# What are some applications of thermodynamics

## Simple thermodynamic applications

Theoretical Physics 4 | Thermodynamics and Statistical Physics pp 103-143 | Cite as

- Matthias Bartelmann
- Bjorn Feuerbacher
- Timm Kruger
- Dieter Lüst
- Anton Rebhan
- Andreas Wipf

### Summary

In chap. 1, thermodynamics was initially justified phenomenologically, i. H. based on such observations associated with the experience of temperature and warmth. We introduced the temperature as a variable of state, formulated the first law and understood how, based on the fundamental experience of irreversible processes, we got to entropy and the second law.

In chap. 2 we have described the statistical justification of thermodynamics based on microphysics, whereby we have mainly seen that entropy can be identified with a logarithmic measure of the phase space volume of a system. In addition, the statistical procedure derived from microphysics made it possible to further sharpen the terms “heat” and “work”.

In this chapter we want to apply this knowledge to simple thermodynamic systems. We begin in Sect. 3.1 with the variety of thermodynamic functions or potentials, which are connected to one another by Legendre transformations and adapted to various external conditions. The enthalpy, which we briefly discussed in Sect. 1.6.4, is an example of such a thermodynamic potential. On the basis of these potentials we discuss in Sect. 3.2 the extreme and stability properties of thermodynamic systems in thermal equilibrium.

### tasks

Occasionally the tasks contain more information than is necessary for the solution. For some others, however, data from general knowledge, other sources, or meaningful estimates are required.

• easy tasks with just a few calculation steps

•• Moderately difficult tasks that require some thinking and possibly a combination of different concepts

••• Challenging tasks that require advanced concepts (possibly also from later chapters) or your own mathematical modeling

### 3.1 ••• Thermodynamics of a rubber band

- (a)
Determine the internal energy \ (U \) and the entropy \ (S \) of the rubber band as functions of \ (T \) and \ (L \) with respect to the point \ ((T_ {0}, L_ {0) }) \) in the state space of the rubber band, with \ (T \ in [T _ {\ mathrm {a}}, T _ {\ mathrm {e}}] \).

- (b)
What work is done on the rubber band with isothermal quasi-static stretching of the rubber band from the length \ (L_ {0} \) to the length \ (L_ {1}> L_ {0} \)? The temperature is \ (T_ {0} \).

- (c)
What change in entropy does the rubber band experience during this process? How does the entropy of the heat reservoir change, which ensures the constant temperature \ (T_ {0} \)?

- (d)
Go from the state \ ((T_ {0}, L_ {1}) \) to the untensioned state, excluding any heat exchange with the environment, by simply letting go of the tensioned belt. Which entropy changes do the band and environment experience? Is this process reversible? What is the temperature of the rubber band when it is not tensioned?

### Solution hint:

First show with the help of the first law that the internal energy only depends on the temperature \ (T \), but not on the length \ (L \). Use the comparable calculation for an ideal gas as a guide.

### 3.2 •• Entropy change of a set of coupled systems

### Solution hint:

Note that \ (T_ {j} \ gtrless T _ {\ mathrm {f}} \) is possible. Estimate the occurring integrals using \ (T _ {\ mathrm {f}} \).

### 3.3 •• Thermodynamics of Electromagnetic Radiation

- (a)Calculate from the first main clause in the so-called Gibbs basic form$$ T \ mathrm {d} S = \ mathrm {d} U + P \ mathrm {d} V $$and the Maxwell relation$$ \ left (\ frac {\ partial S} {\ partial V} \ right) _ {T} = \ left (\ frac {\ partial P} {\ partial T} \ right) _ {V} $$the energy density \ (u (T) \) with \ (u (T = 1 \, \ mathrm {K}) =: \ sigma \).
- (b)
Determine the entropy \ (S = S [T (U, V), V] \) of the radiation. How is the constant of integration for the entropy to be chosen according to the third law?

