Thermodynamic Equations of State

Squeeze a real gas isothermally and its internal energy shifts, not just its pressure — a fact ideal-gas theory cannot explain. That shift, $(\partial U/\partial V)_T$ , is the internal pressure, and it is measurable without ever touching entropy directly. This article shows how the Maxwell relations convert two abstract derivatives, $(\partial U/\partial V)_T$ and $(\partial H/\partial p)_T$ , into combinations of pressure, volume, temperature, and their responses to each other. These "thermodynamic equations of state" matter because they connect a substance's mechanical equation of state, like $pV=nRT$ or the van der Waals form, to its energetics. They let you predict how internal energy and enthalpy respond to compression or expansion using only $p$ , $V$ , $T$ , and thermal expansivity data from a lab bench. This article builds both equations from the Maxwell relations. You will see the derivations in full. Then you will apply them to an ideal gas and to a van der Waals gas. Finally, the same machinery will foreshadow how Gibbs energy itself depends on pressure.

Key takeaways

Internal pressure $\pi_T=(\partial U/\partial V)_T$ measures energy change from volume change at constant $T$ .
The thermodynamic equation of state gives $\pi_T = T(\partial p/\partial T)_V - p$ , built from measurables only.
For enthalpy, $(\partial H/\partial p)_T = V - T(\partial V/\partial T)_p = V(1-\alpha T)$ .
Both relations follow directly from the Maxwell relations applied to $dU$ and $dH$ .
For an ideal gas both quantities vanish, exposing why ideal gases have no cohesive-energy dependence on volume.
Key idea: Maxwell relations turn inaccessible entropy derivatives into measurable $p$ - $V$ - $T$ derivatives, converting energy and enthalpy dependences on volume/pressure into laboratory quantities.

The core insight

Because entropy cannot be measured directly as a mechanical response, the Maxwell relations trade $(\partial S/\partial V)_T$ and $(\partial S/\partial p)_T$ for the measurable slopes $(\partial p/\partial T)_V$ and $(\partial V/\partial T)_p$ , which is precisely what turns the definitions of $\pi_T$ and $(\partial H/\partial p)_T$ into practical, testable equations of state for energy and enthalpy.

The physical picture

Imagine an ideal gas and a real gas, both at the same $T$ , expanding into a larger volume. For the ideal gas, $U$ depends only on $T$ , so nothing happens to the internal energy. For the real gas, molecules that were close enough to attract each other are pulled apart, and since separating attracting particles costs energy, $U$ rises. The rate of that rise with volume, at fixed temperature, is the internal pressure,

\pi_T \equiv \left(\frac{\partial U}{\partial V}\right)_T .

The name is deliberately suggestive: $\pi_T$ behaves like an extra pressure term arising from intermolecular forces, distinct from the mechanical pressure $p$ that a gauge would read. For gases dominated by attractive forces, $\pi_T>0$ ; compressed liquids near their boiling point can show large $\pi_T$ , while a hard-sphere fluid with no attractions has $\pi_T \approx 0$ . A parallel question arises for enthalpy: how does $H=U+pV$ respond to squeezing a system at constant temperature? This is $(\partial H/\partial p)_T$ , essential for understanding the Joule–Thomson effect and for tracking how the enthalpy of a real substance shifts as pressure is dialed up, independent of any phase change. Both quantities look, at first glance, to require knowledge of how entropy changes with volume or pressure — something no thermometer or manometer measures directly. The trick is that the Maxwell relations, derived from the exactness of $dU$ , $dH$ , $dA$ , and $dG$ , convert exactly those awkward entropy derivatives into $(\partial p/\partial T)_V$ and $(\partial V/\partial T)_p$ : slopes read straight off an experimental $p$ - $V$ - $T$ surface.

Figure

Internal pressure as an energy-volume slope, ideal vs. real gas

Figure

Maxwell relations as the bridge from entropy slopes to measurable slopes

The core relations

\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p

where: the left side is the internal pressure $\pi_T$ ; the right side uses only the equation of state $p(V,T)$ .

