第二次作业——Chapter 2.Thermodynamic Concepts and Processes

热统 郭忠智2014301020087

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Problem 2.1. Physiological sensation of temperature
(a) Suppose you are blindfolded and place one hand in a pan of warm water and the other hand in a pan of cold water. Then your hands are placed in another pan of water at room temperature. What temperature would each hand perceive?
答：先前放入热水中的手会感觉到室温的水较冷；先前放入冷水中的手感觉到室温的水较热。
(b) What are some other examples of the subjectivity of our perception of temperature?
答：同样的水温，夏天感觉会很烫，但是冬天感觉不到那么烫了；
洗澡前后对室内的气温的感觉会有不同，在洗完热水澡之后感觉温度会变低。

Problem 2.2. Describe some other properties that also satisfy a law similar to the zeroth law.
答：在数学上，如果a,b,c满足a=b,b=c,则有a=c
Problem 2.3. Why are thermometers relatively small devices in comparison to the system of interest?
答：因为温度计测量物体温度时是两者达到热力学平衡的过程，彼此有能量交换，所以温度计必须做得相对地小，以保证所测量的物体和温度计交换的能量相对于物体本身的能量足够小，以致可以忽略而不影响测量结果。
Problem 2.4. Temperature scales
(a) The Fahrenheit scale is defined such that the ice point is at 32◦F and the steam point is 212◦F. Derive the relation between the Fahrenheit and Celsius temperature scales.
答：假设Fahrenheit温标和Celsius温标有如下关系：

$$T_{Fahrenheit}=aT_{Celsius}+b$$
代入ice point值： $$\left\{ \begin{aligned} T_{Fahrenheit}(ice)=32^{\circ}F\\ \\ T_{Celsius}(ice)=0^{\circ}C \end{aligned} \right. \implies{b=32}$$
又代入steam point值：
$$\left\{ \begin{aligned} T_{Fahrenheit}(steam)=212^{\circ}F\\ \\ T_{Celsius}(ice)=100^{\circ}C \end{aligned} \right. \implies{a={9\over5}}$$
所以有 $$T_{Fahrenheit}={9\over5}T_{Celsius}+32$$
(b) What is body temperature (98.6◦F) on the Celsius and Kelvin scales?
答：$T_{Fahrenheit}=98.6^{\circ}F$，由 $$T_{Fahrenheit}={9\over5}T_{Celsius}+32$$
有 $$T_{Celsius}=37.0^{\circ}C$$
又由 $$T_{Celsius}=T-273.15$$
所以 $$T=310.15K$$
(c) A meteorologist in Canada reports a temperature of 30◦C. How does this temperature compare to 70◦F?
答：由 $$T_{Fahrenheit}={9\over5}T_{Celsius}+32$$
当$T_{Celsius}=30^{\circ}C$ , $T_{Fahrenheit}=77^{\circ}F$ , 所以$30^{\circ}C较大$。

(d) The centigrade temperature scale is defined as

$$T_{centigrade} = (T −T_{ice}){100\over{T_{steam} −T_{ice}}} \tag {2.4}$$
where Tice and Tsteam are the ice and steam points of water (see Table 2.1). By definition, there are 100 centigrade units between the ice and steam points. How does the centigrade unit defined in (2.4) compare to the Kelvin or Celsius unit? The centigrade scale has been superseded by the Celsius scale.
答：代入数值，式(2.4)可写为： $$T_{centigrade} = (T −273.15){100\over{99.97}}$$
又有： $$T_{celsius}=T-273.15 \tag{2.3}$$
可以看出，$T_{celsius}$或者$T$的单位大小比$T_{centigrade}$单位大小要小。

Problem 2.5. What is the range of temperatures that is familiar to you from your everyday experience and from your prior studies?
答：$0—100^{\circ}C$
Problem 2.6. Are the following processes reversible or irreversible?
(a) Squeezing a plastic bottle.
答：不可逆
(b) Ice melting in a glass of water.
答：不可逆
(c) Movement of a real piston (where there is friction) to compress a gas.
答：可逆
(d) Pumping of air into a tire.
答：不可逆

