Chapter 13, Functions and methods Video Solutions, Thermodynamics: A complete undergraduate course

Problem 1

gives expressions for the work done on various simple systems in a reversible change. We have already discussed a simple compressible fluid. An elastic rod is described to good approximation by its length $L$ and the tension $f$. To stretch the rod by $\mathrm{d} L$ against this force, the work done is clearly $f \mathrm{~d} L$. The surface tension in a liquid surface is, by definition, the parameter $\sigma$ such that the work required to increase the area of the surface by $\mathrm{d} A$ is $\sigma \mathrm{d} A$. "These cases are all straightforward. The expressions for electric and magnetic systems are discussed at the end of the chapter (Section 14.5).

Eric Mockensturm

Numerade Educator

06:44

Problem 1

A capacitor is filled with a linear homogenous dielectric slab with constant relative permittivity $\epsilon_r$, area $A$, and thickness $x$. A variable voltage source is connected, and used to raise the charge on the capacitor from $\theta$ to $q$ in a quasistatic adiabatic process. Find the work done by the source and the work done on the dielectric material.
Solution.
When the charge on the capacitor is $q$, the relevant quantities have the following values:
fields in the medium
total dipole moment
voltage between the plates
'applied field'
'applicd ficid'
work done by the source
$D=q / A$
$p=q x\left(1-1 / \epsilon_r\right)$
$\phi=x \mathbb{E}$.
$E_0=q / A \epsilon_0 \quad=\epsilon, E$
$E_{0 \phi}=\phi / x \quad=E$.
$W=\int \phi \mathrm{d} q \quad=x q^2 / 2 A \epsilon_0 \epsilon_r=q \phi / 2$
The last line answers the first part of the question. The second part of the question is ambiguous, so has no single answer. On one model, the energy assigned to the 'applied ficld' is $(1 / 2) \epsilon_0 E_0^2 A x=\epsilon_r q \phi / 2$ and the energy accounted as work on the medium is
$$
W_p=\int-p d E_0=\frac{1}{2} q \phi\left(1-\epsilon_\eta\right)
$$
On the other model, the energy assigned to the 'applied field' is $(1 / 2) \epsilon_0 E_{0,2}^2 A x=4 \phi / 2 \epsilon$, and the energy accounted as work on the medium is
$$
W_\phi=\int E_{0 \phi} \mathrm{d} p=\frac{1}{2} q \phi\left(1-1 / \epsilon_r\right)
$$

Khoobchandra Agrawal

Numerade Educator

01:22

Problem 1

Use the Gibbs-Duhem relationship to show that $(\partial \mu / \partial N)_{D, T}=0$.

Penny Riley

Numerade Educator

04:00

Problem 2

When a long, straight object such as a rod or wire is stretched, most of the work is done against the tension in the rod or wire. There is also a small contribution coming from pressure forces acting on the curved surface, but these can be neglected to good approximation under many circumstances. We can therefore treat a rod of length $L$ as a closed thermodynamic system with fundamental relation
$$
\mathrm{d} U=T \mathrm{dS}+f \mathrm{~d} L
$$
The natural length $L_0$ is the length when the tension falls to zero. The strain is defined as the fractional extension, $\mathrm{d} L / L_0$ when the rod is put into tension. The stress is defined as the force per unit area, $d / / A$. The Young's modulus is defined as the ratio of stress to strain. Hence the isothermal Young's modulus is
$$
E_T=\left.\frac{L}{A} \frac{\partial f}{\partial L}\right|_T .
$$
Thermidymawic: A Cawpleve Lindergnahwer Cowrse Andrew M. Steane.
C Andrew M. Steane 2017. Published 2017 by Oxaford Lniversity Press.

Khoobchandra Agrawal

Numerade Educator

01:57

Problem 2

A closed simple compressible system has work $\Delta W$ done on it in a reversible process without heat exchange. (i) By how much does the internal energy change? (ii) Is it possible to tell, from the given information, the change in the Helmholtz function, enthalpy, and Gibbs function? (iii) Suppose the work were done in conditions of constant temperature. Now what can you say about $U, F, H, G$ ?

