Chapter 28, Thermoelectricity and entropy flow Video Solutions, Thermodynamics: A complete undergraduate course

Problem 1

Another look at Thomson's argument.
(i) Starting from equations (28.15)-(28.18), obtain
$$
\frac{\mathrm{d} S}{\mathrm{~d} t}=f_U \frac{\mathrm{d} U_1}{\mathrm{~d} t}+f_N \frac{\mathrm{d} N_1}{\mathrm{~d} t}
$$ and hence
$$
\frac{\mathrm{d} S}{\mathrm{~d} t}=L_{U U} f_U^2+L_{\mathrm{NN}} f_N^2+\left(L_{U N}+L_{\mathrm{NU}}\right) f_U f_N
$$
(making no assumption about the symmetry of $L_{i j}$ ).
(ii) From (28.18), obtain $f_N=\left(\dot{N}_1-L_{N U} f_U\right) / L_{N N}$ and hence
$$
\begin{aligned}
\frac{\mathrm{d} S}{\mathrm{~d} t}= & \left(L_{U U}-\frac{L_{U N} L_{N U}}{L_{N N}}\right) f_U^2 \\
& +\left(\frac{L_{U N}-L_{N U}}{L_{N N}}\right) f_U \dot{N}_1+\frac{\dot{N}_1^2}{L_{N N}}
\end{aligned}
$$
Applying this to thermoelectricity, we may note that when no current flows, $\dot{N}_1=0$ and in this case only the first term on the right-hand side survives. It follows that this term must describe entropy production associated with ordinary thermal conduction owing to a temperature gradient (the temperature gradient is expressed by $\left.f_U\right)$. Similarly, when $f_U=0$ only the third term survives, so this must be the term describing Ohmic or resistive heating. To derive the second Thomson relation using equation (28.9), Thomson assumed that these two processes account for all the entropy production-that is, the reversible processes do not themselves generate entropy, and this is still true even in the presence of the irreversible processes. We can now see that Thomson's assumption is valid if and only if $L_{U N}=L_{N U}$.

Khoobchandra Agrawal

Numerade Educator

01:14

Problem 2

Entrained entropy. Show that the rate at which entropy enters region 1 in Figure 28.5 is
$$
\frac{\mathrm{d} S_1}{\mathrm{~d} t}=\frac{1}{T}\left[\left(L_{U U}-\frac{L_{U N} L_{N U}}{L_{N N}}\right) f_U+\left(\frac{L_{U N}}{L_{N N}}-\mu\right) \dot{N}_1\right] .
$$
Interpret the first term, and hence show that the second term is the entrained entropy, that is, the extra entropy carried along by the particle current. Note, this is not the same as (and has little relation to) the equilibrium entropy per particle in the fluid.

Ajay Singhal

Numerade Educator

02:51

Problem 3

A thermodynamic argument for the reciprocal relation. Consider a situation of steady-state flow between two reservoirs separated by a thin membrane of thickness $\delta x$. The membrane is not perfectly conducting but does allow heat and particles, etc. to pass. We assume that the flow is slow enough for each reservoir to be characterized by a well-defined temperature, pressure etc., even though it is slow brane is in steady state, so a
Page 433 i 464 other of the reservoirs. (i) Show that the rate of entropy generation in the whole system, per unit volume of the membrane, is
$$
\frac{\mathrm{d} s}{\mathrm{~d} t}=\sum_i\left(\nabla \phi_i\right) \cdot \mathbf{j}_i,
$$
where the potentials $\phi_i$ and currents $\mathbf{j}_i$ are as described in (28.31) and (28.32). This implies that $\dot{s}$ can be regarded as a function of the gradients and the currents:
$$
\dot{s}=\dot{s}\left(\boldsymbol{\phi}_1^{\prime}, \boldsymbol{\phi}_2^{\prime}, \ldots, \mathbf{j}_1, \mathbf{j}_2, \ldots\right),
$$
where $\phi_i^{\prime} \equiv \nabla \phi_i$.
(ii) From (28.36), show that $\mathbf{j}_i=\left(\partial \dot{s} / \partial \phi_i^{\prime}\right)$ and hence (using (28.32)) obtain
$$
L_{i k}=\frac{\partial^2 \dot{s}}{\partial \phi_k^{\prime} \partial \phi_i^{\prime}} .
$$
Similarly,
$$
L_{k i}=\frac{\partial^2 \dot{s}}{\partial \phi_i^{\prime} \partial \phi_k^{\prime}} .
$$
Since $\dot{s}$ is a well-behaved analytic function, it follows that $L_{k i}=L_{i k}$. Thus we appear to have obtained Onsager's reciprocal relation without the need to consider thermal fluctuations, and without the need to invoke the regression hypothesis or the principle of microscopic reversibility.
(iii) The above argument is plausible. However, when Thomson and others first discussed this subject, they were careful to raise questions over whether this type of argument is valid, because equilibrium concepts are being applied to out of equilibrium processes. ${ }^5$ Does the presence of the flow invalidate the use of the concepts being used to describe it? Consider the case of not one thin layer, but many adjacent thin layers. Then temperature, and other intensive properties, are being ascribed to a system with a non-negligible temperature gradient, through which energy and other quantities are flowing, and within which entropy is being generated. Is (28.36) correct, and does it involve any hidden assumptions?

Sana Riaz

Numerade Educator