The $T\,ds$ equations

[We shall -- 2022 -- use only a few pieces of this topic]

We have already covered Chapter 7.1 and 7.2 and 7.4.

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

The 3 '$T\,ds$' equations allow us to calculate changes in entropy using quantities, such as heat capacities, which are easy to measure, instead of the difficult-to-measure reversible heat flow ($\delta q_r$).

We're going to...

• Do a problem illustrating the usefulness of one of the three '$T\,ds$' equations, and then...
• Derive that '$T\,ds$' equation [Not today!].
• Come up with an expression for the entropy change for any change in a simple solid or liquid that has ~constant $\beta$ and $\kappa$.

Mixing equal amounts of water at different $T$'s

One of the three $T\,ds$ equations relates changes in temperature and pressure to the change of entropy of a system: $$T\,ds = c_P\, dT -Tv\beta \,dP.$$ Let's use this to find the entropy change of the universe upon mixing equal amounts of water at different temperatures together in an open container (that is, a constant pressure process).

We'll consider a "path" to the final state that does not explicitly include mixing, but will end up at the same state.

Consider first just changing the temperature of one portion of water by putting it in contact with a thermal reservoir. (See previous notes). Fill in the steps below:

1. Write out the $T\,ds$ equation above for the case that there is no pressure change ($dP=0$).
   
   
   
   
2. Re-arrange the equation into the form $ds=(...)dT$.
   
   
   
   
3. Assuming $c_P$ is a constant, integrate the equation to find the change in entropy of the system (water), $\Delta s$, when the temperature changes from $T_i$ to $T_f$.
   
   $$\Delta s = \int_{T_i}^{T_f} c_P\frac{dT}{T}=c_P\ln(T_f/T_i)$$
   
   
   
   
4. Let's assume that the system had its temperature changed by placing it in contact with a thermal reservoir at $T_f$. In a reversible process, an amount of heat equal to $q_r=c_P\Delta T$ would need to flow into (or out of) the reservoir to bring about a temperature change. What is the change in entropy of the thermal reservoir at $T_f$ if this amount of heat flows out of (or in to, depending on $T_i$ and $T_f$) the heat reservoir?
   
   See equation (2) in the notes about entropy and heat reservoirs. $$\Delta S_\text{res}=c_P\frac{(T_i-T_f)}{T_f}.$$
   
   
   
   

You've calculated the change in entropy of the system, and of a heat reservoir that it's in contact with. Now consider the compound problem of mixing $m$ kilograms of water at initial temperature $T_1$, and $m$ kg of water at initial temperature $T_2$ together. Consider $c_P$ to be given in units of heat energy per kg, rather than heat energy per kilomole.

5. What will be the final temperature of the mixture? (I'm not asking you to derive this, but to answer this based on your intuition or experience about mixing equal quantities of the same liquid (same $c_P$) together.)
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   $$T_f=\frac{T_1+T_2}{2}.$$ Let's assume that $T_1 \lt T_f \lt T_2$.
   
   
   
   
6. Using the calculations you did in 1.-4. above, What is the entropy change, $\Delta S=m\Delta s$ for the system that starts at $T_1$ and ends at $T_f$ through the processes you calculated above?
   

   $$\Delta S=mc_P\ln(T_f/T_1)$$
   
   
   

7. What is the entropy change of the thermal reservoir at $T_f$?
   

   $$\Delta S_\text{res}=-mc_P\frac{(T_f-T_1)}{T_f}=mc_P\left(.$$

   
   
   
   
8. What is the entropy change for the system that starts at $T_2$ and ends at $T_f$ through the processes you calculated above?
   

   
   
   
   

9. What is the entropy change of the thermal reservoir at $T_f$ for this part of the process?
   

   $$\Delta S_\text{res}=mc_P{(T_f-T_2)}{T_f}.$$
   
   
   

10. Put it all together: What is the total change in entropy of: The two partial systems that end up at $T_f$, and the two thermal reservoirs that they are each coupled too? (Compare to problem #8 in Chapter 7).
    
    
    
    
11. What was the total amount of heat that went in (or out) of the two heat reservoirs?
    
    
    

$T\,dS$ equation for $S(T,P)$

The relationship we used, $$T\,ds = c_P\, dT -Tv\beta \,dP.$$ 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. Solve this last relation for $du+P\,dV$, and 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:

$T\,ds =$

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

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

$ds =$

$$ds = \frac{1}{T} \left( \frac{\partial h}{\partial T}\right)_P dT + \frac{1}{T}\left[\left(\frac{\partial h}{\partial P}\right)_T -v\right]dP.$$

The Pfaffian of $s(T,P)$ is: $$ds = \left( \frac{\partial s}{\partial T}\right)_P dT + \left(\frac{\partial s}{\partial P}\right)_T dP.$$ 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...

$\left( \frac{\partial s}{\partial T}\right)_P = $





$\left(\frac{\partial s}{\partial P}\right)_T = $

$$\left( \frac{\partial s}{\partial T}\right)_P = \frac{1}{T} \left( \frac{\partial h}{\partial T}\right)_P {\rm \ \ and \ \ } \left(\frac{\partial s}{\partial P}\right)_T = \frac{1}{T}\left[\left(\frac{\partial h}{\partial P}\right)_T -v\right].$$

Because $ds$ is an exact differential, the cross - partial derivatives will be equal: $$ \left[ \frac{\partial}{\partial P} \left( \frac{\partial s}{\partial T}\right)_P \right]_T = \left[ \frac{\partial}{\partial T} \left( \frac{\partial s}{\partial P}\right)_T \right]_P .$$ 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: $$\frac{1}{T}\frac{\partial^2 h}{\partial P \partial T} = \frac{1}{T}\left[ \frac{\partial^2 h}{\partial T \partial P} -\left(\frac{\partial v}{\partial T}\right)_P \right] -\frac{1}{T^2}\left[ \left(\frac{\partial h}{\partial P}\right)_T -v \right].$$

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 leave this "magic relation"?:

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

The 3 $T\,ds$ equations

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

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

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

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

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 $s=s(T,v)$:

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

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

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

Entropy change for a solid (or liquid)

We developed an equation of state for solids and liquids by arguing that

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

$$v = v_0\left( 1+\beta\left(T-T_0\right)-\kappa \left(P-P_0\right) \right).\label{eos}$$

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

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

According to our solid/liquid equation of state, Eq $\ref{eos}$, for small changes of $T$ and $P$, we have $v\approx v_0$. And therefore $$ds \approx c_P\frac{dT}{T} dT - v_0\beta dP.$$

Now, we can integrate $ds$ to get $$\Delta s = c_P \ln\left( \frac{T}{T_0} \right) - v_0 \beta(P-P_0).$$

This is the change in entropy for a material that has the equation $\ref{eos}$ as its equation of state.