The $T\,ds$ equations

Read Chapter 7!

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.