# 9.4: The Enthalpy of an Ideal Gas

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.