Given Constant Temperature, Why Can Pressure Affect Internal Energy?

At fixed temperature, changing pressure does not change the internal energy of an ideal gas with fixed composition. It can change the internal energy of a real gas or liquid, because pressure changes the material’s volume and therefore the energy associated with interactions between its molecules.

The key distinction is between an idealized model, which neglects intermolecular potential energy, and a real substance, for which that energy can matter. Pressure is not an extra, independent ingredient added to internal energy: at fixed temperature and composition, changing pressure moves the substance to a different state, usually with a different volume or density.

Why an ideal gas has no such pressure effect

For an ideal gas, molecules are modeled as having no interactions except during collisions. Its internal energy is the sum of the molecules’ energies, with no intermolecular interaction energy; for fixed composition, that energy depends only on temperature. For a fixed amount of ideal gas,

$U=U(T).$

So if the temperature is held constant, the internal energy stays constant even when pressure changes. The pressure and volume still change according to the ideal-gas law, $PV=nRT$; they just do not change $U$ in this model.

Why real substances can behave differently

Real molecules attract or repel one another. Their separation affects intermolecular potential energy, which is part of internal energy. Compressing a real fluid changes molecular separations and can therefore change that potential energy even if the average molecular kinetic energy—associated with temperature—stays the same.

For a fixed amount of a simple compressible substance of fixed composition, a useful thermodynamic identity is

$$dU=C_V\,dT+\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right]dV.$$

Here, $C_V$ is the heat capacity at constant volume, and the pressure derivative is evaluated at fixed volume. At constant temperature, $dT=0$, leaving

$dU=\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right]dV.$

For an ideal gas, $P=nRT/V$, so $T(\partial P/\partial T)_V=P$. The bracket is zero, and there is no internal-energy change along an isotherm. For a real substance, the bracket need not vanish. If pressure is used instead of volume to describe the state, the same result can be written as

$$\left(\frac{\partial U}{\partial P}\right)_T= \left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right] \left(\frac{\partial V}{\partial P}\right)_T.$$

This makes clear that the pressure effect depends on the substance’s equation of state. It is not a universal rule that increasing pressure always increases internal energy.

Worked example: an isothermal compression

The van der Waals model for one mole of a real gas gives, at fixed temperature,

$U_m=C_{V,m}T-\frac{a}{V_m}+\text{constant},$

assuming the molar heat capacity is constant over the temperature range considered. Here, $V_m$ is molar volume and $a$ represents the strength of molecular attractions. Consider a model gas with $a=0.137\ \mathrm{Pa\,m^6\,mol^{-2}}$. Compress one mole isothermally from $V_{m,1}=0.020\ \mathrm{m^3/mol}$ to $V_{m,2}=0.010\ \mathrm{m^3/mol}$. The change in molar internal energy is

$$\Delta U_m=-\frac{a}{V_{m,2}}+\frac{a}{V_{m,1}} =a\left(\frac{1}{V_{m,1}}-\frac{1}{V_{m,2}}\right) =0.137(50-100) =-6.85\ \mathrm{J/mol}.$$

On a mechanically stable gas branch, pressure rises as the gas is compressed, but this model predicts that its internal energy falls slightly. The reason is that the attractive-potential-energy term becomes more negative as the molecules are brought closer. This is an illustrative model result, not a general prediction for every substance or pressure range.

What happens to the work and heat?

A constant-temperature process does not necessarily have zero heat transfer or zero work. During compression, work is done on the gas. For the real-gas example, internal energy decreases, so heat must also leave the gas; otherwise the work input would raise its internal energy. The first law keeps track of these transfers, while the equation of state and interaction model determine the state-to-state change in internal energy.

Common mistakes

• Applying the ideal-gas rule to every gas. The result $U=U(T)$ is specific to the ideal-gas model; real gases can have a pressure or volume dependence at fixed temperature.
• Treating pressure and temperature as independent causes. For a fixed amount of material in equilibrium, pressure, temperature, and volume are linked by an equation of state. At fixed temperature, changing pressure usually means changing volume.
• Assuming higher pressure must mean higher internal energy. The sign and size of the change depend on the substance and state. Attractive interactions, repulsive interactions, and phase changes can matter differently.
• Confusing internal energy with heat. Heat is energy transferred across a system boundary; internal energy is a property of the system’s state. An isothermal process can involve both heat transfer and work.