Problem 2.7. Work To refresh your understanding of work in the context of mechanics, look at Figure 2.3 and explain whether the following quantities are positive, negative, or zero:
(a) The work done on the block by the hand.
(b) The work done on the block by the Earth.
(c) The work done on the hand by the block (if there is no such work, state so explicitly).
解：由

$$W=Fx\cos\theta$$
(a)正，因为物体运动方向和手对物体的作用力的方向一致；
(b)负，因为物体运动方向和地球对物体的作用力的方向所夹的角为钝角;
(c)负，因为物体对手的力的方向和手的位移方向相反

Problem 2.8. Work in a cyclic process
Consider the cyclic process as described in Example 2.1.
(a) Because the system was returned to its original pressure and volume, why is the net amount of work done on the system not zero?
答：在P-V图中，状态曲线下方的面积和外界对气体做的功的数值绝对值相等，当初始状态的体积大于末状态的体积时功为正值，反之为负值，在本例中，$1\to2$过程做的功的数值绝对值大于$3\to4$过程，所以当经历$1\to2\to3\to4\to1$整个循环后，外界对系统做的功的值并不为0.
(b) What would be the work done on the gas if the gas were taken from 1 → 2 → 3 and then back to 1 along the diagonal path connecting 3 and 1?
答：改变路径为$1\to2\to3\to1$,$1\to3$为对角线连接，$1\to2$和$2\to3$
过程所做的功和例2.1中$W_{1\to2}$相同,即

$$W_{1\to2}=-P_{high}(V_{high}-V_{low}).$$
$3\to1$过程所做的功为：
$$W_{3\to1}={1\over2}(P_{high}+P_{low})(V_{high}-V_{low})$$
所以整个过程所做的功为： $$W_{net}^{'}=W_{1\to2}+W_{3\to1}=-{1\over2}(P_{high}-P_{low})(V_{high}-V_{low})$$

Problem 2.9. Pumping air
A bicycle pump contains one mole of a gas. The piston fits tightly so that no air escapes and friction is negligible between the piston and the cylinder walls. The pump is thermally insulated from its surroundings. The piston is quickly pressed inward. What happens to the temperature of the gas? Explain your reasoning.
答：气体温度上升。这是一个绝热过程，绝热过程中有外界对系统做的功等于系统内能的改变量，而本系统又不与外界交换热量，内能和气体温度成正比，内能增加，所以气体温度上升。

Problem 2.10. Distinguishing different types of water transfer How could the owner of the pond distinguish between the different types of water transfer assuming that the owner has flow meters, a tarpaulin, and a vertical pole?
答：将流量计安装在池塘和其他储水装置之间的排水沟之间，监测流速；将竖直杆竖直插在池塘中，并在其露出水面的部分做上标记，以及刻度，监测水位的升降。通过观察流量计和竖直杆的示数来确定不同的水的转化。
降雨：流量计的示数增大，竖直杆标记的水位上升
水蒸气液化：流量计的示数缓慢增加，竖直杆标记的水位缓慢上升
蒸发：流量计的示数缓慢减小，竖直杆标记的水位缓慢下降
Problem 2.11. Convert the statement “I am cold, please turn on the heat,” to the precise language of physics.
答：I feel the temperature of the air is too low, please turn on the heating machine to transfer some energy from the hotter machine to the colder air.

