22.4: The Enthalpy of an Ideal Gas is Independent of Pressure

How does pressure affect enthalpy \(H\)? As we showed above we have the following relations of first and second order for \(G\)

\[\left( \dfrac{\partial G}{\partial T} \right)_P = -S \nonumber \]

\[ \left( \dfrac{\partial G}{\partial P} \right)_T = -V \nonumber \]

\[ -\left (\dfrac{\partial S}{\partial P }\right)_T = \left (\dfrac{\partial V}{\partial T} \right)_P \nonumber \]

We also know that by definition:

\[G = H - TS \label{def} \]

Consider an isothermal change in pressure, so taking the partial derivative of each side of Equation \(\ref{def}\), we get:

\[ \left( \dfrac{\partial G}{\partial P}\right)_T = \left( \dfrac{\partial H}{ \partial P}\right)_T -T \left( \dfrac{\partial S}{\partial P}\right)_T \nonumber \]

\[ \left( \dfrac{\partial H}{\partial P}\right)_T = V -T \left( \dfrac{\partial V}{\partial T}\right)_P \label{Eq12} \]

For an ideal gas

\[\dfrac{\partial V}{\partial T} = \dfrac{nR}{P} \nonumber \]

so Equation \(\ref{Eq12}\) becomes

\[ \left( \dfrac{\partial H}{\partial P}\right)_T = V - T \left( \dfrac{nR}{P}\right) = 0 \nonumber \]

As we can see for an ideal gas, there is no dependence of \(H\) on \(P\).

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