91影视

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:

$尾=\frac{\mathrm{螖}V/V}{\mathrm{螖}T}$

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:

$$\begin{gathered} 尾=\frac{\mathrm{螖}V/V}{\mathrm{螖}T} \\ =\frac{1}{V}{\left(\frac{\mathrm{鈭�}V}{\mathrm{鈭�}T}\right)}_p \\ 鈫�{\left(\frac{\mathrm{鈭�}V}{\mathrm{鈭�}T}\right)}_p=尾V \end{gathered}$$

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

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

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

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

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

$d{V}_1=尾VdT$

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

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

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

$$\begin{gathered} {魏}_T=鈭�\frac{1}{V}{\left(\frac{\mathrm{鈭�}V}{\mathrm{鈭�}P}\right)}_T \\ 鈫�{\left(\frac{\mathrm{鈭�}V}{\mathrm{鈭�}P}\right)}_T \\ =鈭�{魏}_{T}V \end{gathered}$$

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

$d{V}_2=鈭�{魏}_{T}VdP$

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

$$\begin{gathered} d{V}_1+d{V}_2=0 \\ 鈫�d{V}_1=鈭�d{V}_2 \end{gathered}$$

Substitute

$尾VdT={魏}_{T}VdP$

$鈫�{\left(\frac{\mathrm{鈭�}P}{\mathrm{鈭�}T}\right)}_V=\frac{尾}{{魏}_T}$

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

${\left(\frac{\mathrm{鈭�}P}{\mathrm{鈭�}T}\right)}_V=\frac{尾}{魏y}=\frac{\frac{1}{V}{\left(\frac{\mathrm{鈭�}V}{\mathrm{鈭�}T}\right)}_{\mathrm{螕}}}{\frac{1}{V}{\left(\frac{\mathrm{鈭�}V}{\mathrm{鈭�}P}\right)}_T}$

$鈫�{\left(\frac{\mathrm{鈭�}P}{\mathrm{鈭�}T}\right)}_V=鈭�\frac{{\left(\frac{\mathrm{鈭�}P}{\mathrm{鈭�}T}\right)}_T}{{\left(\frac{\mathrm{鈭�}V}{\mathrm{鈭�}P}\right)}_T}$

$鈫�{\left(\frac{\mathrm{鈭�}P}{\mathrm{鈭�}T}\right)}_V=鈭�\frac{(\mathrm{鈭�}V/\mathrm{鈭�}T{)}_P}{(\mathrm{鈭�}V/\mathrm{鈭�}P{)}_T}$

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

$尾=\frac{1}{V}{\left(\frac{\mathrm{鈭�}V}{\mathrm{鈭�}T}\right)}_p=\frac{1}{V}{\left(\frac{\mathrm{鈭�}\left(\frac{NkT}{P}\right)}{\mathrm{鈭�}T}\right)}_p$

$鈫�尾=\frac{Nk}{PV}=\frac{Nk}{NkT}=\frac{1}{T}$

$$\begin{gathered} {魏}_T=鈭�\frac{1}{V}{\left(\frac{\mathrm{鈭�}V}{\mathrm{鈭�}P}\right)}_T \\ =鈭�\frac{1}{V}{\left(\frac{\mathrm{鈭�}\left(\frac{NkT}{P}\right)}{\mathrm{鈭�}P}\right)}_T \\ =\frac{NkT}{V{P}^2} \end{gathered}$$

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

Divide

$鈫�\frac{P}{T}=\frac{尾}{{魏}_T}$

Now from equation

${\left(\frac{\mathrm{鈭�}P}{\mathrm{鈭�}T}\right)}_V=\frac{尾}{{魏}_T}$

$鈫�{\left(\frac{\mathrm{鈭�}\left(\frac{NkT}{V}\right)}{\mathrm{鈭�}T}\right)}_V=\frac{尾}{{魏}_T}$

$鈫�\frac{Nk}{V}=\frac{尾}{{魏}_T}$

$鈫�\frac{P}{T}=\frac{尾}{{魏}_T}$

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:

$\begin{array}{c}尾=2.57脳{10}^{鈭�4}{\mathrm{K}}^{鈭�1}\\ 魏=4.52脳{10}^{鈭�10}{\mathrm{Pa}}^{鈭�1}\end{array}$

So the pressure increase

$$\begin{gathered} \frac{P}{T}=\frac{尾}{{魏}_T} \\ 鈫�P=\frac{尾}{{魏}_T}T \\ 鈫�\mathrm{螖}P=\frac{尾}{{魏}_T}\mathrm{螖}T \end{gathered}$$

But the temperature difference is

$$\begin{gathered} \mathrm{螖}T=T_f鈭�T_i \\ =30鈭�20 \\ ={10}^{鈭�}\mathrm{K} \end{gathered}$$

So the pressure difference is

$$\begin{gathered} \mathrm{螖}P=\frac{尾}{{魏}_T}\mathrm{螖}T \\ =\frac{2.57脳{10}^{鈭�4}}{4.52脳{10}^{鈭�10}}脳10 \\ =5.686脳{10}^6\mathrm{Pa} \end{gathered}$$

$鈫�\mathrm{螖}P=5.686脳{10}^6\mathrm{Pa}$

For mercury the temperature range is

$\begin{array}{c}尾=1.81脳{10}^{鈭�4}{\mathrm{K}}^{鈭�1}\\ 魏=4.04脳{10}^{鈭�11}{\mathrm{Pa}}^{鈭�1}\end{array}$

So the pressure increase must be

$\begin{array}{r}\mathrm{螖}P=\frac{尾}{{魏}_T}\mathrm{螖}T=\frac{1.81脳{10}^{鈭�4}}{4.04脳{10}^{鈭�11}}脳10\\ 鈫�\mathrm{螖}P=4.48脳{10}^7\mathrm{Pa}\end{array}$