# The $T\,ds$ equations

We can get a lot of mileage out of the fact that $S$ is a state variable.

We're going to...

• Derive one of the three '$T\,ds$' equations. These relate the difficult-to-measure heat flow ($\delta q_r$) to quantitities, such as heat capacities, that are easier to measure.
• Come up with an expression for the entropy change for any change in a simple solid or liquid that has ~constant $\beta$ and $\kappa$.

## That "magic" relationship

When analyzing the Joule-Thomson coefficient, we used the result that: $$$\frac{\partial h}{\partial P}$_T = v-T$\frac{\partial v}{\partial T}$_P.$$

This relationship can be derived from Gibbs law, $$T\,ds = du + P\,dv.$$

Let's go:

Since $s$ is a state function, we can choose to express it in terms of any two of the thermodynamic variables. Making the choice $s=s(T,P)$, it will turn out to be useful to use the enthalpy $h=u + Pv \Rightarrow dh = du +Pdv + vdP$ once more. Substitute this into Gibbs law:

$T\,ds =$

$$T\,ds = dh -vdP.$$

We can express $h=h(T,P)$, Then... $dh = (\partial h/\partial T)_P\, dT + (\partial h/\partial P)_T\, dP$. Subbing this expression for $dh$ into the equation you had above:

$T\,ds =$

$$T\,ds = $\frac{\partial h}{\partial T}$_P dT + $\frac{\partial h}{\partial P}$_T dP -vdP.$$

Divide this by $T$ and group the terms like $ds=...dT+...dP$

$ds =$

$$ds = \frac{1}{T} $\frac{\partial h}{\partial T}$_P dT + \frac{1}{T}$$\frac{\partial h}{\partial P}$_T -v$dP.$$

The Pfaffian of $s(T,P)$ is: $$ds = $\frac{\partial s}{\partial T}$_P dT + $\frac{\partial s}{\partial P}$_T dP.$$ Since $T$ and $P$ are independent, we can equate the two coefficients of $dT$ in the two previous equations, and likewise with the coefficients of $dP$. Write these out...

$$\frac{\partial s}{\partial T}$_P =$

$$\frac{\partial s}{\partial P}$_T =$

$$$\frac{\partial s}{\partial T}$_P = \frac{1}{T} $\frac{\partial h}{\partial T}$_P {\rm \ \ and \ \ } $\frac{\partial s}{\partial P}$_T = \frac{1}{T}$$\frac{\partial h}{\partial P}$_T -v$.$$

Because $ds$ is an exact differential, the cross - partial derivatives will be equal: $$$\frac{\partial}{\partial P} $\frac{\partial s}{\partial T}$_P$_T = $\frac{\partial}{\partial T} $\frac{\partial s}{\partial P}$_T$_P .$$ Go ahead and calculate the two sides of this equality from the partial derivatives you just wrote...

Using the previous two partial derivatives of $s$, and differentiating: $$\frac{1}{T}\frac{\partial^2 h}{\partial P \partial T} = \frac{1}{T}$\frac{\partial^2 h}{\partial T \partial P} -$\frac{\partial v}{\partial T}$_P$ -\frac{1}{T^2}$$\frac{\partial h}{\partial P}$_T -v$.$$

You should get two 2nd-order partial derivatives among the terms you have. Use the fact that the order of partial derivation doesn't matter. Can you re-arrange what's left to leaving our "magic relation"?:

$$\stackrel{?}{\Rightarrow}$\frac{\partial h}{\partial P}$_T = v-T$\frac{\partial v}{\partial T}$_P.$$

### The 3 $T\,ds$ equations

One of the results along the way in the last derivation was that... $$ds = \frac{1}{T} $\frac{\partial h}{\partial T}$_P dT + \frac{1}{T}$$\frac{\partial h}{\partial P}$_T -v$dP.$$

Our just-proved result is: $$$\frac{\partial h}{\partial P}$_T = v-T$\frac{\partial v}{\partial T}$_P.$$

For a reversible process, $( \partial h / \partial T )_P = c_P$.

Substituting these two relations into the $ds$ equation and multiplying by $T$... $$Tds = c_P dT -T $\frac{\partial v}{\partial T}$_P dP.$$

Since the expansivity (coefficient of volume expansion) is $\beta =(\partial v/\partial T)_P /v$,

$$T\,ds = c_P\, dT -Tv\beta \,dP.$$

This is the first of the three "$T\,ds$" equations, and relates the heat flow in a reversible process to empirically measurable quantities of the material considered.

The other two are derived by the same kind of derivation, only ...

considering $s=s(T,v)$:

$$T\,ds = c_v dT+T$\frac{\partial P}{\partial T}$_v dv =c_v dT+T\frac{\beta}{\kappa} dv.$$

considering $s=s(v,P)$:

$$T\,ds = c_P $\frac{\partial T}{\partial v}$_P dv+c_v$\frac{\partial T}{\partial P}$_v dP =c_P \frac{1}{\beta v} dv+c_v\frac{\kappa}{\beta} dP.$$

### Entropy change for a solid (or liquid)

We developed an equation of state for solids by arguing that

• $\beta$ and $\kappa$ are nearly constant over a wide range of temperature, and
• keeping only first order terms.

$$v = v_0$1+\beta$$T-T_0$-\kappa $P-P_0$$$.$$

Using the $T\,ds$ equation we just came up with, namely... $$T\,ds = c_P dT -Tv\beta dP.$$

rearranging for $ds$: $$ds = \frac{c_P}{T} dT - v\beta dP.$$

Integrating: $$\Delta s = c_P \ln$\frac{T}{T_0}$ - v_0 \beta(P-P_0).$$

This is the change in entropy for a material that has the equation of state above. Entropy is increasing with temperature, and decreasing with pressure.