\left(\frac{\partial H}{\partial p}\right)_T = V - T\left(\frac{\partial V}{\partial T}\right)_p

where: the right side can be rewritten using the thermal expansivity $\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p$ as $V(1-\alpha T)$ .

\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p, \qquad \kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_T

where: $\alpha$ is isobaric thermal expansivity (units K⁻¹) and $\kappa_T$ is isothermal compressibility (units bar⁻¹ or Pa⁻¹), both tabulated for real fluids.

Derivation AThe thermodynamic equation of state for U

Goal. Express $(\partial U/\partial V)_T$ using only $p$ , $V$ , $T$ and their mutual derivatives.

Step 1. Start from the combined first and second law for a closed system doing only $pV$ work,

dU = TdS - pdV .

Step 2. Divide formally by $dV$ at constant $T$ . Because $U=U(S,V)$ with $S$ and $V$ independent variables here, this is a legitimate partial-derivative manipulation:

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

Step 3. The term $(\partial S/\partial V)_T$ is not something a manometer or thermometer measures directly. Recall the Helmholtz-energy Maxwell relation, obtained from the exactness of $dA = -SdT - pdV$ :

\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial p}{\partial T}\right)_V .

Step 4. Substitute this Maxwell relation into the Step 2 result. The unmeasurable entropy slope is replaced entirely by a pressure–temperature slope at constant volume, obtainable from any equation of state or from real $p$ - $V$ - $T$ data:

\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p .

Step 5 (sanity check). For an ideal gas, $p=nRT/V$ , so $(\partial p/\partial T)_V = nR/V = p/T$ . Substituting gives $T(p/T) - p = 0$ , consistent with the known fact that ideal-gas internal energy depends on temperature alone.

\boxed{\;\pi_T = \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p\;}

Go deeper · Derivation B · 1 minThe thermodynamic equation of state for H

Derivation BThe thermodynamic equation of state for H

Goal. Express $(\partial H/\partial p)_T$ using only measurable $p$ - $V$ - $T$ quantities.

Step 1. Start from the enthalpy differential for a closed system with $pV$ work only,

dH = TdS + Vdp .

Step 2. Treat $H=H(S,p)$ and divide by $dp$ at constant $T$ :

\left(\frac{\partial H}{\partial p}\right)_T = T\left(\frac{\partial S}{\partial p}\right)_T + V .

Step 3. The entropy-pressure slope is again inaccessible directly, but the Gibbs-energy Maxwell relation, from the exactness of $dG = -SdT + Vdp$ , gives

\left(\frac{\partial S}{\partial p}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_p .

Step 4. Substitute into Step 2. The entropy derivative is replaced by minus the volume's temperature slope at constant pressure — directly measurable via thermal expansion experiments:

\left(\frac{\partial H}{\partial p}\right)_T = V - T\left(\frac{\partial V}{\partial T}\right)_p .

Step 5. Introduce the thermal expansivity $\alpha \equiv (1/V)(\partial V/\partial T)_p$ to write this compactly:

\left(\frac{\partial H}{\partial p}\right)_T = V(1-\alpha T).

Step 6 (sanity check). For an ideal gas, $V=nRT/p$ , so $(\partial V/\partial T)_p = nR/p = V/T$ , and $\alpha T = 1$ identically. The bracket vanishes, giving zero — consistent with ideal-gas enthalpy depending on temperature alone.

\boxed{\;\left(\frac{\partial H}{\partial p}\right)_T = V - T\left(\frac{\partial V}{\partial T}\right)_p = V(1-\alpha T)\;}

Worked examples

Example 1 — Internal pressure of a van der Waals gas (foundational). The van der Waals equation of state is

p = \frac{nRT}{V-nb} - \frac{an^2}{V^2}.

Differentiate at constant $V$ :

\left(\frac{\partial p}{\partial T}\right)_V = \frac{nR}{V-nb}.