Dheeraj Sharma

Numerade Educator

05:35

Problem 3

Surface tension is the tendency of the surface of a body to oppose being stretched. It can be exhibited by solids and gases, but it is most readily observed at the interface between a liquid and a gas, because such surfaces can be easily deformed. The work required to increase the area of the surface by a small amount $\mathrm{d} A$ is given by
$$
\mathrm{dW}=\sigma \mathrm{d} A
$$
where $\sigma$ is the surface tension. Surface tension is typically a strong function of temperature, but almost independent of pressure. It must vanish at the critical point (where the liquid and vapour become indistinguishable; see Chapters 15 and 22). A reasonable approximation for many liquids (c.g. water, see Figure 14.1) is Eötves's rule:
$$
\sigma \simeq k V_{\mathrm{M}}^{-2 / 3}\left(T_{\mathrm{c}}-T\right),
$$
where $V_M$ is the molar volume, $T_c$ is the critical temperature, and $k$ is a constant which has approximately the same value for many liquids, $k \simeq 2.1 \times$ $10^{-7} \mathrm{JK}^{-1} \mathrm{~mol}^{-2 / 3}$. For water, $V_{\mathrm{M}}=18.0 \mathrm{~cm}^3$ and $T_{\mathrm{c}}=647.096 \mathrm{~K}$.

Consider the situation shown in Figure 14.2. Let $V=(4 / 3) \pi r^2$ be the volume of the spherical drop, $p$ be the pressure inside the drop, and $p_0$ the ambient pressure. When the plunger is pushed down by a small amount, the volume of the spherical drop gets larger by $\mathrm{d} V$ and the plunger does work $p \mathrm{~d} V$ if we assume the liquid is incompressible. The net work done on the system by all the external forces is therefore $$
\mathrm{d} W=p \mathrm{~d} V-p_0 \mathrm{~d} V
$$
If the system is in mechanical equilibrium, this energy is preciscly equal to the work required to increase the surface area, given by $(14,7)$ :
$$
\left(p-p_0\right) \mathrm{d} V=\sigma \mathrm{d} A
$$
For a spherical drop of radius $r$, we have $A=4 \pi r^2$ and $V=(4 / 3) \pi r^3$, so $\mathrm{d} A / \mathrm{d} r=8 \pi r$ and $\mathrm{d} V / \mathrm{d} r=4 \pi r^2$. Therefore the pressure difference is given by
$$
\left(p-p_0\right)=\frac{2 \sigma}{r}
$$
In the above we have neglected heat. This is correct when we are merely interested in a balance of mechanical forces: we may use the principle of virtual work. From a thermodynamic point of view this amounts to exploring the out of equilibrium situations near to the equilibrium point; this will be explained in Chapter 17.
If we now consider instead of a liquid drop, a bubble, then there are two surfaces to account for. Assuming the bubble wall is thin, so that the inner and outer surfaces have approximately the same radii, one has
$$
\left(p-p_0\right)=\frac{4 \sigma}{r}
$$
Although the complete system in these examples consists of both the liquid and its surface, we may choose the surface itself as our thermodynamic system, and write the fundamental relation
$$
\mathrm{d} U=T \mathrm{~d} S+\sigma \mathrm{d} A
$$

Khoobchandra Agrawal

Numerade Educator

05:21

Problem 3

Derive all the Maxwell relations.

Mohammad Mehran

Numerade Educator

05:33

Problem 4

The subject of paramagnetism was introduced in Section 6.2.4. An experimental setup is shown schematically in Fugure 6.6, and with some further details in Figure 14.7. The quantities that typically we are chicfly interested in are the heat emitted or absorbed during an isothermal process, and the change in temperature during an adiabatic process. The first is quantified by
$$
\frac{\mathrm{d} Q_T}{\mathrm{~d} B}=\left.T \frac{\partial S}{\partial B}\right|_T,
$$ $$
\left.\frac{\partial T}{\partial B}\right|_s
$$
where $B$ is the magnetic field. We would like to relate these quantities to the equation of state and the heat capacities.