- (c)
Given are two cavities (1) and (2) with the volumes \ (V_ {1} \) and \ (V_ {2} \). The walls are ideally reflective and rigid, i.e. that is, they do not take up any energy and are impermeable to heat. In the cavity (2) there is initially no radiation, whereas the cavity (1) is filled with radiation of temperature \ (T_ {1} \). Then, by opening a passage in the partition, radiation is allowed to pass from the cavity (1) into the cavity (2). Calculate the final temperature \ (T _ {\ mathrm {f}} \). Is the process reversible or irreversible?

- (d)Derive the relation given in subtask (a)$$ \ left (\ frac {\ partial S} {\ partial V} \ right) _ {T} = \ left (\ frac {\ partial P} {\ partial T} \ right) _ {V} $$from the Gibbs basic form.

### 3.4 •• Possible and permitted state changes

A system consists of three heat reservoirs with the temperatures \ (T_ {1}> T_ {2}> T_ {3} \). Two of them are connected by reversible heat engines. The work gained in one machine is fed to the other machine. The heat quantities \ (Q_ {1} \), \ (Q_ {2} \) and \ (Q_ {3} \) are exchanged with the reservoirs, whereby all \ (Q_ {i} \) can be positive and negative . Which combinations of signs of the \ (Q_ {i} \) are compatible with the main clauses?

### 3.5 • Thermodynamics of a magnetized system

- (a)Refresh your knowledge about the partial derivatives of \ (U \): Which physical quantities do the partial derivatives correspond to?$$ \ left (\ frac {\ partial U} {\ partial S} \ right) _ {V, M, N} \ ,, \ quad \ left (\ frac {\ partial U} {\ partial V} \ right ) _ {S, M, N} \ ,, \ quad \ left (\ frac {\ partial U} {\ partial N} \ right) _ {S, V, M} \;? $$
- (b)
Determine the thermodynamic potential \ (J (T, P, M, \ mu) \) and the corresponding differential \ (\ mathrm {d} J \) using a suitable Legendre transformation from \ (U \). Why is \ (J (T, P, M, \ mu) = BM \)?

- (c)Show that the Maxwell relation$$ \ left (\ frac {\ partial N} {\ partial M} \ right) _ {T, P, \ mu} = - \ left (\ frac {\ partial B} {\ partial \ mu} \ right) _ {T, P, M} $$applies.

### Solution hint:

Often \ ({\ varvec {H}} \) and \ ({\ varvec {M}} \) instead of \ ({\ varvec {B}} \) and \ ({\ varvec {M}} \) are used as fields conjugate to one another are assumed. We understand \ ({\ varvec {B}} \) as an externally applied magnetic field and therefore use \ ({\ varvec {B}} \).