Apply the boxed result:

\pi_T = T\left(\frac{\partial p}{\partial T}\right)_V - p = \frac{nRT}{V-nb} - \left[\frac{nRT}{V-nb} - \frac{an^2}{V^2}\right] = \frac{an^2}{V^2}.

Take 1.00 mol of CO₂ ( $a = 0.3640\ \text{Pa m}^6\text{mol}^{-2}$ , from standard van der Waals tables) confined to $V = 1.00\times10^{-3}\ \text{m}^3$ (1.00 L) at 298 K.

\pi_T = \frac{(0.3640\ \text{Pa m}^6\text{mol}^{-2})(1.00\ \text{mol})^2}{(1.00\times10^{-3}\ \text{m}^3)^2} = 3.64\times10^{5}\ \text{Pa} = 3.64\ \text{atm (equivalent pressure units)}.

Sanity check: $\pi_T$ depends only on $a$ , not $b$ or $T$ — exactly as expected, since attractive forces (captured by $a$ ) drive the energy-volume dependence, while $b$ only affects the mechanical equation of state through excluded volume. Note also that $\pi_T$ is positive, confirming that expanding this gas at constant $T$ raises $U$ , consistent with pulling attracting molecules apart costing energy. Compare with the actual pressure at this state:

p = \frac{(1.00)(8.314)(298)}{1.00\times10^{-3}-(1.00)(4.27\times10^{-5})} - \frac{(0.3640)(1.00)^2}{(1.00\times10^{-3})^2} \approx 2.22\times10^{6}\ \text{Pa},

so $\pi_T$ is roughly 16% of $p$ here — a real, measurable correction, not a rounding artifact.

Example 2 — Enthalpy-pressure slope for liquid benzene (applied, exam-level). At 298 K, liquid benzene has molar volume $V_m = 89.4\times10^{-6}\ \text{m}^3\,\text{mol}^{-1}$ and isobaric expansivity $\alpha = 1.24\times10^{-3}\ \text{K}^{-1}$ (standard tabulated values). Find $(\partial H_m/\partial p)_T$ and estimate the enthalpy change on compressing 1.00 mol of benzene from 1.00 bar to 100 bar at constant $T = 298$ K, assuming $V_m$ and $\alpha$ stay roughly constant over this range. Step 1 — compute the bracket:

\alpha T = (1.24\times10^{-3}\ \text{K}^{-1})(298\ \text{K}) = 0.3695.

1 - \alpha T = 1 - 0.3695 = 0.6305.

Step 2 — apply the boxed relation:

\left(\frac{\partial H_m}{\partial p}\right)_T = V_m(1-\alpha T) = (89.4\times10^{-6}\ \text{m}^3\text{mol}^{-1})(0.6305) = 5.64\times10^{-5}\ \text{m}^3\text{mol}^{-1}.

Step 3 — estimate $\Delta H_m$ over $\Delta p = 99\ \text{bar} = 99\times10^5\ \text{Pa} = 9.9\times10^{6}\ \text{Pa}$ :

\Delta H_m \approx \left(\frac{\partial H_m}{\partial p}\right)_T \Delta p = (5.64\times10^{-5}\ \text{m}^3\text{mol}^{-1})(9.9\times10^{6}\ \text{Pa}) = 558\ \text{J mol}^{-1}.

Sanity check: the units work out to $\text{m}^3\,\text{Pa} = \text{J}$ , correct for a molar enthalpy change. The value, roughly half a kilojoule per mole, is small compared to typical vaporization enthalpies (~30 kJ/mol) — reasonable, since liquids are nearly incompressible and this is a purely isothermal pressure effect, not a phase change. Notice too that because $\alpha T < 1$ for this liquid, $(\partial H/\partial p)_T > 0$ : compressing benzene isothermally slightly raises its enthalpy, unlike an ideal gas where the effect is exactly zero.