Before proceeding we must note that the above expressions are ambiguous as they stand, because there are several magnetic fields in the problem. These are:
1. The ficld $\mathbf{B}_{\text {i inside the sample. }}$
2. The field $\mathbf{H}_i=\left(1 / \mu_0\right) \mathbf{B}_i-\mathbf{M}$ inside the sample, where $\mathbf{M}$ is the magnetization.
3. The fields $\mathbf{B}^{\prime}$ and $\mathbf{H}^{\prime}$ outside the sample.
4. The field $\mathbf{B}_0$ that would be present in the solenoid if the sample were removed while keeping the total flux $\Phi$ in the solenoid constant.
5. The field $\mathbf{B}_{\mathrm{o}}$ that would be present in the solenoid if the sample were removed while keeping the current $I$ in the solenoid constant.

In the limit of small magnetization, all of these agree (apart from the constant factor $\mu_0$ between $\mathbf{B}$ and $\mathbf{H}$ ), and therefore the distinction between them does not need to be made when treating weakly magnetized materials. More generally, all the above may feature in the treatment of magnetic systems, though $\mathbf{B}^{\prime}$ and $\mathbf{H}$ are of little interest and seldom used.

A given sample will respond to the field present in the sample, so the magnetic susceptibility $\chi$ is defined by
$$
x=\frac{M}{H_1},
$$
where $M=|\mathbf{M}|$ and $H_i=\left|\mathbf{H}_i\right|$. The fields $\mathbf{B}_0$ and $\mathbf{B}_{0 l}$ are useful for simplifying expressions for magnetic work, as explained in Section 14.5. In the simplest geometry they are related by
$$
B_0=\mu_r B_{0 /}
$$
where $\mu_r=1+\chi$ is the relative permeability of the medium, and then $H_i=H_{O O}$. This is why one often sees equation (14.19) quoted with $H_0 l$ instead of $H_i$ on the right-hand side. For a more general geometry, see Exercises $14.8-14.10$. Either of the fields $\mathbf{B}_0$ and $\mathbf{B}_{0 /}$ may be termed 'the external field' or 'the applied field' but for a precise treatment one must specify which is under consideration. Unfortunately, both these descriptive phrases are somewhat misleading, because $\mathbf{B}_0$ and $\mathbf{B}_{0 /}$ do not refer to fields anywhere in the apparatus when the sample is present. However in the remainder of this section we will limit ourselves to the

Keshav Singh

Numerade Educator

03:02

Problem 4

Sketch an isotherm and an adiabat passing through a given point $p, V$ on the indicator diagram for a gas. Hence show that more work energy can be extracted from a gas in an isothermal expansion than in an adiabatic one. Where has the energy for this extra work come from?

Supratim Pal

Numerade Educator

01:28

Problem 5

Give an example of a physical process which can take place at constant pressure and temperature. What thermodynamic potential is unchanged in such a process?

Jerrah Biggerstaff

Numerade Educator

08:40

Problem 5

In this section we consider the performance of work on two types of system: a dielectric medium polarized by an electric field, and a magnetic medium magnetized by a magnetic field., The subject has given rise to very much confusion in the scientific community for a long period, and there remain misleading or ambiguous treatments in some recent textbooks. The reason for this confusion will become apparent as we go along, but in the end there should be no remaining doubt as to what are the correct expressions and what are the physical circumstances to which they apply.
To begin, consider the following puzzle.
Take an ordinary parallel plate capacitor with plates of area $A$ and separation $x$, with charge $q$ on one plate, and charge $-q$ on the other (Figure 14.8). By using Gauss' law one finds that the electric field between the plates is $E=q / \epsilon_0 A$. An electric field in vacuum has an energy density $(1 / 2) \epsilon_0 E^2$, where $\epsilon_0$ is the permittivity of free space, so the total energy of the field between the capacitor plates (treating it as uniform for simplicity) is
$$
U=\frac{1}{2} \epsilon_0 E^2 A x=\frac{q^2}{2 \epsilon_0 A} x
$$
Suppose one of the plates is displaced by dx. The energy of the system rises, and we can use this to find the force on the plate:
$$
\frac{\mathrm{d} U}{\mathrm{~d} x}=\frac{q^2}{2 \epsilon_0 A}=\frac{1}{2} q E
$$
Since $U$ increases with $x$, this is an attractive force.
Now consider the electric potential (or voltage) $\phi$ between the plates. It is given by $\phi=E x$, so we can write
$$
U=\frac{1}{2} \epsilon_0 \frac{\phi^2}{x^2} A x=\frac{\epsilon_0 \phi^2 A}{2 x}
$$
which gives
$$
\frac{\mathrm{d} U}{\mathrm{~d} x}=-\frac{\epsilon_0 \phi^2 A}{2 \mathrm{x}^2}=-\frac{1}{2} q E .
$$
Now it seems as if the force between the plates is repulsive, because the energy falls as $x$ increases. But this contradicts our previous conclusion, so the puzzle is: what is really going on here?

