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Helmholtz energy of a hydrogen atom

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.20.

The Helmholtz energy is defined as

\displaystyle F\equiv U-TS \ \ \ \ \ (1)

We can apply this to a hydrogen atom, at least in an approximate form. The ground state of hydrogen has quantum numbers {n=1}, {\ell=m=0} and is non-degenerate. The first excited state is four-fold degenerate, as its quantum numbers are {n=2} with {\ell=m=0} or {\ell=1} and {m=\pm1,0}. We can therefore say that the entropy of the {n=2} state is

\displaystyle S=k\ln4 \ \ \ \ \ (2)

with {k=8.62\times10^{-5}\mbox{ eV K}^{-1}}.

If we take the energy of the ground state to be zero, then the first excited state has {U=10.2\mbox{ eV}}. The Helmholtz energy is therefore zero at a temperature of

\displaystyle T_{0}=\frac{U}{S}=\frac{10.2}{\left(8.62\times10^{-5}\right)\ln4}=8.54\times10^{4}\mbox{ K} \ \ \ \ \ (3)

For {T>T_{0}}, {F<0} so the excited state is actually the preferred state, and a hydrogen atom in the ground state would spontaneously make the transition to the excited state. However, {T_{0}} is so large that virtually all hydrogen atoms would be ionized at that temperature, as we saw earlier when discussing the Saha equation for stellar atmospheres. We saw there that the fraction of hydrogen atoms that are ionized is essentially 1.0 for temperatures above around 12000 K.

Helmholtz energy as a function of volume

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.19.

From the thermodynamic identity for the Helmholtz free energy we can obtain the derivative

\displaystyle \left(\frac{\partial F}{\partial V}\right)_{T,N}=-P \ \ \ \ \ (1)

 

The Helmholtz energy is defined as

\displaystyle F\equiv U-TS \ \ \ \ \ (2)

To get an intuitive feel for 1, we recall the definition of temperature in terms of the derivative of entropy. The idea there was that if we have two interacting systems, the equilibrium state is the one with the maximum overall entropy. To apply this idea to the current case, we can use the fact that if we have two interacting systems, the equilibrium state is the one with the minimum Helmholtz energy. So suppose we have two systems that each have a fixed number of molecules and that both systems are in contact with a thermal reservoir so their temperatures are equal and constant. Also suppose that these two systems are allowed to exchange volume (say, by means of a moveable boundary between them), but that the total volume is constant.

The condition that the total free energy is minimum is then

\displaystyle \left(\frac{\partial F_{total}}{\partial V}\right)_{T,N} \displaystyle = \displaystyle 0\ \ \ \ \ (3)
\displaystyle \displaystyle = \displaystyle \left(\frac{\partial F_{A}}{\partial V}\right)_{T,N}-\left(\frac{\partial F_{B}}{\partial V}\right)_{T,N}\ \ \ \ \ (4)
\displaystyle \left(\frac{\partial F_{A}}{\partial V}\right)_{T,N} \displaystyle = \displaystyle \left(\frac{\partial F_{B}}{\partial V}\right)_{T,N} \ \ \ \ \ (5)

where the minus sign in the second line is because an increase in volume {A} means a decrease in volume {B}. Thus we end up with an equilibrium condition that the magnitudes of the slopes of a graph of {F} versus {V} are equal. Since the units of this slope are {\mbox{(energy)}/\mbox{(volume)}=\mbox{(pressure)}} we can postulate that this derivative is proportional to the pressure.

A qualitative plot of {F} versus {V} for given values of {T} and {N} looks like this (for two different values of {N}):

With {N} and {T} fixed, an increase in volume usually implies an increase in entropy, so we’d expect {F} to decrease as volume is increased, which gives the general shape of the curve. Thus in general {\left(\frac{\partial F}{\partial V}\right)_{T,N}<0} which explains the minus sign in 1.

With a higher value of {N}, at a given temperature and volume, both {U} and {S} increase (in roughly the same proportion) so {F} would increase with {N}, thus the blue curve is for higher {N}. Note that the slope of the blue curve is steeper than that of the red curve, which also makes sense since increasing {N} within a fixed volume increases the pressure.

Helmholtz and Gibbs energies are minimum at equilibrium

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.18.

The Helmholtz and Gibbs free energies both tend to decrease for a system that is allowed to come to equilibrium while being in contact with a large thermal reservoir. Schroeder gives the derivation for a system at constant volume {V} in thermal contact with a reservoir at temperature {T} and shows that the total entropy change (of system + reservoir) is

\displaystyle  dS_{total}=-\frac{1}{T}dF \ \ \ \ \ (1)

For a system at constant pressure, the derivation is similar, so here goes. The total entropy change is

\displaystyle  dS_{total}=dS+dS_{R} \ \ \ \ \ (2)

where {dS} refers to the system and {dS_{R}} to the reservoir. Using the thermodynamic identity for energy {U}:

\displaystyle  dU=T\;dS-P\;dV+\mu\;dN \ \ \ \ \ (3)

we have for a reservoir at constant pressure {P} and particle number {N}:

\displaystyle  dS_{R}=\frac{1}{T}dU_{R}+\frac{P}{T}dV_{R} \ \ \ \ \ (4)

Since the entire system is at constant pressure, {P} is the same for the system and the reservoir and, since the whole system is at thermal equilibrium, {T} is the same as well. Thus the energy lost (or gained) by the system is gained (or lost) by the reservoir (by conservation of energy) and likewise for the volume, so {dU_{R}=-dU} and {dV_{R}=-dV}. Therefore

\displaystyle   dS_{R} \displaystyle  = \displaystyle  -\frac{1}{T}dU-\frac{P}{T}dV\ \ \ \ \ (5)
\displaystyle  dS_{total} \displaystyle  = \displaystyle  dS-\frac{1}{T}dU-\frac{P}{T}dV\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{T}\left(dU-T\;dS+P\;dV\right)\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{T}dG \ \ \ \ \ (8)

Note that these results have expressed the total entropy change of the system + reservoir in terms of the state variables of the system alone.

