Question

The partial-derivative relations derived in Problems 1.46,3.33, and 5.12, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between $C_P$and$C_V$.

(a) With the heat capacity expressions from Problem 3.33 in mind, first consider$S$to be a function of $T$and$V.$Expand $dS$in terms of the partial derivatives $(\partial S/\partial T{)}_V$and $(\partial S/\partial V{)}_T$. Note that one of these derivatives is related to$C_V$

(b) To bring in $C_P$, considerlocalid="1648430264419" $V$to be a function of$T$and P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set $dP=0$and note that you have derived a nontrivial expression for $(\partial S/\partial T{)}_P$. This derivative is related to $C_P$, so you now have a formula for the difference $C_P-C_V$

(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 1.46. Your final result should be

$C_P=C_V+\frac{TV{\beta }^2}{{\kappa }_T}$

(d) Check that this formula gives the correct value of $C_P-C_V$for an ideal gas.

(e) Use this formula to argue that $C_P$cannot be less than $C_V$.

(f) Use the data in Problem 1.46 to evaluate$C_P-C_V$for water and for mercury at room temperature. By what percentage do the two heat capacities differ?

(g) Figure 1.14 shows measured values of $C_P$for three elemental solids, compared to predicted values of $C_V$. It turns out that a graph of $\beta$vs.T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why $C_P$and $C_V$agree at low temperatures but diverge in the way they do at higher temperatures.

Short answer

a) $dS={\left(\frac{\partial S}{\partial V}\right)}_{T}dV+{\left(\frac{\partial S}{\partial T}\right)}_{V}dT$

b) $C_P=T{\left(\frac{\partial S}{\partial V}\right)}_T{\left(\frac{\partial V}{\partial T}\right)}_P+C_V$

c) $C_P=\frac{{\beta }^{2}TV}{{\kappa }_T}+C_V$

d) $C_P=C_V+Nk$

e) $C_P>C_V$

f) $C_P-C_V=0.0435\mathrm{J}/\mathrm{K},C_P-C_V=3.576\mathrm{J}/\mathrm{K}$

g)$C_P-C_V\propto T{\beta }^2$

Step-by-step solution

Part (a) - Step 1: Solution

Let S be a function of V and T, that is $S=S(V,T)$Then, based on the derivative's definition, we get

role="math" localid="1648427685193" $dS={\left(\frac{\partial S}{\partial V}\right)}_{T}dV+{\left(\frac{\partial S}{\partial T}\right)}_{V}dT......\left(1\right)$

the heat capacity at constant volume and at constant pressure are given by:

role="math" localid="1648427702735" $C_V=T{\left(\frac{\partial S}{\partial T}\right)}_V{C}_P=T{\left(\frac{\partial S}{\partial T}\right)}_P......\left(2\right)$

so the first term of equation (1), is $C_V/T$.

Part(b) Step 2: Solution

Let V be a function of P and T, that is $V=V(P,T)$, then from the definition of the derivative we get:

$dV={\left(\frac{\partial V}{\partial P}\right)}_{T}dP+{\left(\frac{\partial V}{\partial T}\right)}_{P}dT$

at constant pressure $dp$the above equation becomes

$dV={\left(\frac{\partial V}{\partial T}\right)}_{P}dT$

Substitute in equation (1) to get:

$$\begin{gathered} (dS{)}_P={\left(\frac{\partial S}{\partial V}\right)}_T{\left(\frac{\partial V}{\partial T}\right)}_{P}dT+{\left(\frac{\partial S}{\partial T}\right)}_{V}dT \\ {\left(\frac{\partial S}{\partial T}\right)}_P={\left(\frac{\partial S}{\partial V}\right)}_T{\left(\frac{\partial V}{\partial T}\right)}_P+{\left(\frac{\partial S}{\partial T}\right)}_V \end{gathered}$$

Multiply T on the both side we get:

$T{\left(\frac{\partial S}{\partial T}\right)}_P=T{\left(\frac{\partial S}{\partial V}\right)}_T{\left(\frac{\partial V}{\partial T}\right)}_P+T{\left(\frac{\partial S}{\partial T}\right)}_V$

Substitute from equation (2) we get:

role="math" localid="1648428387185" $C_P=T{\left(\frac{\partial S}{\partial V}\right)}_T{\left(\frac{\partial V}{\partial T}\right)}_P+C_V......\left(3\right)$

