# Enthalphy and phase transitions

### Enthalpy and heat of formation

As
a substance goes through a *phase transition* it absorbs (or gives off) heat *even though its temperature does not change*. This goes by names like "heat of formation", or "latent
heat".

Phase transitions are processes that:

- are isothermic
- are isobaric
- involve a volume change $\rightarrow$ work is done.

How much work? $$\Delta w = P(v_2-v_1).$$

The 1st Law says... $$du = \delta q -P\,dv.$$

So, the heat change (latent heat$\equiv l$) for a macroscopic change in phase is: $$(u_2-u_1) = l - P(v_2 - v_1).$$

Solving for $l$: $$l = (u_2+Pv_2) - (u_1+Pv_1).$$

Let's define the combination in parenthesis as the specific **enthalpy**:

$$h \equiv u+Pv.$$

So we say that the latent heat associated with the phase transition is just the difference in enthalpies between the phases. $$l = h_2-h_1.$$

### Enthalpy and $c_p$

With enthalpy per kilomole, $h$, defined as: $$h \equiv u +Pv.$$

$P$, $u$, and $v$ are all state functions (have exact differentials).
Therefore, **enthalpy, $h$, is also a state function**.

The differential form of this definition of $h$ is: $$dh = du + P\,dv + v\,dP.$$

Since $h$ is a state function, we can choose to write its Pfaffian in terms of any two thermodynamic parameters of our choosing. Choosing $h=h(T,P)$, $$dh = \( \frac{\partial h}{\partial T} \)_P dT + \( \frac{\partial h}{\partial P} \)_T dP.$$

The 1st Law says, for a reversible process, that: $$du = \delta q - \delta w= \delta q-P\,dv.$$

Substituting in the differential form of the enthalpy into this... $$\begineq \delta q&=&du +P\,dv =dh -P\,dv-v\,dP+P\,dv\\ &=& dh - v\,dP. \endeq$$

Subbing that Pfaffian expression for $dh$ in to this... $$\begineq \delta q&=&\( \frac{\partial h}{\partial T} \)_P dT + \[ \( \frac{\partial h}{\partial P} \)_T -v \]_T dP. \endeq$$

So, for the particular case of constant pressure processes ($dP=0$), we have... $$\( \delta q \)_P = \( \frac{\partial h}{\partial T} \)_P dT.$$

Or... $$ \( \frac{\partial h}{\partial T} \)_P=\( \frac{\delta q}{\partial T} \)_P\equiv c_P .$$

So

$$c_p = \( \frac{\partial h}{\partial T} \)_P.$$

We shall see that, **for an ideal gas**,
$$\( \frac{\partial h}{\partial P} \)_T = 0,$$

So that, for an ideal gas, $h(P,T) = h(T)$, and we can write the partial differential as an exact differential: $$c_p = \frac{dh}{dT} \text{ (ideal gas)}.$$

Integrating this to find the overall enthalpy: $$\int dh = h-h_0 = \int_{T_0}^{T} c_p dT \text{ (ideal gas)}.$$

### Mayer's equation

A relationship *for ideal gases* between $c_p$ and $c_v$ can now be obtained as
follows. We had two different expressions for the heat transfer $\delta q$
for reversible processes in terms of the internal energy or the enthalpy of
the system. These were...
$$\delta q = du + P\,dv, $$
and
$$\delta q = dh -v\,dP.$$

Setting these equal, and re-arranging just a bit... $$dh - du = P\,dv + v\,dP=d(Pv).$$

Dividing by $dT$, $$\frac{dh}{dT} - \frac{du}{dT} = \frac{d(Pv)}{dT}.$$

*What is $\frac{d(Pv)}{dT}$ for an ideal gas?*

For an ideal gas the two first exact differentials are $c_p$ and $c_v$, and we can calculate the differential on the rate from the ideal gas law: $$\frac{d(Pv)}{dT} = \frac{d(RT)}{dT} = R.$$

Putting this all together

$$c_p - c_v = R.$$

**Problem (Carter 4-6 b)**: *Pretend Oxygen is an ideal gas with $c_v=(5/2)R$.
Suppose that the temperature of 2 kilomoles of O${}_2$ is raised from 27 C
to 227 C. What is the increase in enthalpy?*