Question

Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep $V$fixed as $T$increases, as follows.

(a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume,$d{V}_1$, in terms of $dT$and the thermal expansion coefficient $\beta$introduced in Problem 1.7.

(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, $d{V}_2$, in terms of $dP$and the isothermal compressibility ${\kappa }_T$, defined as

${\kappa }_T\equiv -\frac{1}{V}{\left(\frac{\mathrm{\partial }V}{\mathrm{\partial }P}\right)}_T$

(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of $dP\text{to}dT$is equal to $(\mathrm{\partial }P/\mathrm{\partial }T{)}_V$, since there is no net change in volume. Express this partial derivative in terms of $\beta \text{and}{\kappa }_T$. Then express it more abstractly in terms of the partial derivatives used to define $\beta \text{and}{\kappa }_T$. For the second expression you should obtain

${\left(\frac{\mathrm{\partial }P}{\mathrm{\partial }T}\right)}_V=-\frac{(\mathrm{\partial }V/\mathrm{\partial }T{)}_P}{(\mathrm{\partial }V/\mathrm{\partial }P{)}_T}$

This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.

(d) Compute $\beta ,{\kappa }_T,\text{and}(\mathrm{\partial }P/\mathrm{\partial }T{)}_V$for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).

(e) For water at ${25}^{\circ }\mathrm{C},\beta =2.57\times {10}^{-4}{\mathrm{K}}^{-1}\text{and}{\kappa }_T=4.52\times {10}^{-10}{\mathrm{Pa}}^{-1}$. Suppose you increase the temperature of some water from ${20}^{\circ }\mathrm{C}\text{to}{30}^{\circ }\mathrm{C}$. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which $\text{(at}{25}^{\circ }\mathrm{C}\text{)}\beta =1.81\times {10}^{-4}{\mathrm{K}}^{-1}$and${\kappa }_T=4.04\times {10}^{-11}{\mathrm{Pa}}^{-1}$

Given the choice, would you rather measure the heat capacities of these substances at constant $v$or at constant $p$?

Short answer

(A) The change in volume in $d{V}_1$and thermal coefficient is $d{V}_1=\beta VdT$

(B) The change in volume of $d{V}_2$is $d{V}_2=-{\kappa }_{T}VdP$

(C) The second expression is ${\left(\frac{\mathrm{\partial }P}{\mathrm{\partial }T}\right)}_V=-\frac{(\mathrm{\partial }V/\mathrm{\partial }T{)}_P}{(\mathrm{\partial }V/\mathrm{\partial }P{)}_T}$

(D) An ideal gas of three expression is $\beta =\frac{1}{T},{\kappa }_T=\frac{1}{P}.\frac{P}{T}=\frac{\beta }{{\kappa }_T}$

(E) The heat capacities of substances constant is$\mathrm{\Delta }{P}_{\text{water}}=5.686\times {10}^6\mathrm{Pa},\mathrm{\Delta }{P}_{\text{mercury}}=4.48\times {10}^7\mathrm{Pa}$

Step-by-step solution

Step :1  The thermal expansion coefficient (part a)

Substances' heat capacity can be determined at constant volume or constant pressure. It's relatively simple to assess a gas's heat capacity by enclosing it in a sealed container and keeping it at a constant volume. However, measuring heat capacity at constant pressure is much easier for solids and liquids. We can calculate how much pressure must be increased to prevent a solid or liquid from expanding when heated.

(a) The thermal expansion coefficient is:

$\beta =\frac{\mathrm{\Delta }V/V}{\mathrm{\Delta }T}$

Imagine that the substrate atmospheric temperature slightly at constant pressure, and the thermal expansion coefficient, which is a measure of the relative volume change with temperature at constant pressure, is:

$$\begin{gathered} \beta =\frac{\mathrm{\Delta }V/V}{\mathrm{\Delta }T} \\ =\frac{1}{V}{\left(\frac{\mathrm{\partial }V}{\mathrm{\partial }T}\right)}_p \\ \to {\left(\frac{\mathrm{\partial }V}{\mathrm{\partial }T}\right)}_p=\beta V \end{gathered}$$

However, because$V$is a function of $T\text{and}P,V(T,P)$, the volume change due to the differential:

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

Equation (2) becomes: $dp=0$at constant pressure.