Sarah Mccrumb

Numerade Educator

Problem 6

A certain simple compressible system has Helmholtz function given by $F=-k_{\mathrm{B}} T \ln \left(z_1^N / N !\right)$, where $z_1(T, V)$ is independent of $N$. (i) Assuming $N$ is large so that the Stirling approximation $\ln N ! \simeq N \ln N-N$ is valid, show that the equation of state is $p V=N k_{\mathrm{B}} T$.

Check back soon!

02:53

Problem 7

Prove the following homogencity relations:
$$
\begin{array}{r}
F(T, \lambda V, \lambda N)=\lambda F(T, V, N) \\
H(\lambda S, p, \lambda N)=\lambda H(S, p, N) \\
G(T, p, \lambda N)=\lambda G(T, p, N) .
\end{array}
$$
Hence show that $F(T, V, N)=\left(N / N_0\right) F\left(T, N_0 V / N, N_0\right)$.

Amany Waheeb

Numerade Educator

01:51

Problem 8

(a) Show that, if only the equation of state of a system is known, then it is possible to calculate the work done in an isothermal process, but not in an adiabatic one.
(b) If both the equation of state and $C_V$ are known, show how the work done in an adiabatic process could be obtained.

Prem Bijarniya

Numerade Educator

01:30

Problem 9

Show that the rate of change of $p$ with $T$ in an adiabatic compression may be expressed as $(\partial P / \partial T)_s=\left(C_p / T\right)(\partial T / \partial V)_p$.

Sheh Lit Chang

University of Washington

Problem 10

Obtain equation (13.35).

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Problem 11

Compare some of the measured values in Table 9.1 with the prediction from equation $(13.50)$.

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02:19

Problem 12

Confirm that (13.42) may equally be written
$$
U=\frac{N^\gamma}{(a V)^{\gamma-1}} e^{(\gamma-1) S / N k_{\mathrm{B}}}
$$
and use this to obtain the equation of state.

Narayan Hari

Numerade Educator

02:48

Problem 13

Show that the Helmholtz function of an ideal gas of constant heat capacity is given by
$$
F(T, V, N)=\frac{N k_{\mathrm{B}} T}{\gamma-1}\left(1-\ln \frac{k_{\mathrm{B}} T}{\gamma-1}\right)-N h_{\mathrm{B}} T \ln a \frac{V}{N}
$$

Narayan Hari

Numerade Educator

02:41

Problem 14

Starting from the Sackur-Tetrode equation, obtain an expression for the pressure of an ideal gas as a function of temperature and chemical potential. Hence confirm that $(\partial p / \partial T)_\mu=S / V$ for this system (an example of the Gibbs-Duhem relation).

Robert Zaballa

Numerade Educator

16:43

Problem 15

A piece of rubber of length $L$ is subject to work by hydrostatic pressure and a tensional force $f$.
(i) Construct an expression for $\mathrm{d} U$.
(ii) Generate the potentials which have as proper variables $(\mathrm{S}, \mathrm{V}, \mathrm{f})$ and $(\mathrm{S}, \mathrm{p}, \mathrm{f})$.
(iii) Derive the Maxwell relation (first developing any potential you may need)
$$
\left.\frac{\partial S}{\partial L}\right|_{T, p}=-\left.\frac{\partial f}{\partial T}\right|_{R, L}
$$

João Gabriel Alencar Caribé

Numerade Educator