Problem 2.12. Heat capacities large and small
Give some examples of materials that have either a small or a large heat capacity relative to that of water. You can find values of the heat capacity in books on materials science and on the internet.
答：常见气体的比热容

(单位:kJ/(kg·K))

物质	Cp	Cv
氧气	0.909	0.649
氢气	14.05	9.934
水蒸气	1.842	1.381
氮气	1.038	0.741

Problem 2.13. In Example 2.1 we showed that the net work done on the gas in the cyclic process shown in Figure 2.4 is nonzero. Assume that the gas is ideal with N particles and calculate the energy transfer by heating in each step of the process. Then explain why the net work done on the gas is negative and show that the net change of the internal energy is zero.
解：由

$$Q=\int{CdT}=\int{Cd(PV)\over{Nk}}={C\over{Nk}}\int{(PdV+VdP)}$$
所以 $$Q_{1\to2}={C_P\over{Nk}}\int_{V_{low}}^{V_{high}}P_{high}dV={{5\over2}P_{high}(V_{high}-V_{low})}$$
$$Q_{2\to3}={C_V\over{Nk}}\int_{P_{high}}^{P_{low}}V_{high}dP={{3\over2}V_{high}(-P_{high}+P_{low})}$$
$$Q_{3\to4}={C_P\over{Nk}}\int_{V_{high}}^{V_{low}}P_{low}dV={{5\over2}P_{low}(V_{low}-V_{high})}$$
$$Q_{4\to1}={C_V\over{Nk}}\int_{P_{low}}^{P_{high}}V_{high}dP={{3\over2}V_{high}(P_{high}-P_{low})}$$
因为历经一个循环后，气体回到原来的状态，其内能不变，整个循环过程中系统从环境中吸收的热量和系统对环境做的功的值相等，有以上四个式子可以得到，交换的热量不为零，故做功也不为零。且可以得到： $$\Delta{E}=Q+W=0$$
Problem 2.14. Give some examples of adiabatic processes.
答：绝热压缩：（1）给自行车打气时，可以感觉到气筒温度上升，这正是因为气体压强上升的足够快到可视为绝热过程的缘故，热量没有逃逸，因而温度上升。（2）柴油机在压缩冲程时正是靠绝热压缩原理来给燃烧室内的混合气体点火的。
绝热膨胀：给轮胎放气时，可以明显感觉到放出的气体比较凉，这正是因为气体压强下降的足够快到可视为绝热过程的缘故，气体内能转化为机械能，温度下降。

Problem 2.15. Use (2.44) and the ideal gas pressure equation of state in (2.8) to show that in a quasistatic adiabatic processes P and V are related as

$$PV^ γ = constant. \tag{2.46}$$
Also show that T and P are related as $$TP^{(1−γ)/γ} = constant\tag{2.47}$$
解：（1）由 $$\left\{ \begin{aligned} PV=NkT \\ TV^{γ-1}=constant\\ \end{aligned} \right.$$
两式相乘有 $$PV^γ=constantNk=newconstant=C^{'}$$
（2）
$$(PV)^{γ-1} = (NkT)^{γ-1} \implies \frac{P^{γ-1}}T=\frac{(Nk)^{γ-1}T^{γ-1}}{constant}$$
$$\implies P^{γ-1}=C^{'}T^γ$$
$$\implies TP^{γ\over {γ-1}}=C^{"}$$

Problem 2.16. Although we do work on an ideal gas when we compress it isothermally, why does the energy of the gas not increase?
答：因为

$$E= {3\over2}NkT（单原子理想气体）,$$ 所以理想气体的能量只与温度和粒子数有关，等温过程不改变气体温度，故气体能量不变。

Problem 2.17. Compression of air Air initially at 20◦C is compressed by a factor of 15.
(a) What is the final temperature assuming that the compression is adiabatic and γ ≈ 1.4,17 the value of γ for air in the relevant range of temperatures? By what factor does the pressure increase?
解：绝热过程有

$$TP^{γ\over {γ-1}}=constant\implies\frac{T_1}T_2=(\frac{P_1}P_2)^{{γ-1}\over γ}\tag1$$
$$PV^γ=constant\implies\frac{P_1}P_2=(\frac{V_2}V_1)^γ\tag2$$
代入$\frac{V_2}V_1={1\over15}、γ=1.4$得到：
$$T_2=866K$$
(b) By what factor does the pressure increase if the compression is isothermal?
解：等温过程有 $$PV=constant\implies P_1V_1=P_2V_2$$
所以 $$\frac{P_2}P_1=\frac{V_1}V_2=15$$
所以压强增大为原来的15倍。
(c) For which process does the pressure change more?
答：绝热过程的压强比由（2）可得，为 $$\frac{P_2}P_1=15^γ>15$$ 所以绝热过程压强增量更大。