This derivation seems quite dodgy to me, since {dU_{R}=-dU} and {dV_{R}=-dV} that {dS=-dS_{R}} and therefore {dS_{total}=dG=0}. A derivation that makes more sense (to me, anyway) goes like this. Rather than consider differentials, we’ll consider the actual entropies and energies of the system and reservoir. Then the entropy of the universe is now

\displaystyle  S_{total}=S+S_{R} \ \ \ \ \ (9)

At equilibrium, {S_{total}} must be a maximum, from the second law. The energies obey the relation

\displaystyle  U_{total}=U+U_{R} \ \ \ \ \ (10)

And the volumes:

\displaystyle  V_{total}=V+V_{R} \ \ \ \ \ (11)

Taking the particle numbers {N} and {N_{R}} to be constants, the entropy of the reservoir is, for constant pressure:

\displaystyle   S_{R} \displaystyle  = \displaystyle  \frac{H_{R}}{T}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{U_{R}}{T}+\frac{PV_{R}}{T}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{U_{total}-U}{T}+\frac{P}{T}\left(V_{total}-V\right) \ \ \ \ \ (14)

The total entropy is then

\displaystyle  S_{total}=S-\frac{1}{T}\left(U+PV\right)+\frac{1}{T}\left(U_{total}+PV_{total}\right) \ \ \ \ \ (15)

The last term is a constant, so if we want to maximize {S_{total}}, we need maximize only the first two terms on the RHS. That is

\displaystyle  \max\left[S_{total}\right]=\frac{1}{T}\max\left[TS-U-PV\right]+\frac{1}{T}\left(U_{total}+PV_{total}\right) \ \ \ \ \ (16)

Since {G=U-TS+PV}, this is equivalent to

\displaystyle  \frac{1}{T}\max\left[TS-U-PV\right]=-\frac{\min\left[G\right]}{T} \ \ \ \ \ (17)

In other words, for a system at constant number, temperature and pressure, maximizing the total entropy of the universe requires minimizing {G}.

A similar analysis at constant number, temperature and volume results in minimizing {F=U-TS}, since in that case the second term in 13 becomes a constant and we must maximize {TS-U=-F} instead of {TS-U-PV}.

To minimize {F}, we must reduce {U} and/or increase {S}. For example, if we drop a brick onto the ground from some height, then as it is falling, the brick’s energy (potential + kinetic, neglecting air resistance) remains constant, but when it hits the ground (ground = reservoir), its kinetic energy is reduced to zero while its potential energy remains constant at the value it has at ground level. From this point of view, the brick has certainly lost energy, but the kinetic energy just gets redistributed into thermal motion of the molecules that make up the brick and the patch of ground where the brick landed. Thus an amount of energy {dU} is transmitted from the brick to the reservoir.

Isothermal and isentropic compressibilities

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.16.

An expression similar to that relating the heat capacities can be derived to relate the isothermal and isentropic compressibilities {\kappa_{T}} and {\kappa_{S}}, defined as

\displaystyle   \kappa_{T} \displaystyle  \equiv \displaystyle  -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}\ \ \ \ \ (1)
\displaystyle  \kappa_{S} \displaystyle  \equiv \displaystyle  -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{S} \ \ \ \ \ (2)

These quantities measure the fractional change in volume of a substance in response to a change in pressure. To obtain the relation between them, we use a method similar to that for heat capacities {C_{V}} and {C_{P}}.

If we write {S=S\left(P,T\right)} then

\displaystyle  dS=\left(\frac{\partial S}{\partial P}\right)_{T}dP+\left(\frac{\partial S}{\partial T}\right)_{P}dT \ \ \ \ \ (3)

Also, starting with {V=V\left(P,S\right)} we have

\displaystyle  dV=\left(\frac{\partial V}{\partial P}\right)_{S}dP+\left(\frac{\partial V}{\partial S}\right)_{P}dS \ \ \ \ \ (4)

Substituting 3 into 4 we get

\displaystyle  dV=\left[\left(\frac{\partial V}{\partial S}\right)_{P}\left(\frac{\partial S}{\partial P}\right)_{T}+\left(\frac{\partial V}{\partial P}\right)_{S}\right]dP+\left(\frac{\partial V}{\partial T}\right)_{P}dT \ \ \ \ \ (5)