### Solutions to the tasks

### 3.1

### Detailed solutions to the tasks

### 3.1

- (a)Entropy \ (S \) and internal energy \ (U \) are both sought as functions of \ (T \) and \ (L \). The first law says$$ \ delta Q = \ mathrm {d} U + \ delta W \ ,, $$what in the case of the rubber band$$ \ begin {aligned} \ displaystyle & \ displaystyle T \ left [\ left (\ frac {\ partial S} {\ partial L} \ right) _ {T} \ mathrm {d} L + \ left (\ frac {\ partial S} {\ partial T} \ right) _ {L} \ mathrm {d} T \ right] \ \ displaystyle & \ displaystyle \ quad = \ left (\ frac {\ partial U} {\ partial L} \ right ) _ {T} \ mathrm {d} L + \ left (\ frac {\ partial U} {\ partial T} \ right) _ {L} \ mathrm {d} TF \ mathrm {d} L \ end {aligned} $$follows. From this we first get the two relationships$$ \ begin {aligned} \ displaystyle \ left (\ frac {\ partial S} {\ partial L} \ right) _ {T} & \ displaystyle = \ frac {1} {T} \ left (\ frac {\ partial U} {\ partial L} \ right) _ {T} - \ frac {F} {T} \ ,, \ \ displaystyle \ left (\ frac {\ partial S} {\ partial T} \ right) _ {L} & \ displaystyle = \ frac {1} {T} \ left (\ frac {\ partial U} {\ partial T} \ right) _ {L} \,. \ End {aligned} $$Derivation of the first with respect to \ (T \) using the second yields$$ \ frac {\ partial} {\ partial T} \ left [\ frac {1} {T} \ left (\ frac {\ partial U} {\ partial L} \ right) _ {T} \ right] - \ frac {\ partial} {\ partial T} \ left (\ frac {F} {T} \ right) = \ frac {1} {T} \ frac {\ partial ^ {2} U} {\ partial L \ partial T} \,. $$Since \ (F \) was measured proportionally to the temperature, the derivative from \ (F / T \) to \ (T \) vanishes. Hence (3.234) becomes$$ - \ frac {1} {T ^ {2}} \ left (\ frac {\ partial U} {\ partial L} \ right) _ {T} = 0 \ ,, $$which shows that the internal energy can only depend on the temperature, but not on the length, \ (U = U (T) \). It then results from the heat capacity at constant length,$$ U = U_ {0} + C_ {L, 0} (T-T_ {0}) \ ,, $$as well as the entropy dependence on the temperature at constant length:$$ \ left (\ frac {\ partial S} {\ partial T} \ right) _ {L_ {0}} = \ frac {C_ {L, 0}} {T} \ quad \ Rightarrow \ quad \ Updelta S_ {L_ {0}} = C_ {L, 0} \ ln \ frac {T} {T_ {0}} \,. $$Since the internal energy does not depend on \ (L \), is$$ \ left (\ frac {\ partial S} {\ partial L} \ right) _ {T} = - \ frac {F} {T} = - b \ left (\ frac {L} {L_ {0} } - \ frac {L_ {0} ^ {2}} {L ^ {2}} \ right) \ ,, $$from what$$ \ begin {aligned} \ displaystyle \ Updelta S_ {T} & \ displaystyle = -b \ left [\ frac {1} {2} \ left (\ frac {L ^ {2}} {L_ {0}} -L_ {0} \ right) + \ left (\ frac {L_ {0} ^ {2}} {L} -L_ {0} \ right) \ right] \ \ displaystyle & \ displaystyle = b \ left (\ frac {3L_ {0}} {2} - \ frac {L ^ {2}} {2L_ {0}} - \ frac {L_ {0} ^ {2}} {L} \ right) \ end {aligned} $$follows. Overall, therefore, are the internal energy and entropy$$ \ begin {aligned} \ displaystyle U & \ displaystyle = U_ {0} + C_ {L, 0} (T-T_ {0}) \ ,, \ \ displaystyle S & \ displaystyle = S_ {0} + C_ { L, 0} \ ln \ frac {T} {T_ {0}} + b \ left (\ frac {3L_ {0}} {2} - \ frac {L ^ {2}} {2L_ {0}} - \ frac {L_ {0} ^ {2}} {L} \ right) \,. \ end {aligned} $$
- (b)In the case of isothermal-quasi-static expansion at the temperature \ (T_ {0} \), work has to be done on the rubber band. Because of our convention for the sign of work, the work done on the rubber band is the negative work done by the rubber band, the$$ \ begin {aligned} \ displaystyle \ Updelta W & \ displaystyle = -bT_ {0} \ int_ {L_ {0}} ^ {L_ {1}} \ left (\ frac {L} {L_ {0}} - \ frac {L_ {0} ^ {2}} {L ^ {2}} \ right) \ mathrm {d} L \ \ displaystyle & \ displaystyle = -bT_ {0} \ left (\ frac {3L_ {0} } {2} - \ frac {L_ {1} ^ {2}} {2L_ {0}} - \ frac {L_ {0} ^ {2}} {L_ {1}} \ right) \ end {aligned} $$amounts.
- (c)Since the temperature is constantly equal to \ (T_ {0} \) during this expansion, the internal energy cannot change in this case, because it only depends on the temperature, but not on the length. With \ (\ mathrm {d} U = 0 \) it follows from the first law that \ (\ delta Q = \ Updelta W \). Therefore the entropy changes$$ \ Updelta S = \ frac {\ Updelta W} {T_ {0}} <0 \ ,, $$d. i.e., with isothermal expansion, the entropy decreases! The entropy of the heat reservoir, which ensures the constant temperature \ (T_ {0} \), must increase to the same extent.
- (d)No heat is exchanged with the environment, and no work is done on the rubber band. Therefore its internal energy cannot change, and with it its temperature either: At the end the rubber band has the same temperature \ (T_ {0} \) as at the beginning. The change in entropy then simply results in$$ \ Updelta S = -b \ left (\ frac {3L_ {0}} {2} - \ frac {L_ {1} ^ {2}} {2L_ {0}} - \ frac {L_ {0} ^ {2}} {L_ {1}} \ right) \,. $$The process is not reversible, but the change in entropy as a change in a state variable can be calculated using a reversible comparison process.

