# Why is change in enthalpy zero for isothermal processes?

Jul 15, 2017

The CHANGE in enthalpy is zero for isothermal processes consisting of ONLY ideal gases.

For ideal gases, enthalpy is a function of only temperature. Isothermal processes are by definition at constant temperature. Thus, in any isothermal process involving only ideal gases, the change in enthalpy is zero.

The following is a proof that this is true.

From the Maxwell Relation for the enthalpy for a reversible process in a thermodynamically-closed system,

$\mathrm{dH} = T \mathrm{dS} + V \mathrm{dP}$, $\text{ } \boldsymbol{\left(1\right)}$

where $T$, $S$, $V$, and $P$ are temperature, entropy, volume, and pressure, respectively.

If we modify $\left(1\right)$ by infinitesimally varying the pressure at constant temperature, we get:

${\left(\frac{\partial H}{\partial P}\right)}_{T} = T {\left(\frac{\partial S}{\partial \textcolor{red}{P}}\right)}_{\textcolor{red}{T}} + V {\cancel{{\left(\frac{\partial P}{\partial P}\right)}_{T}}}^{1}$ $\text{ } \boldsymbol{\left(2\right)}$

Now, examine the entropy term, which changes due to the change in pressure at constant temperature.

The Gibbs' free energy is a function of temperature and pressure from its Maxwell Relation for a reversible process in a thermodynamically-closed system:

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$ $\text{ } \boldsymbol{\left(3\right)}$

Since the Gibbs' free energy (as with any thermodynamic function) is a state function, its cross-derivatives are equal

${\left(\frac{\partial S}{\partial P}\right)}_{T} = - {\left(\frac{\partial V}{\partial T}\right)}_{P}$, $\text{ } \boldsymbol{\left(4\right)}$.

Utilizing $\left(4\right)$ in $\left(2\right)$, we get:

$\textcolor{g r e e n}{\overline{| \underline{\text{ "((delH)/(delP))_T = -T((delV)/(delT))_P + V" }} |}}$ $\text{ } \boldsymbol{\left(5\right)}$

This relation, which is entirely general, describes the variation of the enthalpy due to a change in pressure in an isothermal process.

The ideality assumption comes in when we use the ideal gas law, $\boldsymbol{P V = n R T}$.

Thus, $V = \frac{n R T}{P}$, and $\left(5\right)$ becomes:

$\textcolor{b l u e}{{\left(\frac{\partial {H}^{\text{id}}}{\partial P}\right)}_{T}} = - T \frac{\partial}{\partial T} {\left[\frac{n R T}{P}\right]}_{P} + \frac{n R T}{P}$

$= - \frac{n R T}{P} {\cancel{\frac{d}{\mathrm{dT}} {\left[T\right]}_{P}}}^{1} + \frac{n R T}{P}$

$= \textcolor{b l u e}{0}$

Thus, we have shown that for ideal gases at constant temperature, their enthalpy does not change. In other words, we've shown that for ideal gases, the enthalpy is only a function of temperature.