At constant temperature {dT=0} and we get

\displaystyle   \left(\frac{\partial V}{\partial P}\right)_{T} \displaystyle  = \displaystyle  \left(\frac{\partial V}{\partial S}\right)_{P}\left(\frac{\partial S}{\partial P}\right)_{T}+\left(\frac{\partial V}{\partial P}\right)_{S}\ \ \ \ \ (6)
\displaystyle  -V\kappa_{T} \displaystyle  = \displaystyle  \left(\frac{\partial V}{\partial S}\right)_{P}\left(\frac{\partial S}{\partial P}\right)_{T}-V\kappa_{S} \ \ \ \ \ (7)

From the Maxwell relation from the Gibbs energy

\displaystyle  \left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P} \ \ \ \ \ (8)

Also, from the definition of the thermal expansion coefficient {\beta}

\displaystyle  \beta\equiv\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P} \ \ \ \ \ (9)

Combining these last two equations gives

\displaystyle  -V\kappa_{T}=-\beta V\left(\frac{\partial V}{\partial S}\right)_{P}-V\kappa_{S} \ \ \ \ \ (10)

To get rid of the last partial derivative, we observe that the volume change {dV} due to a temperature change {dT} at constant pressure is

\displaystyle  dV=\beta V\;dV \ \ \ \ \ (11)

The entropy change due to an influx of heat {dQ} at constant pressure at temperature {T} is

\displaystyle   dS \displaystyle  = \displaystyle  \frac{dQ}{T}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  C_{P}\frac{dT}{T} \ \ \ \ \ (13)

Dividing these two relations gives

\displaystyle  \left(\frac{\partial V}{\partial S}\right)_{P}=\frac{TV\beta}{C_{P}} \ \ \ \ \ (14)

Inserting this into 10 and cancelling off a factor of {-V} gives the final result

\displaystyle  \kappa_{T}=\kappa_{S}+\frac{TV\beta^{2}}{C_{P}} \ \ \ \ \ (15)

For an ideal gas, we can use this equation to work out {\kappa_{S}}:

\displaystyle   \beta \displaystyle  = \displaystyle  \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}=\frac{Nk}{PV}=\frac{1}{T}\ \ \ \ \ (16)
\displaystyle  \kappa_{T} \displaystyle  = \displaystyle  -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}=\frac{NkT}{P^{2}V}=\frac{1}{P}\ \ \ \ \ (17)
\displaystyle  C_{P} \displaystyle  = \displaystyle  C_{V}+Nk\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  Nk\left(1+\frac{f}{2}\right)\ \ \ \ \ (19)
\displaystyle  \kappa_{S} \displaystyle  = \displaystyle  \frac{1}{P}-\frac{V}{NkT\left(1+\frac{f}{2}\right)}\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{P}\frac{f}{\left(f+2\right)} \ \ \ \ \ (21)

where in the third line, we’ve used Schroeder’s equation 1.48, and {f} is the number of degrees of freedom of each gas molecule.

To check this, recall that for an isentropic (adiabatic) process in an ideal gas

\displaystyle   PV^{\gamma} \displaystyle  = \displaystyle  K\ \ \ \ \ (22)
\displaystyle  V \displaystyle  = \displaystyle  \left(\frac{K}{P}\right)^{1/\gamma} \ \ \ \ \ (23)

where {\gamma=\left(f+2\right)/f} and {K} is a constant. So

\displaystyle   \kappa_{S} \displaystyle  = \displaystyle  -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{S}\ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  -\left(\frac{P}{K}\right)^{1/\gamma}\left(-\frac{1}{\gamma}\right)\left(\frac{K}{P}\right)^{1/\gamma}\frac{1}{P}\ \ \ \ \ (25)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{P\gamma}=\frac{1}{P}\frac{f}{\left(f+2\right)} \ \ \ \ \ (26)

which is the same as 21, so equation 15 checks out for an ideal gas.

Heat capacities using Maxwell relations

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 5.13 – 5.15.

We can use the Maxwell relations to derive some formulas relating to heat capacities. First, recall the thermal expansion coefficient

\displaystyle  \beta\equiv\frac{\Delta V/V}{\Delta T} \ \ \ \ \ (1)

which is the fractional change in volume per Kelvin, assumed to be a constant pressure. For small changes, we can write this as a partial derivative:

\displaystyle  \beta=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P} \ \ \ \ \ (2)

From the Maxwell relation derived from the Gibbs energy:

\displaystyle  \left(\frac{\partial V}{\partial T}\right)_{P}=-\left(\frac{\partial S}{\partial P}\right)_{T} \ \ \ \ \ (3)

so

\displaystyle  \beta=-\frac{1}{V}\left(\frac{\partial S}{\partial P}\right)_{T} \ \ \ \ \ (4)

The third law of thermodynamics says that as {T\rightarrow0}, {S\rightarrow0}. I’m not quite sure how we can use this to show that {\beta\rightarrow0} as {T\rightarrow0} since it’s not clear how entropy depends on pressure at low temperatures. However, if {S\rightarrow0} for any system as {T\rightarrow0}, then presumably it must be true no matter what the pressure is, so in that sense, {S} is independent of pressure and then {\left(\frac{\partial S}{\partial P}\right)_{T}=0}. I’m not sure that constitutes a ‘proof’ as requested in Schroeder’s problem, though.