### 3.2

### 3.3

- (a)With the help of the specifications, the basic Gibbs form in$$ \ begin {aligned} \ displaystyle \ mathrm {d} S & \ displaystyle = \ frac {1} {T} \ left (\ frac {\ partial U} {\ partial T} \ right) _ {V} \ mathrm {d} T + \ frac {1} {T} \ left (\ frac {\ partial U} {\ partial V} \ right) _ {T} \ mathrm {d} V + \ frac {P} {T} \ mathrm {d} V \ \ displaystyle & \ displaystyle = \ frac {V} {T} \ left (\ frac {\ partial u} {\ partial T} \ right) _ {V} \ mathrm {d} T + \ frac { u} {T} \ mathrm {d} V + \ frac {u} {3T} \ mathrm {d} V \ end {aligned} $$be rewritten. The following applies to the derivatives of entropy$$ \ left (\ frac {\ partial S} {\ partial T} \ right) _ {V} = \ frac {V} {T} \ left (\ frac {\ partial u} {\ partial T} \ right ) _ {V} \ ,, \ quad \ left (\ frac {\ partial S} {\ partial V} \ right) _ {T} = \ frac {u} {T} + \ frac {u} {3T} = \ frac {4u} {3T} \,. $$Using the given Maxwell relation, it follows from the second equation$$ \ left (\ frac {\ partial S} {\ partial V} \ right) _ {T} = \ left (\ frac {\ partial P} {\ partial T} \ right) _ {V} = \ frac {1} {3} \ left (\ frac {\ partial u} {\ partial T} \ right) _ {V} = \ frac {4u} {3T} $$and from it through integration$$ \ ln u = 4 \ ln T + {\ mathrm {const}} \,. $$The constant has to be chosen so that \ (u (T = 1 \, \ mathrm {K}) = \ sigma \, \ mathrm {K} ^ {4} \) becomes, which finally leads to.
- (b)For the entropy \ (S [T (U, V), V] \) we first get from Gibbs’s basic form$$ \ begin {aligned} \ displaystyle \ left (\ frac {\ partial S} {\ partial T} \ right) _ {V} & \ displaystyle = \ frac {V} {T} \ left (\ frac {\ partial u} {\ partial T} \ right) _ {V} = 4 \ sigma VT ^ {2} \ ,, \ \ displaystyle \ left (\ frac {\ partial S} {\ partial V} \ right) _ {T} & \ displaystyle = \ frac {4 \ sigma T ^ {3}} {3} \ ,, \ end {aligned} $$from that then$$ S = S_ {0} + \ frac {4 \ sigma VT ^ {3}} {3} \ ,, $$and since the entropy must vanish at \ (T = 0 \), \ (S_ {0} = 0 \) must also be set:$$ S = \ frac {4} {3} \ sigma VT ^ {3} \,. $$
- (c)Under the given conditions, the entire internal energy remains unchanged,$$ \ sigma V_ {1} T ^ {4} = U = U_ {1} + U_ {2} = \ sigma \ left (V_ {1} + V_ {2} \ right) T _ {\ mathrm {f} } ^ {4} \ ,, $$from which the final temperature$$ T _ {\ mathrm {f}} = \ left (\ frac {V_ {1}} {V_ {1} + V_ {2}} \ right) ^ {1/4} T $$results. The process is irreversible because the radiation will not spontaneously withdraw into cavity 1 after it has filled both cavities. The entropy increases by the amount$$ \ begin {aligned} \ displaystyle \ Updelta S & \ displaystyle = \ frac {4} {3} \ sigma \ left [(V_ {1} + V_ {2}) \ left (\ frac {V_ {1}} {V_ {1} + V_ {2}} \ right) ^ {3/4} -V_ {1} \ right] T ^ {3} \ \ displaystyle & \ displaystyle = S \ left [\ left (\ frac { V_ {1} + V_ {2}} {V_ {1}} \ right) ^ {1/4} -1 \ right] \,. \ End {aligned} $$
- (d)We finally get the specified Maxwell relation as follows: We write out Gibbs’s basic form,$$ \ begin {aligned} \ displaystyle & \ displaystyle T \ left [\ left (\ frac {\ partial S} {\ partial T} \ right) _ {V} \ mathrm {d} T + \ left (\ frac {\ partial S} {\ partial V} \ right) _ {T} \ mathrm {d} V \ right] \ \ displaystyle & \ displaystyle \ quad = \ left (\ frac {\ partial U} {\ partial T} \ right ) _ {V} \ mathrm {d} T + \ left (\ frac {\ partial U} {\ partial V} \ right) _ {T} \ mathrm {d} V + P \ mathrm {d} V \ ,, \ end {aligned} $$and get from it$$ T \ left (\ frac {\ partial S} {\ partial T} \ right) _ {V} = \ left (\ frac {\ partial U} {\ partial T} \ right) _ {V} \, , \ quad T \ left (\ frac {\ partial S} {\ partial V} \ right) _ {T} = \ left (\ frac {\ partial U} {\ partial V} \ right) _ {T} + P \,. $$We derive the first equation according to \ (V \), the second according to \ (T \) and eliminate the mixed second derivative of the internal energy \ (U \) between the two resulting equations. This gives$$ T \ frac {\ partial ^ {2} S} {\ partial V \ partial T} = \ left (\ frac {\ partial S} {\ partial V} \ right) _ {T} + T \ frac { \ partial ^ {2} S} {\ partial T \ partial V} - \ left (\ frac {\ partial P} {\ partial T} \ right) _ {V} \ ,, $$from which the given Maxwell relation can already be read off.