Problem 2.18. Work done in a quasistatic adiabatic process
(a) Use the result that we derived in (2.53) to obtain the alternative form (2.54).
解：

$$W=C_V(T_2-T_1)=\frac{C_V}{Nk}(P_2V_2-P_1V_1)$$ $$=\frac{C_V}{C_P-C_V}(P_2V_2-P_1V_1)=\frac{(P_2V_2-P_1V_1)}{γ-1}$$
(b) Show that another way to derive (2.54) is to use the relations (2.14) and (2.46).
解： $$W_{1\to2}=-\int_{V_1}^{V_2}P(T,V)dV\tag{2.14}$$
$$PV^γ=constant\tag{2.46}\\\implies P=\frac{P_1V_1^γ}{V^γ}$$
所以 $$W_{1\to2}=-\int_{V_1}^{V_2}P(T,V)dV=-\int_{V_1}^{V_2}\frac{P_1V_1^γ}{V^γ}dV\\=\frac{(P_2V_2-P_1V_1)}{γ-1}$$

Problem 2.19. A Carnot refrigerator
A refrigerator cools a closed container and heats the outside room surrounding the container. According to the second law of thermodynamics, work must be done by an external body for this process to occur. Suppose that the refrigerator extracts the amount Qcold from the container at temperature Tcold and transfers energy Qhot at temperature Thot to the room. The external work supplied is W (see Figure 2.10). We define the coefficient of performance (COP) as

$$COP =\frac{ what\ you\ get }{what\ you\ pay\ for} =\frac{Q_{cold}}W\tag{2.88}$$
Show that the maximum value of the COP corresponds to a reversible refrigerator and is given by
$$COP =\frac{T_{cold}}{T_{hot}-T_{cold}}\tag{2.89}$$
Note that a refrigerator is more efficient for smaller temperature differences.
解： $$\Delta S_{total}=\Delta S_{cold}+\Delta S_{hot}\\=-\frac{Q_{cold}}{T_{cold}}+ \frac{Q_{hot}}{T_{hot}}\ge 0$$
即 $$\frac{Q_{hot}}{T_{hot}}\ge \frac{Q_{cold}}{T_{cold}}$$
亦即 $$\frac{Q_{hot}}{Q_{cold}}\ge \frac{T_{hot}}{T_{cold}}$$
所以有：
$$COP =\frac{Q_{cold}}W=\frac{Q_{cold}}{Q_{hot}-Q_{cold}}=\frac{1}{\frac{Q_{hot}}{Q_{cold}}-1}\le \frac{1}{\frac{T_{hot}}{T_{cold}}-1}$$
而 $$\frac{1}{\frac{T_{hot}}{T_{cold}}-1} =\frac{T_{cold}}{T_{hot}-T_{cold}}$$
故得证。

Problem 2.20. Heat pump
A heat pump works on the same principle as a refrigerator, but the goal is to heat a room by cooling its cooler surroundings. For example, we could heat a building by cooling a nearby body of water. If we extract energy Qcold from the surroundings at Tcold, do work W, and deliver Qhot to the room at Thot, the coefficient of performance is given by

$$COP =\frac{what\ you\ get}{ what\ you\ pay\ for} =\frac{Q_{hot}}{ W}\tag{2.90}$$