We’re on firmer ground when we wish to derive a relation between the heat capacities {C_{V}} (constant volume) and {C_{P}} (constant pressure). In terms of entropy, they are

\displaystyle   C_{V} \displaystyle  = \displaystyle  T\left(\frac{\partial S}{\partial T}\right)_{V}\ \ \ \ \ (5)
\displaystyle  C_{P} \displaystyle  = \displaystyle  T\left(\frac{\partial S}{\partial T}\right)_{P} \ \ \ \ \ (6)

If we write {S=S\left(V,T\right)} then

\displaystyle  dS=\left(\frac{\partial S}{\partial V}\right)_{T}dV+\left(\frac{\partial S}{\partial T}\right)_{V}dT \ \ \ \ \ (7)

Also, starting with {V=V\left(P,T\right)} we have

\displaystyle  dV=\left(\frac{\partial V}{\partial P}\right)_{T}dP+\left(\frac{\partial V}{\partial T}\right)_{P}dT \ \ \ \ \ (8)

Inserting this into 7 and setting {dP=0} (constant pressure) gives

\displaystyle   dS \displaystyle  = \displaystyle  \left[\left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}+\left(\frac{\partial S}{\partial T}\right)_{V}\right]dT\ \ \ \ \ (9)
\displaystyle  \left(\frac{\partial S}{\partial T}\right)_{P} \displaystyle  = \displaystyle  \left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}+\left(\frac{\partial S}{\partial T}\right)_{V}\ \ \ \ \ (10)
\displaystyle  C_{P} \displaystyle  = \displaystyle  T\left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V} \ \ \ \ \ (11)

where we’ve used 5 and 6 in the last line.

From the Helmholtz energy Maxwell relation

\displaystyle  \left(\frac{\partial S}{\partial V}\right)_{T}=\left(\frac{\partial P}{\partial T}\right)_{V} \ \ \ \ \ (12)

we have

\displaystyle  C_{P}=T\left(\frac{\partial P}{\partial T}\right)_{V}\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V} \ \ \ \ \ (13)

Using 2 with its relation to the isothermal compressibility:

\displaystyle  \frac{\beta}{\kappa_{T}}=\left(\frac{\partial P}{\partial T}\right)_{V} \ \ \ \ \ (14)

we have

\displaystyle  C_{P}=C_{V}+\frac{TV\beta^{2}}{\kappa_{T}} \ \ \ \ \ (15)

For an ideal gas, {PV=NkT}, so 13 becomes

\displaystyle   C_{P} \displaystyle  = \displaystyle  T\frac{Nk}{P}\frac{Nk}{V}+C_{V}\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \frac{NkT}{PV}Nk+C_{V}\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  C_{V}+Nk \ \ \ \ \ (18)

which agrees with Schroeder’s equation 1.48.

From 15, we can see that {C_{P}>C_{V}} provided that the term {\frac{TV\beta^{2}}{\kappa_{T}}} is always positive. The numerator is certainly positive, and from the definition of the isothermal compressibility

\displaystyle  \kappa_{T}\equiv-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T} \ \ \ \ \ (19)

we see that it is positive provided that an increase in pressure causes a decrease in volume, which is pretty well always true for any realistic material. Since {\beta\rightarrow0} for low temperatures and as we wouldn’t expect {V} or {\kappa_{T}} to change much for very low temperatures (where we’d expect most substances to be solid, or possibly liquid, such as helium), then we’d expect {C_{P}\approx C_{V}} for low temperatures, with {C_{P}>C_{V}} as temperature increases. This agrees with Schroeder’s Figure 1.14.

To put in some numbers, we can use the data from the earlier problem. For water, the values given by Schroeder are (at {25^{\circ}\mbox{ C}=298\mbox{ K}}):

\displaystyle   \beta \displaystyle  = \displaystyle  2.57\times10^{-4}\mbox{ K}^{-1}\ \ \ \ \ (20)
\displaystyle  \kappa_{T} \displaystyle  = \displaystyle  4.52\times10^{-10}\mbox{ Pa}^{-1} \ \ \ \ \ (21)

For one mole of water, the volume is {18.068\times10^{-6}\mbox{ m}^{3}}, so

\displaystyle   C_{P}-C_{V} \displaystyle  = \displaystyle  \frac{\left(298\right)\left(18.068\times10^{-6}\right)\left(2.57\times10^{-4}\right)^{2}}{4.52\times10^{-10}}\ \ \ \ \ (22)
\displaystyle  \displaystyle  = \displaystyle  0.787\mbox{ J K}^{-1} \ \ \ \ \ (23)

{C_{P}=75.29\mbox{ J K}^{-1}} for one mole of water so the difference between the heat capacities is around 1% of {C_{P}}.

For mercury we have

\displaystyle   \beta \displaystyle  = \displaystyle  1.81\times10^{-4}\mbox{ K}^{-1}\ \ \ \ \ (24)
\displaystyle  \kappa_{T} \displaystyle  = \displaystyle  4.04\times10^{-11}\mbox{ Pa}^{-1} \ \ \ \ \ (25)

One mole of mercury has a volume of {14.81\times10^{-6}\mbox{ m}^{3}} so we get

\displaystyle   C_{P}-C_{V} \displaystyle  = \displaystyle  \frac{\left(298\right)\left(14.81\times10^{-6}\right)\left(1.81\times10^{-4}\right)^{2}}{4.04\times10^{-11}}\ \ \ \ \ (26)
\displaystyle  \displaystyle  = \displaystyle  3.58\mbox{ J K}^{-1} \ \ \ \ \ (27)

The heat capacity for one mole of mercury (from the appendix to Schroeder’s book) is {27.98\mbox{ J K}^{-1}} so the difference is around 12.8% of {C_{P}}.

