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

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

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

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

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:

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

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

For an ideal gas

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

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

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

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

The enthalpy of an ideal Gas Is independent of pressure.