What is the maximum value of COP for a heat pump in terms of Tcold and Thot? What is the COP when the outside temperature is 0◦C and the interior temperature is 23◦C? Is it more effective to operate a heat pump during the winters in New England where the winters are cold or in the Pacific Northwest where the winters are relatively mild? (It is too bad that the maximum efficiency of a heat pump occurs when it is needed least.)
解：

$$\Delta S_{total}=\Delta S_{cold}+\Delta S_{hot}\\=-\frac{Q_{cold}}{T_{cold}}+ \frac{Q_{hot}}{T_{hot}}\ge 0$$
即 $$\frac{Q_{hot}}{T_{hot}}\ge \frac{Q_{cold}}{T_{cold}}$$
亦即 $$\frac{Q_{cold}}{Q_{hot}}\le \frac{T_{cold}}{T_{hot}}$$
所以有：
$$COP =\frac{Q_{hot}}W=\frac{Q_{hot}}{Q_{hot}-Q_{cold}}=\frac{1}{1-\frac{Q_{cold}}{Q_{hot}}}\\\le\frac{1}{1-\frac{T_{cold}}{T_{hot}}}=\frac{T_{hot}}{T_{hot}-T_{cold}}$$
带入$T_{cold}=273.15K、T_{hot}=23+273.15=296.15K.$

得到：

$$COP=\frac{296.15}{23}=12.876.$$
从推得式子知道寒冷的冬天泵的COP比在温和的冬天更小。

Problem 2.21. Water in contact with two heat bathsin succession
The temperature of 1kg of water at 0◦C is increased to 50◦C by first bringing it into contact with a heat bath at 25◦C and then with a heat bath at 50◦C. What is the change in entropy of the entire system? How does this change in entropy compare with the change that was found in Example 2.15?
解：第一次接触，水的熵变为

$$\Delta S_{A1}=Cln\frac{T_A}{T_{B1}}=4184J/Kln\frac{273+25}{273+0}=366.6J/K$$
从第一个热缸传入水的热量为 $$Q=C(T_{B1}-T_A)=4181J/K\times 25K=104600J.$$
第一个热缸的熵变为 $$\Delta S_{B1}=-\frac{Q}{T_{B1}}=-{104600J\over 298K}=-351J/K.$$
所以第一次热接触的熵变为 $$\Delta {S1}=\Delta S_{A1}+\Delta S_{B1}=366.6J/K-351J/K=15.6J/K.$$
对于第二次热接触，
水的熵变为 $$\Delta S_{A2}=Cln\frac{T_{B2}}{T_{B1}}=4184J/Kln\frac{273+50}{298}=337J/K$$
第二个热缸传入水的能量为 $$Q=C(T_{B2}-T_{B1})=4181J/K\times 25K=104600J.$$
第二个热缸的熵变为 $$S_{B2}=-\frac{Q}{T_{B2}}=-{104600J\over 323K}=-324J/K.$$
所以第二次热接触的熵变为 $$\Delta {S2}=\Delta S_{A2}+\Delta S_{B2}=337J/K-324J/K=13J/K.$$
所以整个过程中系统的熵变为： $$\Delta S=\Delta {S1}+\Delta {S2}=15.6J/K+13J/K=28.6J/K.$$
这样的过程熵变小于Example1.15中的情况。

Problem 2.22. Use the numerical values for various quantities from Example 2.15 to show that (2.105) gives the same numerical result as (2.98).
解：

$$\Delta {S}=\Delta S_{A}+\Delta S_{B}=704J/K-648J/K=56J/K$$ $$\tag{2.98}$$
$$\Delta S=C_A [ ln\frac{T_B}{T_A}+\frac{T_A}{T_B} -1]\tag{2.105}$$
带入数值$T_A=273K、T_B=323K$得：
$$\Delta S=4184J/K[ ln\frac{323}{273}+\frac{273}{323} -1]=55.998J/K$$ 可见两个式子结果几乎相同。

Problem 2.23. More work
(a) Show that the work performed by the heat engine in Example 2.19 is given by W = CA(TA −T) + CB(TB −T), (2.110) where CA and CB are constants and T is given by (2.109) if the process is reversible. (Recall that our convention is to consider the work done on a system, except when we are discussing heat engines.)
解：释放的最大功应为整个系统热量变化的负值，即：

$$W=-(Q_A+Q_B)=-\int_{T_A}^TC_AdT-\int_{T_B}^TC_BdT\\=C_A(T_A-T)+C_B(T_B-T).$$
(b) Suppose that CA = CB = C (a constant independent of T) in (2.93) and (2.109). Compare the form of the expressions for the final temperature.