### 3.4

### 3.5

- (a)The given partial derivatives correspond to the following physical quantities:$$ \ begin {aligned} \ displaystyle \ left (\ frac {\ partial U} {\ partial S} \ right) _ {V, M, N} & \ displaystyle = T \ ,, \ \ displaystyle \ left ( \ frac {\ partial U} {\ partial V} \ right) _ {S, M, N} & \ displaystyle = -P \ ,, \ \ displaystyle \ left (\ frac {\ partial U} {\ partial N } \ right) _ {S, V, M} & \ displaystyle = \ mu \ ,, \ end {aligned} $$i.e. the absolute temperature, the negative pressure and the chemical potential.
- (b)The grand-canonical potential \ (J \), which is introduced in Sect. 4.3, should depend on \ ((T, P, M, \ mu) \), while the internal energy \ (U \) depends on \ ((S, V, M, N) \) depends. Accordingly, the transition from \ (U \) to \ (J \) requires the Legendre transformations \ (S \ to T \), \ (V \ to P \) and \ (N \ to \ mu \), where the Signs are given by the partial derivatives of \ (U \): The differential \ (\ mathrm {d} J \) is$$ \ begin {aligned} \ displaystyle \ mathrm {d} J & \ displaystyle = \ mathrm {d} US \ mathrm {d} TT \ mathrm {d} S + P \ mathrm {d} V + V \ mathrm {d } P- \ mu \ mathrm {d} NN \ mathrm {d} \ mu \ \ displaystyle & \ displaystyle = \ left (T \ mathrm {d} SP \ mathrm {d} V + B \ mathrm {d} M + \ mu \ mathrm {d} N \ right) \ \ displaystyle & \ displaystyle \ quad-S \ mathrm {d} TT \ mathrm {d} S + P \ mathrm {d} V + V \ mathrm {d} P- \ mu \ mathrm {d} NN \ mathrm {d} \ mu \ \ displaystyle & \ displaystyle = -S \ mathrm {d} T + V \ mathrm {d} PN \ mathrm {d} \ mu + B \ mathrm {d } M \,. \ End {aligned} $$According to Euler's theorem about homogeneous functions, must for internal energy$$ \ begin {aligned} \ displaystyle U & \ displaystyle = S \ left (\ frac {\ partial U} {\ partial S} \ right) _ {V, M, N} + V \ left (\ frac {\ partial U} {\ partial V} \ right) _ {S, M, N} \ \ displaystyle & \ displaystyle \ quad + M \ left (\ frac {\ partial U} {\ partial M} \ right) _ {S, V, N} + N \ left (\ frac {\ partial U} {\ partial N} \ right) _ {S, V, M} \ end {aligned} $$hold, since \ (U \) is homogeneous of degree 1 in the extensive quantities \ ((S, V, M, N) \). With the results from subtask (a) and the specification it follows and is therefore$$ J = U-TS + PV- \ mu N = BM \,. $$
- (c)From the differential \ (\ mathrm {d} J \) we first get$$ \ left (\ frac {\ partial J} {\ partial M} \ right) _ {T, P, \ mu} = B \ ,, \ quad \ left (\ frac {\ partial J} {\ partial \ mu} \ right) _ {T, P, M} = - N \,. $$Derivation of the first equation according to \ (\ mu \), the second according to \ (M \) and elimination of the mixed second derivative$$ \ frac {\ partial ^ {2} J} {\ partial \ mu \ partial M} $$immediately delivers the desired result:$$ \ begin {aligned} \ displaystyle \ left (\ frac {\ partial B} {\ partial \ mu} \ right) _ {T, P, M} & \ displaystyle = \ frac {\ partial} {\ partial \ mu} \ left (\ frac {\ partial J} {\ partial M} \ right) _ {T, P, \ mu} \ \ displaystyle & \ displaystyle = \ frac {\ partial} {\ partial M} \ left ( \ frac {\ partial J} {\ partial \ mu} \ right) _ {T, P, M} \ \ displaystyle & \ displaystyle = - \ left (\ frac {\ partial N} {\ partial M} \ right) _ {T, P, \ mu} \,. \ End {aligned} $$

### literature

- Maslow, V. P .: Zeroth-Order Phase Transitions. Mathematical Notes 76/5, 697-710 (2004) CrossRefGoogle Scholar
- Schwabl, F .: Statistical Mechanics. Springer, Berlin, Heidelberg (2006) zbMATHGoogle Scholar

### Copyright information

### Authors and Affiliations

- Matthias BartelmannEmail author
- Bjorn Feuerbacher
- Timm Kruger
- Dieter Lüst
- Anton Rebhan
- Andreas Wipf

- 1.University of Heidelberg-Heidelberg-Germany
- 2. Heidenheim Germany
- 3.University of EdinburghEdinburghUnited Kingdom
- 4. Ludwig Maximilians University MunichMunichGermany
- 5. Technical University of Vienna, Vienna, Austria
- 6. Friedrich Schiller University JenaJenaGermany

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