Limiting cases & common traps

Situation	Behavior	Why
Ideal gas	$\pi_T = 0$ , $(\partial H/\partial p)_T = 0$	$p=nRT/V$ makes $T(\partial p/\partial T)_V = p$ exactly, and $\alpha T = 1$ exactly
Van der Waals gas	$\pi_T = an^2/V^2 > 0$	attractive term $a$ alone sets the volume-dependence of $U$
Perfect incompressible solid/liquid ( $\alpha\to 0$ )	$(\partial H/\partial p)_T \to V$	with no thermal expansion, compressing raises $H$ by $V\,dp$ , matching $H=U+pV$ intuition directly
Near a liquid-vapor critical point	$\pi_T$ can become very large	compressibility diverges, amplifying the pressure–temperature slope

The chemist behind it

James Prescott Joule and William Thomson (Lord Kelvin) jointly investigated how real gases cool or warm on expansion through a porous plug — the Joule–Thomson effect — precisely because ideal-gas theory predicted no temperature change at all. Their careful throttling experiments on air, carbon dioxide, and hydrogen in the 1850s revealed the nonzero $(\partial H/\partial p)_T$ that the thermodynamic equation of state now explains from first principles. Their apparatus, refined over years, gave the first quantitative evidence that real gases retain a "memory" of intermolecular attraction, encoded exactly in the $a$ -dependent internal pressure derived above.

Origins · 1 minWilliam Thomson, 1st Baron Kelvin and how this was discovered

William Thomson, 1st Baron Kelvin

1824–1907 · Co-discovered the Joule–Thomson effect and helped establish the absolute temperature scale underlying every equation of state here

Ask the tutor: How does this same Maxwell-relation strategy show that $(\partial G/\partial p)_T = V$ , setting up the pressure dependence of Gibbs energy for gases?

Why can't we just measure (∂S/∂V)_T directly in a lab?

Entropy has no direct mechanical or thermal probe analogous to a thermometer or manometer; it must be inferred from heat measurements combined with a reversible path, or from statistical models. The Maxwell relations sidestep this by using the mathematical fact that $dU$ , $dH$ , $dA$ , and $dG$ are exact differentials. Mixed second partial derivatives of these state functions must be equal regardless of order, which forces $(\partial S/\partial V)_T$ to equal $(\partial p/\partial T)_V$ — a quantity read straight off a $p$ - $V$ - $T$ dataset.

Is internal pressure the same as the pressure term "a" in the van der Waals equation?

Not quite. The parameter $a$ is a fitted constant characterizing the strength of attractive interactions in the model equation of state. Internal pressure $\pi_T$ is the actual thermodynamic derivative $(\partial U/\partial V)_T$ , which for the van der Waals model happens to equal $an^2/V^2$ . For other equations of state, such as the virial expansion, $\pi_T$ takes a different, more complex functional form, though it still reduces to zero for an ideal gas in every case.

Does (∂H/∂p)_T relate to the Joule–Thomson coefficient?

Yes, closely. The Joule–Thomson coefficient is defined as $\mu_{JT} = (\partial T/\partial p)_H$ , and it can be shown using the chain rule and heat capacity that $\mu_{JT} = -\frac{1}{C_p}(\partial H/\partial p)_T$ . So the same thermodynamic equation of state derived here for $(\partial H/\partial p)_T$ directly predicts the sign and rough magnitude of Joule–Thomson cooling or heating in real gases, linking this topic to refrigeration technology.

Why does this topic lead naturally into the pressure dependence of Gibbs energy?

The same Maxwell relation used for $H$ , namely $(\partial S/\partial p)_T = -(\partial V/\partial T)_p$ , comes from the exactness of $dG = -SdT + Vdp$ . Once you have practiced extracting measurable quantities from this relation for enthalpy, the identical algebra shows directly that $(\partial G/\partial p)_T = V$ , which is the starting point for deriving how Gibbs energy, and hence chemical potential, varies with pressure for gases and condensed phases.