# What is the enthalpy change for an isothermal process?

Feb 4, 2018

$\Delta H = {\int}_{{P}_{1}}^{{P}_{2}} {\left(\frac{\partial H}{\partial P}\right)}_{T} \mathrm{dP} = {\int}_{{P}_{1}}^{{P}_{2}} V - T {\left(\frac{\partial V}{\partial T}\right)}_{P} \mathrm{dP}$

Now decide what gas law to use, or what $\alpha$ corresponds to your substance.

Well, from the total differential at constant temperature,

$\mathrm{dH} = {\cancel{{\left(\frac{\partial H}{\partial T}\right)}_{P} \mathrm{dT}}}^{0} + {\left(\frac{\partial H}{\partial P}\right)}_{T} \mathrm{dP}$,

so by definition of integrals and derivatives,

$\Delta H = {\int}_{{P}_{1}}^{{P}_{2}} {\left(\frac{\partial H}{\partial P}\right)}_{T} \mathrm{dP}$ $\text{ } \boldsymbol{\left(1\right)}$

The natural variables are $T$ and $P$, which are given in the Gibbs' free energy Maxwell relation.

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$$\text{ } \boldsymbol{\left(2\right)}$

This is also related, obviously, by the well-known isothermal Gibbs relation

$\mathrm{dG} = \mathrm{dH} - T \mathrm{dS}$$\text{ } \boldsymbol{\left(3\right)}$

Differentiating $\left(3\right)$ at constant temperature,

${\left(\frac{\partial G}{\partial P}\right)}_{T} = {\left(\frac{\partial H}{\partial P}\right)}_{T} - T {\left(\frac{\partial S}{\partial P}\right)}_{T}$

From $\left(2\right)$,

${\left(\frac{\partial G}{\partial P}\right)}_{T} = V$

and also from $\left(2\right)$,

${\left(\frac{\partial S}{\partial P}\right)}_{T} = - {\left(\frac{\partial V}{\partial T}\right)}_{P}$

since the Gibbs' free energy is a state function and its cross-derivatives must be equal. Thus from $\left(3\right)$ we get

$V = {\left(\frac{\partial H}{\partial P}\right)}_{T} + T {\left(\frac{\partial V}{\partial T}\right)}_{P}$

or we thus go back to $\left(1\right)$ to get:

$\overline{\underline{|}} \stackrel{\text{ ")(" "DeltaH = int_(P_1)^(P_2) ((delH)/(delP))_TdP = int_(P_1)^(P_2)V - T((delV)/(delT))_PdP" }}{|}$

And what remains is to distinguish between the last term for gases, liquids and solids...

GASES

Use whatever gas law you want to find it. If for whatever reason your gas is ideal, then

${\left(\frac{\partial V}{\partial T}\right)}_{P} = \frac{n R}{P}$

and that just means

${\left(\frac{\partial H}{\partial P}\right)}_{T} = V - \frac{n R T}{P}$

$= V - V = 0$

which says that ideal gases have changes in enthalpy as a function of only temperature. One would get

$\textcolor{b l u e}{\Delta H = {\int}_{{P}_{1}}^{{P}_{2}} 0 \mathrm{dP} = 0}$.

Not very interesting.

Of course, if your gas is not ideal, this isn't necessarily true.

LIQUIDS AND SOLIDS

These data are tabulated as coefficients of volumetric thermal expansion $\alpha$,

$\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P}$

at VARIOUS temperatures for VARIOUS condensed phases. Some examples at ${20}^{\circ} \text{C}$:

• ${\alpha}_{{H}_{2} O} = 2.07 \times {10}^{- 4} {\text{K}}^{- 1}$
• ${\alpha}_{A u} = 4.2 \times {10}^{- 5} {\text{K}}^{- 1}$ (because that's REAL useful, right?)
• ${\alpha}_{E t O H} = 7.50 \times {10}^{- 4} {\text{K}}^{- 1}$
• ${\alpha}_{P b} = 8.7 \times {10}^{- 5} {\text{K}}^{- 1}$

In that case,

${\left(\frac{\partial H}{\partial P}\right)}_{T} = V - T V \alpha$

$= V \left(1 - T \alpha\right)$

Thus,

$\textcolor{b l u e}{\Delta H = {\int}_{{P}_{1}}^{{P}_{2}} V \left(1 - T \alpha\right) \mathrm{dP} \approx V \left(1 - T \alpha\right) \Delta P}$

since liquids and solids are very incompressible and require a large change in pressure.