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 \(P\).