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Q. 1.46

Expert-verifiedFound in: Page 32

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

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 $p$ ?

(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 $\beta =\frac{1}{T},\phantom{\rule{1em}{0ex}}{\kappa}_{T}=\frac{1}{P}.\phantom{\rule{1em}{0ex}}\frac{P}{T}=\frac{\beta}{{\kappa}_{T}}$

(E) The heat capacities of substances constant is $\mathrm{\Delta}{P}_{\text{water}}=5.686\times {10}^{6}\mathrm{Pa},\phantom{\rule{1em}{0ex}}\mathrm{\Delta}{P}_{\text{mercury}}=4.48\times {10}^{7}\mathrm{Pa}$

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:

$\beta =\frac{\mathrm{\Delta}V/V}{\mathrm{\Delta}T}\phantom{\rule{0ex}{0ex}}=\frac{1}{V}{\left(\frac{\mathrm{\partial}V}{\mathrm{\partial}T}\right)}_{p}\phantom{\rule{0ex}{0ex}}\to {\left(\frac{\mathrm{\partial}V}{\mathrm{\partial}T}\right)}_{p}=\beta V$

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:

$d{V}_{1}=\beta VdT$

(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:

${\kappa}_{T}=-\frac{1}{V}{\left(\frac{\mathrm{\partial}V}{\mathrm{\partial}P}\right)}_{T}\phantom{\rule{0ex}{0ex}}\to {\left(\frac{\mathrm{\partial}V}{\mathrm{\partial}P}\right)}_{T}\phantom{\rule{0ex}{0ex}}=-{\kappa}_{T}V$

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

$d{V}_{2}=-{\kappa}_{T}VdP$

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

$d{V}_{1}+d{V}_{2}=0\phantom{\rule{0ex}{0ex}}\to d{V}_{1}=-d{V}_{2}$

Substitute

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

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

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

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

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

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

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

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

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

${\kappa}_{T}=-\frac{1}{V}{\left(\frac{\mathrm{\partial}V}{\mathrm{\partial}P}\right)}_{T}\phantom{\rule{0ex}{0ex}}=-\frac{1}{V}{\left(\frac{\mathrm{\partial}\left(\frac{NkT}{P}\right)}{\mathrm{\partial}P}\right)}_{T}\phantom{\rule{0ex}{0ex}}=\frac{NkT}{V{P}^{2}}$

$\to {\kappa}_{T}=\frac{NkT}{V{P}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{NkT}{\left(VP\right)P}\phantom{\rule{0ex}{0ex}}=\frac{NkT}{\left(NkT\right)P}\phantom{\rule{0ex}{0ex}}=\frac{1}{P}$

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.

(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

$\frac{P}{T}=\frac{\beta}{{\kappa}_{T}}\phantom{\rule{0ex}{0ex}}\to P=\frac{\beta}{{\kappa}_{T}}T\phantom{\rule{0ex}{0ex}}\to \mathrm{\Delta}P=\frac{\beta}{{\kappa}_{T}}\mathrm{\Delta}T$

But the temperature difference is

$\mathrm{\Delta}T={T}_{f}-{T}_{i}\phantom{\rule{0ex}{0ex}}=30-20\phantom{\rule{0ex}{0ex}}={10}^{\circ}\mathrm{K}$

So the pressure difference is

$\mathrm{\Delta}P=\frac{\beta}{{\kappa}_{T}}\mathrm{\Delta}T\phantom{\rule{0ex}{0ex}}=\frac{2.57\times {10}^{-4}}{4.52\times {10}^{-10}}\times 10\phantom{\rule{0ex}{0ex}}=5.686\times {10}^{6}\mathrm{Pa}$

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

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

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

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