Part (c) - Step 3: Solution

Consider the Helmholtz energy Maxwell relation we have:

${\left(\frac{\partial S}{\partial V}\right)}_T={\left(\frac{\partial P}{\partial T}\right)}_V$

substitute into (3) we get:

$C_P=T{\left(\frac{\partial P}{\partial T}\right)}_V{\left(\frac{\partial V}{\partial T}\right)}_P+C_V$

The following is the relationship between thermal expansion and isothermal compressibility:

$\frac{\beta }{{\kappa }_T}={\left(\frac{\partial P}{\partial T}\right)}_V$

Substitute in above expression:

$C_P=\frac{\beta T}{{\kappa }_T}{\left(\frac{\partial V}{\partial T}\right)}_P+C_V$

The thermal expansion is determined by:

$\beta =\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_P$

Substitute value in above equation we get:

$C_P=\frac{{\beta }^{2}TV}{{\kappa }_T}+C_V......\left(4\right)$

Part (d) - Step 4: Solution

$PV=NkT$is the ideal gas relation, hence the thermal expansion and isothermal compressibility are given by:

$$\begin{gathered} \beta =\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_P=\frac{1}{V}\frac{\partial }{\partial T}{\left(\frac{NkT}{P}\right)}_P=\frac{Nk}{PV}=\frac{1}{T} \\ {\kappa }_T=-\frac{1}{V}{\left(\frac{\partial V}{\partial P}\right)}_T=\frac{1}{V}\frac{\partial }{\partial P}{\left(\frac{NkT}{P}\right)}_T=\frac{NkT}{P^{2}V} \end{gathered}$$

substitute into (4) to get:

$$\begin{gathered} C_P=\frac{TV}{T^2}\frac{P^{2}V}{NkT}+C_V \\ {C}_P=\frac{P^2{V}^2}{Nk{T}^2}+C_V \end{gathered}$$

To use $PV=NKT$we get:

$$\begin{gathered} C_P=\frac{N^2{k}^2{T}^2}{Nk{T}^2}+C_V \\ {C}_P=C_V+Nk \end{gathered}$$

Part (e) - Step 5: Solution

The thermal expansion can be negative, but it is squared, it will always be positive; however, the thermal compressibility cannot be negative because adding pressure to the system will increase its volume, which is impossible; hence, $C_P$ is always greater than$C_V$.

Part (f) - Step 6: Solution

Using the data from problem 1.46e The difference between the heat capacities at constant pressure and constant volume for one gram of water at room temperature is:

$$\begin{gathered} C_P-C_V=\frac{{\beta }^{2}TV}{{\kappa }_T} \\ =\frac{{\left(2.57\times {10}^{-4}{\mathrm{K}}^{-1}\right)}^2(298\mathrm{K})\left({10}^{-6}{\mathrm{m}}^3\right)}{4.52\times {10}^{-10}{\mathrm{Pa}}^{-1}} \\ =0.0435\mathrm{J}/\mathrm{K} \end{gathered}$$

role="math" localid="1648429880211" $$\begin{gathered} \text{this is just}1\%\text{of the heat capacity}4.186\mathrm{J}/\mathrm{K}\text{. Now for one mole of mercury (from the data at page 405)} \\ \text{we have:} \end{gathered}$$

$$\begin{gathered} C_P-C_V=\frac{{\beta }^{2}TV}{{\kappa }_T} \\ =\frac{{\left(1.81\times {10}^{-4}{\mathrm{K}}^{-1}\right)}^2(298\mathrm{K})\left(14.8\times {10}^{-6}{\mathrm{m}}^3\right)}{4.04\times {10}^{-11}{\mathrm{Pa}}^{-1}} \\ =3.576\mathrm{J}/\mathrm{K} \end{gathered}$$

$\text{this is just under}13\%\text{of the heat capacity}{C}_P=28.0\mathrm{J}/\mathrm{K}$

Part (g) - Step 7: Solution

Since the volume of a solid is not totally dependent by its temperature,

$C_P-C_V\propto T{\beta }^2$

As a result, the difference in heat capacity as the temperature approaches zero is zero, and the two curves of $C_P$and $C_V$ diverge at high temperatures.