Finally, we can derive 15 by starting with {U} and {H} instead of {S} and {V}. The heat capacities are defined in terms of internal energy and enthalpy as

\displaystyle   C_{V} \displaystyle  \equiv \displaystyle  \left(\frac{\partial U}{\partial T}\right)_{V}\ \ \ \ \ (28)
\displaystyle  C_{P} \displaystyle  \equiv \displaystyle  \left(\frac{\partial H}{\partial T}\right)_{P} \ \ \ \ \ (29)

Writing {U=U\left(V,T\right)} we have

\displaystyle  dU=\left(\frac{\partial U}{\partial V}\right)_{T}dV+\left(\frac{\partial U}{\partial T}\right)_{V}dT \ \ \ \ \ (30)

The enthalpy is defined as {H=U+PV}, so at constant pressure

\displaystyle   dH \displaystyle  = \displaystyle  dU+P\;dV\ \ \ \ \ (31)
\displaystyle  \displaystyle  = \displaystyle  \left[\left(\frac{\partial U}{\partial V}\right)_{T}+P\right]dV+\left(\frac{\partial U}{\partial T}\right)_{V}dT \ \ \ \ \ (32)

Dividing through by {dT} at constant pressure gives

\displaystyle   \left(\frac{\partial H}{\partial T}\right)_{P} \displaystyle  = \displaystyle  \left[\left(\frac{\partial U}{\partial V}\right)_{T}+P\right]\left(\frac{\partial V}{\partial T}\right)_{P}+\left(\frac{\partial U}{\partial T}\right)_{V}\ \ \ \ \ (33)
\displaystyle  C_{P} \displaystyle  = \displaystyle  \left[\left(\frac{\partial U}{\partial V}\right)_{T}+P\right]\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V} \ \ \ \ \ (34)

The Helmholtz free energy is defined as {F=U-TS} from which we can derive the relation

\displaystyle  P=-\left(\frac{\partial F}{\partial V}\right)_{T} \ \ \ \ \ (35)

Using this, we have

\displaystyle   C_{P} \displaystyle  = \displaystyle  \left[\left(\frac{\partial U}{\partial V}\right)_{T}-\left(\frac{\partial F}{\partial V}\right)_{T}\right]\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V}\ \ \ \ \ (36)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{\partial\left(U-F\right)}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V}\ \ \ \ \ (37)
\displaystyle  \displaystyle  = \displaystyle  T\left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}+C_{V} \ \ \ \ \ (38)

This is the same as 11, so from here on the derivation is the same as before, and we get 15 again.

Maxwell relations from thermodynamic identities

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.12.

The various partial derivatives that can be obtained from the four thermodynamic identities can be used to derive yet more partial derivatives known as the Maxwell relations (although Maxwell’s main claim to fame is, of course, his unification of electricity and magnetism and subsequent prediction of electromagnetic radiation, he also made important contributions to thermal and statistical physics).

Starting with the original thermodynamic identity:

\displaystyle  dU=T\;dS-P\;dV+\mu\;dN \ \ \ \ \ (1)

we have

\displaystyle   \frac{\partial U}{\partial S} \displaystyle  = \displaystyle  T\ \ \ \ \ (2)
\displaystyle  \frac{\partial U}{\partial V} \displaystyle  = \displaystyle  -P \ \ \ \ \ (3)

[To save writing, all partial derivatives implicitly assume that only the variable with which the derivative is being taken varies, with the other 2 variables appearing in the differential in 1 being held constant. Thus {\frac{\partial U}{\partial S}} assumes {V} and {N} are constant. The molecule number {N} is always constant in what follows.]

Using the fact that mixed second-order partial derivatives do not depend on the order in which the two derivatives are taken (for continuous, differentiable functions, which represents pretty well all functions of physical interest), we can take second derivatives of these equations to get

\displaystyle  \frac{\partial}{\partial V}\left(\frac{\partial U}{\partial S}\right)=\left(\frac{\partial T}{\partial V}\right)_{S}=\frac{\partial}{\partial S}\left(\frac{\partial U}{\partial V}\right)=-\left(\frac{\partial P}{\partial S}\right)_{V} \ \ \ \ \ (4)

In taking the second derivative, we assume that the variable in the first derivative is held constant, which is indicated by the subscripts. This gives us the first Maxwell relation:

\displaystyle  \left(\frac{\partial T}{\partial V}\right)_{S}=-\left(\frac{\partial P}{\partial S}\right)_{V} \ \ \ \ \ (5)

From the enthalpy identity:

\displaystyle  dH=T\;dS+V\;dP+\mu\;dN \ \ \ \ \ (6)

we get

\displaystyle   \frac{\partial H}{\partial S} \displaystyle  = \displaystyle  T\ \ \ \ \ (7)
\displaystyle  \frac{\partial H}{\partial P} \displaystyle  = \displaystyle  V \ \ \ \ \ (8)

Taking second derivatives gives

\displaystyle  \left(\frac{\partial T}{\partial P}\right)_{S}=\left(\frac{\partial V}{\partial S}\right)_{P} \ \ \ \ \ (9)

From the Helmholtz free energy:

\displaystyle  dF=-S\;dT-P\;dV+\mu\;dN \ \ \ \ \ (10)

we get

\displaystyle   \frac{\partial F}{\partial T} \displaystyle  = \displaystyle  -S\ \ \ \ \ (11)
\displaystyle  \frac{\partial F}{\partial V} \displaystyle  = \displaystyle  -P \ \ \ \ \ (12)

Taking second derivatives gives

\displaystyle  \left(\frac{\partial S}{\partial V}\right)_{T}=\left(\frac{\partial P}{\partial T}\right)_{V} \ \ \ \ \ (13)

Finally, from the Gibbs free energy:

\displaystyle  dG=-S\;dT+V\;dP+\mu\;dN \ \ \ \ \ (14)

we get

\displaystyle   \frac{\partial G}{\partial T} \displaystyle  = \displaystyle  -S\ \ \ \ \ (15)
\displaystyle  \frac{\partial G}{\partial P} \displaystyle  = \displaystyle  V \ \ \ \ \ (16)

Taking second derivatives gives

\displaystyle  \left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P} \ \ \ \ \ (17)

Hydrogen fuel cell at higher temperatures

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.11.

We’ll revisit the hydrogen fuel cell, which we considered originally operating at room temperature ({25^{\circ}\mbox{C}=298\mbox{ K}}) and atmospheric pressure. The reaction is

\displaystyle  \mbox{H}_{2}+\frac{1}{2}\mbox{O}_{2}\rightarrow\mbox{H}_{2}\mbox{O} \ \ \ \ \ (1)

We can use the relation derived from the Gibbs thermodynamic identity to estimate the behaviour of a hydrogen fuel cell at the higher temperature of {75^{\circ}\mbox{C}=348\mbox{ K}} (and still at atmospheric pressure. The relation is

\displaystyle  S=-\left(\frac{\partial G}{\partial T}\right)_{V,N} \ \ \ \ \ (2)

If we take the values of {\Delta G} from Schroeder’s book, where the values for pure hydrogen and oxygen gas are taken as zero (to set a reference point), then {\Delta G=-237.13\mbox{ kJ mol}^{-1}} and {S=69.91\mbox{ J K}^{-1}} for water at 298 K. For water, if we assume that {S} is constant over the temperature range in question (I’m not sure how good an approximation this is, but since the water remains liquid over this range, it’s probably not too bad), we can estimate the change in {G} as we go from 298 K to 348 K:

\displaystyle   \Delta G \displaystyle  = \displaystyle  -S\;\Delta T\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  -69.91\times10^{-3}\mbox{kJ K}^{-1}\mbox{mol}^{-1}\times50\mbox{ K}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  -3.496\mbox{ kJ mol}^{-1} \ \ \ \ \ (5)

We can do similar calculations for hydrogen and oxygen, using the entropy values given in Schroeder. Here, we’re dealing with gases at constant pressure, so the volume does change as we increase the temperature, so we’d expect the entropy to increase as the temperature is raised. However, as the problem asks us to use only the data at 298 K to estimate the changes, I suppose we can approximate things by assuming the entropy remains constant. The results are, for 1 mole

{S\mbox{ (J K}^{-1}\mbox{)}} {G_{25}\mbox{ (kJ)}} {\Delta G\mbox{ (kJ)}} {G_{75}\mbox{ (kJ)}}
{\mbox{H}_{2}\mbox{O}} {69.91} {-237.13} {-3.496} {-240.63}
{\mbox{H}_{2}} {130.68} {0} {-6.534} {-6.534}
{\mbox{O}_{2}} {205.14} {0} {-10.257} {-10.257}

The net {\Delta G} for the reaction at {75^{\circ}\mbox{ C}} for one mole of hydrogen is therefore

\displaystyle  \Delta G=-240.63-\left(-\frac{10.257}{2}-6.534\right)=-228.97\mbox{ kJ} \ \ \ \ \ (6)

Thus the maximum amount of electrical work we can get from the fuel cell at {75^{\circ}\mbox{ C}} is 228.97 kJ, slightly lower than the 237 kJ we got at {25^{\circ}\mbox{ C}}.

Thermodynamic identity for Gibbs & Helmholtz free energies

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 5.8 – 5.10.

The thermodynamic identity for energy is

\displaystyle dU=T\;dS-P\;dV+\mu\;dN \ \ \ \ \ (1)

With the definitions of Helmholtz and Gibbs free energies {F} and{G} and enthalpy {H} we can derive analogues to this identity for these other three energies.

The Helmholtz energy is

\displaystyle F=U-TS \ \ \ \ \ (2)

Taking differentials, we get

\displaystyle dF \displaystyle = \displaystyle dU-TdS-SdT\ \ \ \ \ (3)
\displaystyle \displaystyle = \displaystyle -S\;dT-P\;dV+\mu\;dN \ \ \ \ \ (4)

For enthalpy, we get

\displaystyle H \displaystyle = \displaystyle U+PV\ \ \ \ \ (5)
\displaystyle dH \displaystyle = \displaystyle dU+P\;dV+V\;dP\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle T\;dS+V\;dP+\mu\;dN \ \ \ \ \ (7)

Finally, for the Gibbs energy we have

\displaystyle G \displaystyle = \displaystyle U-TS+PV\ \ \ \ \ (8)
\displaystyle dG \displaystyle = \displaystyle dU-T\;dS-S\;dT+P\;dV+V\;dP\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle -S\;dT+V\;dP+\mu\;dN \ \ \ \ \ (10)

