring that the overall enthalpy (H) remains constant.

If attractive forces are strong, the expansion results in cooling as work is done to overcome these forces, decreasing internal kinetic energy. However, if repulsive forces predominate at high temperatures or pressures, these can impart additional kinetic energy upon expansion, leading to an increase in temperature.

Distinguishing isenthalpic adiabatic expansion from adiabatic expansion is crucial because in a general adiabatic expansion that is not isenthalpic, work done on the gas can change the internal energy, and by extension, enthalpy. However, in an isenthalpic adiabatic process, the enthalpy remains constant despite the adiabatic conditions.