 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q. 1.46

Expert-verified Found in: Page 32 ### An Introduction to Thermal Physics

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 $\left(\mathrm{\partial }P/\mathrm{\partial }T{\right)}_{V}$, 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${\left(\frac{\mathrm{\partial }P}{\mathrm{\partial }T}\right)}_{V}=-\frac{\left(\mathrm{\partial }V/\mathrm{\partial }T{\right)}_{P}}{\left(\mathrm{\partial }V/\mathrm{\partial }P{\right)}_{T}}$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 $\beta ,{\kappa }_{T},\text{and}\left(\mathrm{\partial }P/\mathrm{\partial }T{\right)}_{V}$ for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).(e) For water at ${25}^{\circ }\mathrm{C},\beta =2.57×{10}^{-4}{\mathrm{K}}^{-1}\text{and}{\kappa }_{T}=4.52×{10}^{-10}{\mathrm{Pa}}^{-1}$. 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 $\text{(at}{25}^{\circ }\mathrm{C}\text{)}\beta =1.81×{10}^{-4}{\mathrm{K}}^{-1}$ and${\kappa }_{T}=4.04×{10}^{-11}{\mathrm{Pa}}^{-1}$ 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 ${\left(\frac{\mathrm{\partial }P}{\mathrm{\partial }T}\right)}_{V}=-\frac{\left(\mathrm{\partial }V/\mathrm{\partial }T{\right)}_{P}}{\left(\mathrm{\partial }V/\mathrm{\partial }P{\right)}_{T}}$

(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 $\mathrm{\Delta }{P}_{\text{water}}=5.686×{10}^{6}\mathrm{Pa},\phantom{\rule{1em}{0ex}}\mathrm{\Delta }{P}_{\text{mercury}}=4.48×{10}^{7}\mathrm{Pa}$

See the 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:

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

However, because$V$ is a function of $T\text{and}P,V\left(T,P\right)$, 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:

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

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

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

## 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:

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

Substitute

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

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

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

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

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

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

## 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:

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

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

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

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

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.

## Step :5  For water (part e)

(e) For water, the given values are:

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

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

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

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

For mercury the temperature range is

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

So the pressure increase must be

$\begin{array}{r}\mathrm{\Delta }P=\frac{\beta }{{\kappa }_{T}}\mathrm{\Delta }T=\frac{1.81×{10}^{-4}}{4.04×{10}^{-11}}×10\\ \to \mathrm{\Delta }P=4.48×{10}^{7}\mathrm{Pa}\end{array}$ ### Want to see more solutions like these? 