From the latter identity we can derive a few partial derivatives:

\displaystyle S \displaystyle = \displaystyle -\left(\frac{\partial G}{\partial T}\right)_{V,N}\ \ \ \ \ (11)
\displaystyle V \displaystyle = \displaystyle \left(\frac{\partial G}{\partial P}\right)_{T,N}\ \ \ \ \ (12)
\displaystyle \mu \displaystyle = \displaystyle \left(\frac{\partial G}{\partial N}\right)_{T,P} \ \ \ \ \ (13)

From 11, we can get an idea of how {G} varies with temperature for a pure substance as it starts from a solid, then melts to a liquid and finally boils to a gas. We expect the entropy to be lowest when the substance is a solid, then become larger when it’s a liquid and finally to be largest when it’s a gas. This is illustrated in the following plot (qualitative; don’t pay any attention to the values on the axes).

Because of the minus sign, we therefore expect {G} to decrease with temperature, starting off with a fairly shallow slope in the solid phase (red), then changing to a steeper slope in the liquid phase (green) and finally to the steepest slope in the gas phase (blue). The various segments would probably not be straight lines, especially in the gas phase, since the entropy would tend to increase with temperature, so the curves in the liquid and gas phases would probably be concave downwards.

Over relatively small changes in the state variables, we can use 10 to estimate changes in {G} as we vary one of the variables, holding the others constant. For example, if we start with a mole of water at {25^{\circ}\mbox{ C}} and 1 bar and increase its temperature to {30^{\circ}\mbox{ C}} at constant pressure (and constant molecule number {N}), the change in {G} is approximately (using the value for {S} from Schroeder’s book):

\displaystyle \Delta G \displaystyle \approx \displaystyle -S\;\Delta T\ \ \ \ \ (14)
\displaystyle \displaystyle = \displaystyle -69.91\times5\ \ \ \ \ (15)
\displaystyle \displaystyle = \displaystyle -349.55\mbox{ J} \ \ \ \ \ (16)

If we want {G} to remain constant, we can compensate the decrease by increasing the pressure so that

\displaystyle V\;\Delta P=S\;\Delta T \ \ \ \ \ (17)

One mole of water has a volume of around 18 ml, so the required pressure change is

\displaystyle \Delta P=\frac{349.55\mbox{ J}}{18\times10^{-6}\mbox{ m}^{3}}=1.94\times10^{7}\mbox{ N m}^{-2}=194\mbox{ bar} \ \ \ \ \ (18)

Thus we’d need to increase the pressure by a large amount. These are only rough estimates, since we’re assuming that the behaviour of the variables is linear over the changes in {T} and {P}.

Muscle as a fuel cell

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 5.6 – 5.7.

As another example of a fuel cell, we’ll look at the metabolism of glucose in an animal’s muscle cells. The overall reaction is

\displaystyle  \mbox{C}_{6}\mbox{H}_{12}\mbox{O}_{6}+6\mbox{O}_{2}\rightarrow6\mbox{H}_{2}\mbox{O}+6\mbox{CO}_{2} \ \ \ \ \ (1)

The Gibbs free energy {\Delta G} and enthalpy {\Delta H} changes for this reaction can be obtained from the {\Delta G} and {\Delta H} values in Schroeder’s book (all values for 1 mole at 298 K and 1 bar). As the reaction occurs at room temperature and pressure, we’ll assume that the water product appears as a liquid rather than as a gas.

{\Delta G} (kJ) {\Delta H} (kJ) {S\mbox{ J K}^{-1}}
{\mbox{C}_{6}\mbox{H}_{12}\mbox{O}_{6}} {-910} {-1273} {212}
{\mbox{O}_{2}} {0} {0} {205.14}
{\mbox{H}_{2}\mbox{O}} {-237.13} {-285.83} {69.91}
{\mbox{CO}_{2}} {-394.36} {-393.51} {213.74}

The {\Delta G} for the reaction is the sum of the values for the products minus the sum for the reactants:

\displaystyle  \Delta G=6\left(-237.13-394.36\right)-\left(6\times0-910\right)=-2878.94\mbox{ kJ mol}^{-1} \ \ \ \ \ (2)

The value is per mole of glucose molecules.

The corresponding {\Delta H} is found the same way:

\displaystyle  \Delta H=6\left(-285.83-393.51\right)-\left(6\times0-1273\right)=-2803.04\mbox{ kJ mol}^{-1} \ \ \ \ \ (3)

The Gibbs energy represents the maximum energy that may be extracted as ‘other’ (that is, not due to volume changes) work, in this case chemical work. Thus we may extract up to 2878.94 kJ of electric work per mole of glucose metabolized.

As the reaction occurs at constant pressure, {\Delta H} represents the total energy difference between the reactants and products. Since the enthalpy drop is less than the amount of work extracted, the difference must be absorbed as heat. The amount of heat is

\displaystyle  Q=2878.94-2803.04=75.9\mbox{ kJ mol}^{-1} \ \ \ \ \ (4)

The entropy increase resulting from absorbing this heat is

\displaystyle  \Delta S=\frac{Q}{T}=\frac{75.9\times10^{3}}{298}=254.7\mbox{ J K}^{-1}\mbox{mol}^{-1} \ \ \ \ \ (5)

If we work out the entropy change {\Delta S} for this reaction using the values in the table above, we find

\displaystyle  \Delta S=6\left(69.91+213.74\right)-\left(6\times205.14+212\right)=259.06\mbox{ J K}^{-1}\mbox{mol}^{-1} \ \ \ \ \ (6)

The values are roughly the same, so the absorption of heat can be explained by the entropy of the products being greater than that of the reactants.

