# Why can enthalpy not be measured directly?

Apr 12, 2015

Because it is a function of variables that are not all called Natural Variables. The Natural Variables are those that we can measure easily from direct measurements, like volume , pressure , and temperature.

T: Temperature
V: Volume
P: Pressure
S: Entropy
G: Gibbs' Free Energy
H: Enthalpy

Below is a somewhat rigorous derivation showing how we CAN measure Enthalpy, even indirectly. Eventually we get to an expression that lets us measure enthalpy at a constant temperature!

Enthalpy is a function of Entropy, Pressure, Temperature, and Volume, with Temperature, Pressure, and Volume as its natural variables under this Maxwell relation:

$H = H \left(S , P\right)$
$\mathrm{dH} = T \mathrm{dS} + V \mathrm{dP}$ (Eq. 1) -- Maxwell relation

We do not need to use this equation here; the point is, we cannot directly measure Entropy either (we don't have a "heat-flow-o-meter"). So, we have to find a way to measure Enthalpy using other variables.

Since Enthalpy is commonly defined in the context of temperature and pressure, consider the common equation for Gibbs' free energy (a function of temperature and pressure) and its Maxwell relation:

$\Delta G = \Delta H - T \Delta S$ (Eq. 2)
$\mathrm{dG} = \mathrm{dH} - T \mathrm{dS}$ (Eq. 3) -- Differential form

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$ (Eq. 4) -- Maxwell relation

From here we can write the partial derivative with respect to pressure at a constant temperature using Eq. 3:

${\left(\frac{\delta G}{\delta P}\right)}_{T} = {\left(\frac{\delta H}{\delta P}\right)}_{T} - T {\left(\frac{\delta S}{\delta P}\right)}_{T}$ (Eq. 5)

Using Eq. 4, we can take the first partial derivative we see in Eq. 5 (for Gibbs). $- S \mathrm{dT}$ becomes 0 since $\Delta T = 0$, and $\delta P$ gets divided out.

${\left(\frac{\delta G}{\delta P}\right)}_{T} = V$ (Eq. 6)

And another thing we can write, since G is a state function, are the cross-derivatives from the Maxwell relation to figure out the entropy half of Eq. 5:

$- {\left(\frac{\delta S}{\delta P}\right)}_{T} = {\left(\frac{\delta V}{\delta T}\right)}_{P}$ (Eq. 7)

Finally, we can plug in Eqs. 6 and 7 into Eq. 5:

$V = {\left(\frac{\delta H}{\delta P}\right)}_{T} + T {\left(\frac{\delta V}{\delta T}\right)}_{P}$ (Eq. 8-1)

And further simplify it:
${\left(\frac{\delta H}{\delta P}\right)}_{T} = V - T {\left(\frac{\delta V}{\delta T}\right)}_{P}$ (Eq. 8-2)

There we go! We've got a function that describes how to measure enthalpy "directly".

What this says is, we can begin by measuring the change in volume of a gas as its temperature changes in a constant-pressure environment (such as a vacuum). Then, we've got ${\left(\frac{\delta V}{\delta T}\right)}_{P}$.

Afterwards, to take it further, you could multiply by $\mathrm{dP}$ and integrate from the first to second pressure. Then you can get enthalpy change at a specific temperature by varying the pressure of the vessel.

$\Delta H = {\int}_{{P}_{1}}^{{P}_{2}} \left[V - T {\left(\frac{\delta V}{\delta T}\right)}_{P}\right] \mathrm{dP}$ (Eq. 9)

And as an example, you could apply the ideal gas law and get ${\left(\frac{\delta V}{\delta T}\right)}_{P} = {\left(\frac{\delta}{\delta T} \left(\frac{n R T}{P}\right)\right)}_{P} = \frac{n R}{P}$

You can tell that the ideal gas then makes it that

$\Delta H = {\int}_{{P}_{1}}^{{P}_{2}} V - V \mathrm{dP} = 0$

meaning that Enthalpy is only dependent on temperature for an ideal gas! Neat.