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

Substitute $\left(2\right)$for $\left(1\right)$to get the following volume change:

$d{V}_1=\beta VdT$

Step :2  Constant temperature (part b)

(b) Assume we compress a solid (or a liquid) slightly at constant temperature $dT=0$, resulting in equation $\left(2\right)$:

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

Given that the isothermal compressibility is the reciprocal of the bulk modulus:

$$\begin{gathered} {\kappa }_T=-\frac{1}{V}{\left(\frac{\mathrm{\partial }V}{\mathrm{\partial }P}\right)}_T \\ \to {\left(\frac{\mathrm{\partial }V}{\mathrm{\partial }P}\right)}_T \\ =-{\kappa }_{T}V \end{gathered}$$

Substituting equation $\left(5\right)$into equation $\left(4\right)$, the volume change is:

$d{V}_2=-{\kappa }_{T}VdP$

Step :3 Change in volume after two action (part c)

(c) The net change in volume after the two actions in $\left(a\right)$and $\left(b\right)$is zero:

$$\begin{gathered} d{V}_1+d{V}_2=0 \\ \to d{V}_1=-d{V}_2 \end{gathered}$$

Substitute

$\beta VdT={\kappa }_{T}VdP$

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

From $\left(1\right)and\left(5\right)$

${\left(\frac{\mathrm{\partial }P}{\mathrm{\partial }T}\right)}_V=\frac{\beta }{\kappa y}=\frac{\frac{1}{V}{\left(\frac{\mathrm{\partial }V}{\mathrm{\partial }T}\right)}_{\mathrm{\Gamma }}}{\frac{1}{V}{\left(\frac{\mathrm{\partial }V}{\mathrm{\partial }P}\right)}_T}$

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

$\to {\left(\frac{\mathrm{\partial }P}{\mathrm{\partial }T}\right)}_V=-\frac{(\mathrm{\partial }V/\mathrm{\partial }T{)}_P}{(\mathrm{\partial }V/\mathrm{\partial }P{)}_T}$

Step :4  Ideal gas law (part d)

(d) From the ideal gas law, $PV=NkT$, and using equation $\left(5\right)$and $\left(1\right)$we have:

$\beta =\frac{1}{V}{\left(\frac{\mathrm{\partial }V}{\mathrm{\partial }T}\right)}_p=\frac{1}{V}{\left(\frac{\mathrm{\partial }\left(\frac{NkT}{P}\right)}{\mathrm{\partial }T}\right)}_p$

$\to \beta =\frac{Nk}{PV}=\frac{Nk}{NkT}=\frac{1}{T}$

$$\begin{gathered} {\kappa }_T=-\frac{1}{V}{\left(\frac{\mathrm{\partial }V}{\mathrm{\partial }P}\right)}_T \\ =-\frac{1}{V}{\left(\frac{\mathrm{\partial }\left(\frac{NkT}{P}\right)}{\mathrm{\partial }P}\right)}_T \\ =\frac{NkT}{V{P}^2} \end{gathered}$$

$$\begin{gathered} \to {\kappa }_T=\frac{NkT}{V{P}^2} \\ =\frac{NkT}{\left(VP\right)P} \\ =\frac{NkT}{\left(NkT\right)P} \\ =\frac{1}{P} \end{gathered}$$

Divide

$\to \frac{P}{T}=\frac{\beta }{{\kappa }_T}$

Now from equation

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

$\to {\left(\frac{\mathrm{\partial }\left(\frac{NkT}{V}\right)}{\mathrm{\partial }T}\right)}_V=\frac{\beta }{{\kappa }_T}$

$\to \frac{Nk}{V}=\frac{\beta }{{\kappa }_T}$

$\to \frac{P}{T}=\frac{\beta }{{\kappa }_T}$

We can conclude that the results from $\left(d\right)$, equation , and $\left(c\right)$equation , are identical.

Step :5  For water (part e)

(e) For water, the given values are:

$\begin{array}{c}\beta =2.57\times {10}^{-4}{\mathrm{K}}^{-1}\\ \kappa =4.52\times {10}^{-10}{\mathrm{Pa}}^{-1}\end{array}$

So the pressure increase

$$\begin{gathered} \frac{P}{T}=\frac{\beta }{{\kappa }_T} \\ \to P=\frac{\beta }{{\kappa }_T}T \\ \to \mathrm{\Delta }P=\frac{\beta }{{\kappa }_T}\mathrm{\Delta }T \end{gathered}$$

But the temperature difference is

$$\begin{gathered} \mathrm{\Delta }T=T_f-T_i \\ =30-20 \\ ={10}^{\circ }\mathrm{K} \end{gathered}$$

So the pressure difference is

$$\begin{gathered} \mathrm{\Delta }P=\frac{\beta }{{\kappa }_T}\mathrm{\Delta }T \\ =\frac{2.57\times {10}^{-4}}{4.52\times {10}^{-10}}\times 10 \\ =5.686\times {10}^6\mathrm{Pa} \end{gathered}$$

$\to \mathrm{\Delta }P=5.686\times {10}^6\mathrm{Pa}$

For mercury the temperature range is

$\begin{array}{c}\beta =1.81\times {10}^{-4}{\mathrm{K}}^{-1}\\ \kappa =4.04\times {10}^{-11}{\mathrm{Pa}}^{-1}\end{array}$

So the pressure increase must be

$\begin{array}{r}\mathrm{\Delta }P=\frac{\beta }{{\kappa }_T}\mathrm{\Delta }T=\frac{1.81\times {10}^{-4}}{4.04\times {10}^{-11}}\times 10\\ \to \mathrm{\Delta }P=4.48\times {10}^7\mathrm{Pa}\end{array}$