Internal pressure is a measure of how theinternal energy of a system changes when it expands or contracts at constanttemperature. It has the same dimensions aspressure, theSI unit of which is thepascal.

Internal pressure is usually given the symbol${\displaystyle {\pi }_T}$. It is defined as apartial derivative of internal energy with respect tovolume at constant temperature:

${\displaystyle {\pi }_T={\left(\frac{\mathrm{\partial }U}{\mathrm{\partial }V}\right)}_T}$

Internal pressure can be expressed in terms of temperature, pressure and their mutual dependence:

${\displaystyle {\pi }_T=T{\left(\frac{\mathrm{\partial }p}{\mathrm{\partial }T}\right)}_V-p}$

This equation is one of the simplestthermodynamic equations. More precisely, it is a thermodynamic property relation, since it holds true for any system and connects the equation of state to one or more thermodynamic energy properties. Here we refer to it as a "thermodynamic equation of state."

Thefundamental thermodynamic equation states for theexact differential of theinternal energy:

${\displaystyle \mathrm{d}U=T\mathrm{d}S-p\mathrm{d}V}$

Dividing this equation by${\displaystyle \mathrm{d}V}$ at constant temperature gives:

${\displaystyle {\left(\frac{\mathrm{\partial }U}{\mathrm{\partial }V}\right)}_T=T{\left(\frac{\mathrm{\partial }S}{\mathrm{\partial }V}\right)}_T-p}$

And using one of theMaxwell relations:

${\displaystyle {\left(\frac{\mathrm{\partial }S}{\mathrm{\partial }V}\right)}_T={\left(\frac{\mathrm{\partial }p}{\mathrm{\partial }T}\right)}_V}$, this gives

${\displaystyle {\pi }_T=T{\left(\frac{\mathrm{\partial }p}{\mathrm{\partial }T}\right)}_V-p}$

In aperfect gas, there are nopotential energy interactions between the particles, so any change in the internal energy of the gas is directly proportional to the change in thekinetic energy of its constituent species and therefore also to the change in temperature:

${\displaystyle \mathrm{d}U\propto \mathrm{d}T}$.

The internal pressure is taken to be at constant temperature, therefore

${\displaystyle dT=0}$, which implies${\displaystyle dU=0}$ and finally${\displaystyle {\pi }_T=0}$,

i.e. the internal energy of a perfect gas is independent of the volume it occupies. The above relation can be used as a definition of a perfect gas.

The relation${\displaystyle {\pi }_T=0}$ can be proved without the need to invoke any molecular arguments. It follows directly from the thermodynamic equation of state if we use theideal gas law${\displaystyle pV=nRT}$. We have

${\displaystyle {\pi }_T=T{\left(\frac{\mathrm{\partial }p}{\mathrm{\partial }T}\right)}_V-p=T(\frac{nR}{V})-(\frac{nRT}{V})=0}$

Real gases have non-zero internal pressures because their internal energy changes as the gases expand isothermally - it can increase on expansion (${\displaystyle {\pi }_T>0}$, signifying presence of dominant attractive forces between the particles of the gas) or decrease (, dominant repulsion).

In the limit of infinite volume these internal pressures reach the value of zero:

${\displaystyle \underset{V\to \mathrm{\infty }}{lim}{\pi }_T=0}$,

corresponding to the fact that all real gases can be approximated to be perfect in the limit of a suitably large volume. The above considerations are summarized on the graph on the right.

If a real gas can be described by thevan der Waals equation

${\displaystyle p=\frac{nRT}{V-nb}-a\frac{n^2}{V^2}}$

it follows from the thermodynamic equation of state that

${\displaystyle {\pi }_T=a\frac{n^2}{V^2}}$

Since the parameter${\displaystyle a}$ is always positive, so is its internal pressure: internal energy of a van der Waals gas always increases when it expands isothermally.

The${\displaystyle a}$ parameter models the effect of attractive forces between molecules in the gas. However, real non-ideal gases may be expected to exhibit a sign change between positive and negative internal pressures under the right environmental conditions if repulsive interactions become important, depending on the system of interest. Loosely speaking, this would tend to happen under conditions of temperature and pressure such that${\displaystyle Z}$ thecompression factor of the gas, is greater than 1.

In addition, through the use of the Euler chain relation it can be shown that

${\displaystyle {\left(\frac{\mathrm{\partial }U}{\mathrm{\partial }V}\right)}_T=-{\left(\frac{\mathrm{\partial }U}{\mathrm{\partial }T}\right)}_V{\left(\frac{\mathrm{\partial }T}{\mathrm{\partial }V}\right)}_U}$

Defining${\displaystyle {\mu }_J={\left(\frac{\mathrm{\partial }T}{\mathrm{\partial }V}\right)}_U}$ as the "Joule coefficient"^[1] and recognizing${\displaystyle {\left(\frac{\mathrm{\partial }U}{\mathrm{\partial }T}\right)}_V}$ as the heat capacity at constant volume${\displaystyle =C_V}$, we have

${\displaystyle {\pi }_T=-C_V{\mu }_J}$

The coefficient${\displaystyle {\mu }_J}$ can be obtained by measuring the temperature change for a constant-${\displaystyle U}$ experiment, i.e., anadiabatic free expansion (see below). This coefficient is often small, and usually negative at modest pressures (as predicted by the van der Waals equation).

James Joule tried to measure the internal pressure of air in hisexpansion experiment byadiabatically pumping high pressure air from one metal vessel into another evacuated one. The water bath in which the system was immersed did not change its temperature, signifying that no change in the internal energy occurred. Thus, the internal pressure of the air was apparently equal to zero and the air acted as a perfect gas. The actual deviations from the perfect behaviour were not observed since they are very small and thespecific heat capacity ofwater is relatively high.

Much later, in 1925Frederick Keyes andFrancis Sears published measurements of the Joule effect forcarbon dioxide at${\displaystyle T_1}$= 30 °C,${\displaystyle P_1}$= (13.3-16.5) atm using improved measurement techniques and better controls. Under these conditions the temperature dropped when the pressure was adiabatically lowered, which indicates that${\displaystyle {\mu }_J}$ is negative. This is consistent with the van der Waals gas prediction that${\displaystyle {\pi }_T}$ is positive.^[2]

• Peter Atkins and Julio de Paula,Physical Chemistry 8th edition, pp. 60–61 (2006).

• Thermodynamic properties
• Pressure