This model assumes that the muscle is ideal, in the sense that all of the available {\Delta G} is converted into chemical work in the muscle. If the muscle is less than ideal, then the amount of work performed is less than {\Delta G}, so less heat is absorbed. However, the entropy difference between the reactants and products remains the same, so some of this entropy must be provided by means other than heat flow. This makes sense since a non-ideal muscle would use an irreversible process to perform its motion, resulting in an increase of entropy from other means.

The actual process by which glucose is metabolized is much more complicated than the simple reaction 1. In the process, 38 ATP (adensoine triphosphate) molecules are synthesized. When an ATP molecule splits into ADP (adenosine diphosphate) and a phosphate ion, it releases energy that is used in a variety of processes, including muscle contraction. The splitting of one ATP molecule provides energy for a molecule of myosin (an enzyme) to contract with a force of {4\times10^{-12}\mbox{ N}} over a distance of {1.1\times10^{-8}\mbox{ m}}. Thus one glucose molecule provides the energy for an amount of work

\displaystyle  W=38\times4\times10^{-12}\times1.1\times10^{-8}=1.672\times10^{-18}\mbox{ J} \ \ \ \ \ (7)

The maximum amount of energy provided by one glucose molecule is obtained from 2 as

\displaystyle  W_{max}=\frac{\left|\Delta G\right|}{6.02\times10^{23}}=4.78\times10^{-18}\mbox{ J} \ \ \ \ \ (8)

Thus the efficiency of muscle contraction is

\displaystyle  e=\frac{W}{W_{max}}=0.35 \ \ \ \ \ (9)

Methane fuel cell

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.5.

As another example of a fuel cell, suppose we burn methane in oxygen according to the reaction

\displaystyle  \mbox{CH}_{4}+2\mbox{O}_{2}\rightarrow2\mbox{H}_{2}\mbox{O}+\mbox{CO}_{2} \ \ \ \ \ (1)

The Gibbs free energy {\Delta G} and enthalpy {\Delta H} changes for this reaction can be obtained from the {\Delta G} and {\Delta H} values in Schroeder’s book (all values for 1 mole at 298 K and 1 bar). As the reaction occurs at room temperature and pressure, we’ll assume that the water product appears as a liquid rather than as a gas.

{\Delta G} (kJ) {\Delta H} (kJ)
{\mbox{CH}_{4}} {-50.72} {-74.81}
{\mbox{O}_{2}} {0} {0}
{\mbox{H}_{2}\mbox{O}} {-237.13} {-285.83}
{\mbox{CO}_{2}} {-394.36} {-393.51}

The {\Delta G} for the reaction is the sum of the values for the products minus the sum for the reactants:

\displaystyle  \Delta G=\left(-394.36-2\times237.13\right)-\left(2\times0-50.72\right)=-817.9\mbox{ kJ mol}^{-1} \ \ \ \ \ (2)

The value is per mole of methane molecules.

A negative {\Delta G} means that the energy of the products is lower than that of the reactants, so the tendency is for the reaction to ‘run downhill’, that is, it will occur spontaneously in the direction shown.

The corresponding {\Delta H} is found the same way:

\displaystyle  \Delta H=\left(-393.51-2\times285.83\right)-\left(2\times0-74.81\right)=-890.36\mbox{ kJ mol}^{-1} \ \ \ \ \ (3)

The Gibbs energy represents the maximum energy that may be extracted as ‘other’ (that is, not due to volume changes) work, in this case electrical work. Thus we may extract up to 817.9 kJ of electric work per mole of methane burned.

As the reaction occurs at constant pressure, {\Delta H} represents the total energy difference between the reactants and products. Since the enthalpy drop is greater than the amount of electrical work extracted, the excess is discarded as heat. The amount of heat is

\displaystyle  Q=890.36-817.9=72.46\mbox{ kJ mol}^{-1} \ \ \ \ \ (4)

The overall reaction occurs in two phases, one at each electrode. At the negative electrode we have

\displaystyle  \mbox{CH}_{4}+2\mbox{H}_{2}\mbox{O}\rightarrow\mbox{CO}_{2}+8\mbox{H}^{+}+8e^{-} \ \ \ \ \ (5)

At the positive electrode, we have

\displaystyle  2\mbox{O}_{2}+8\mbox{H}^{+}+8e^{-}\rightarrow4\mbox{H}_{2}\mbox{O} \ \ \ \ \ (6)

As 8 electrons are pushed around the circuit for each methane molecule, the energy {\Delta G} is distributed over 8 moles of electrons, so the energy per electron is

\displaystyle  E=\frac{\left|\Delta G\right|}{8\times6.02\times10^{23}}=\frac{817.9\times10^{3}}{8\times6.02\times10^{23}}=1.70\times10^{-19}\mbox{ J}=1.06\mbox{ eV} \ \ \ \ \ (7)

Thus the voltage of the fuel cell is 1.06 volts.