91Ó°ÊÓ

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:

$\mathrm{β}=\frac{\mathrm{Δ}V/V}{\mathrm{Δ}T}$

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:

$$\begin{gathered} \mathrm{β}=\frac{\mathrm{Δ}V/V}{\mathrm{Δ}T} \\ =\frac{1}{V}{\left(\frac{\mathrm{∂}V}{\mathrm{∂}T}\right)}_p \\ →{\left(\frac{\mathrm{∂}V}{\mathrm{∂}T}\right)}_p=\mathrm{β}V \end{gathered}$$

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

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

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

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

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

$d{V}_1=\mathrm{β}VdT$

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

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

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

$$\begin{gathered} {\mathrm{κ}}_T=−\frac{1}{V}{\left(\frac{\mathrm{∂}V}{\mathrm{∂}P}\right)}_T \\ →{\left(\frac{\mathrm{∂}V}{\mathrm{∂}P}\right)}_T \\ =−{\mathrm{κ}}_{T}V \end{gathered}$$

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

$d{V}_2=−{\mathrm{κ}}_{T}VdP$

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

$$\begin{gathered} d{V}_1+d{V}_2=0 \\ →d{V}_1=−d{V}_2 \end{gathered}$$

Substitute

$\mathrm{β}VdT={\mathrm{κ}}_{T}VdP$

$→{\left(\frac{\mathrm{∂}P}{\mathrm{∂}T}\right)}_V=\frac{\mathrm{β}}{{\mathrm{κ}}_T}$

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

${\left(\frac{\mathrm{∂}P}{\mathrm{∂}T}\right)}_V=\frac{\mathrm{β}}{\mathrm{κ}y}=\frac{\frac{1}{V}{\left(\frac{\mathrm{∂}V}{\mathrm{∂}T}\right)}_{\mathrm{Γ}}}{\frac{1}{V}{\left(\frac{\mathrm{∂}V}{\mathrm{∂}P}\right)}_T}$

$→{\left(\frac{\mathrm{∂}P}{\mathrm{∂}T}\right)}_V=−\frac{{\left(\frac{\mathrm{∂}P}{\mathrm{∂}T}\right)}_T}{{\left(\frac{\mathrm{∂}V}{\mathrm{∂}P}\right)}_T}$

$→{\left(\frac{\mathrm{∂}P}{\mathrm{∂}T}\right)}_V=−\frac{(\mathrm{∂}V/\mathrm{∂}T{)}_P}{(\mathrm{∂}V/\mathrm{∂}P{)}_T}$

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

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

$→\mathrm{β}=\frac{Nk}{PV}=\frac{Nk}{NkT}=\frac{1}{T}$

$$\begin{gathered} {\mathrm{κ}}_T=−\frac{1}{V}{\left(\frac{\mathrm{∂}V}{\mathrm{∂}P}\right)}_T \\ =−\frac{1}{V}{\left(\frac{\mathrm{∂}\left(\frac{NkT}{P}\right)}{\mathrm{∂}P}\right)}_T \\ =\frac{NkT}{V{P}^2} \end{gathered}$$

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

Divide

$→\frac{P}{T}=\frac{\mathrm{β}}{{\mathrm{κ}}_T}$

Now from equation

${\left(\frac{\mathrm{∂}P}{\mathrm{∂}T}\right)}_V=\frac{\mathrm{β}}{{\mathrm{κ}}_T}$

$→{\left(\frac{\mathrm{∂}\left(\frac{NkT}{V}\right)}{\mathrm{∂}T}\right)}_V=\frac{\mathrm{β}}{{\mathrm{κ}}_T}$

$→\frac{Nk}{V}=\frac{\mathrm{β}}{{\mathrm{κ}}_T}$

$→\frac{P}{T}=\frac{\mathrm{β}}{{\mathrm{κ}}_T}$

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

(e) For water, the given values are:

$\begin{array}{c}\mathrm{β}=2.57×{10}^{−4}{\mathrm{K}}^{−1}\\ \mathrm{κ}=4.52×{10}^{−10}{\mathrm{Pa}}^{−1}\end{array}$

So the pressure increase

$$\begin{gathered} \frac{P}{T}=\frac{\mathrm{β}}{{\mathrm{κ}}_T} \\ →P=\frac{\mathrm{β}}{{\mathrm{κ}}_T}T \\ →\mathrm{Δ}P=\frac{\mathrm{β}}{{\mathrm{κ}}_T}\mathrm{Δ}T \end{gathered}$$

But the temperature difference is

$$\begin{gathered} \mathrm{Δ}T=T_f−T_i \\ =30−20 \\ ={10}^{∘}\mathrm{K} \end{gathered}$$

So the pressure difference is

$$\begin{gathered} \mathrm{Δ}P=\frac{\mathrm{β}}{{\mathrm{κ}}_T}\mathrm{Δ}T \\ =\frac{2.57×{10}^{−4}}{4.52×{10}^{−10}}×10 \\ =5.686×{10}^6\mathrm{Pa} \end{gathered}$$

$→\mathrm{Δ}P=5.686×{10}^6\mathrm{Pa}$

For mercury the temperature range is

$\begin{array}{c}\mathrm{β}=1.81×{10}^{−4}{\mathrm{K}}^{−1}\\ \mathrm{κ}=4.04×{10}^{−11}{\mathrm{Pa}}^{−1}\end{array}$

So the pressure increase must be

$\begin{array}{r}\mathrm{Δ}P=\frac{\mathrm{β}}{{\mathrm{κ}}_T}\mathrm{Δ}T=\frac{1.81×{10}^{−4}}{4.04×{10}^{−11}}×10\\ →\mathrm{Δ}P=4.48×{10}^7\mathrm{Pa}\end{array}$