解：若$C_A=C_B=C$,则

$$T=\frac{C_AT_A+C_BT_B}{C_A+C_B}\tag{2.93}\\=\frac{T_A+T_B}2$$
$$T=T_A^{\frac{C_A}{C_A+C_B}}T_B^{\frac{C_B}{C_A+C_B}}\tag{2.109}\\=(T_AT_B)^{1\over2}\le\frac{T_A+T_B}2$$

(c) Suppose that TA = 256K and TB = 144K. What are the relative values of the final temperatures in (2.93) and (2.109) assuming that the heat capacities of the two bodies are equal? For which process is the final temperature lower? Why?

解：$T_A=256K,T_B=144K$

则

$$T_1=\frac{T_A+T_B}2=200K$$
$$T_2=(T_AT_B)^{1\over2}=196K$$

可以看出，（2.109）中算出的最终温度更低，因为系统的一部分能量被转化成了功。

(d) Suppose that the heat capacities of both bodies depend linearly on the temperature T rather than being constant; that is, CA = AT and CB = BT, where A and B are constants. What is the final temperature assuming that the two bodies are placed in thermal contact? What is the final temperature for the case when the maximum work is extracted? What is the maximum work done?
解：（1）

$$\Delta E=Q_A+Q_B \\\implies \int_{T_A}^TATdT+\int_{T_B}^TBTdT=0\\ \implies T=\sqrt{\frac{AT^2+BT^2}{A+B}}$$
（2）能做最大功时的系统的过程可逆，则有
$$\Delta S=\Delta {S_A}+\Delta {S_B}=0\\\implies \int_{T_A}^T{\frac{AT}T}dT+\int_{T_B}^T{\frac{BT}T}dT=0\\\implies A(T-T_A)+B(T-T_B)=0\\\implies T=\frac{AT_A+BT_B}{A+B}.$$
（3） $$W=-(Q_A+Q_B)=-\int_{T_A}^TATdT-\int_{T_B}^TBTdT\\={A\over2}(T_A^2-T^2)+{B\over2}(T_B^2-T^2).$$

Problem 2.24. Applications of (2.133)
(a) Use (2.133) to derive the relation (2.44) between T and V for a quasistatic adiabatic process.
解：

$$\Delta S={3\over2}Nkln\frac{T_2}{T_1}+Nkln\frac{V_2}{V_1}=0\\\implies (\frac{T_2}{T_1})^{3\over2}=\frac{V_1}{V_2}\\\implies T_2V_2^{2\over3}=T_1V_1^{2\over3}\\\implies TV^{γ-1}=constant.$$
(b) An ideal gas of N particles is confined to a box of chamber V1 at temperature T. The gas is then allowed to expand freely into a vacuum to fill the entire container of volume V2. The container is thermally insulated. What is the change in entropy of the gas?
解： $$\Delta S=Nkln\frac{V_2}{V_1}.$$ 这相当于式（2.133）中右边第一项取0的情况，因为整个过程气体能量不变，温度不变。
(c) Find ∆S(T,P) for an ideal classical gas.
解： $$\Delta S={3\over2}Nkln\frac{T_2}{T_1}+Nkln\frac{\frac{NkT_2}{P_2}}{\frac{NkT_1}{P_1}}\\={5\over2}Nkln\frac{T_2}{T_1}+Nkln\frac{P_1}{P_2}$$

Problem 2.25. The enthalpy
(a) Given the definition of the enthalpy in (2.29) show that

$$dH = TdS + V dP + µdN, \tag{2.159}$$
and
$$T =(\frac{∂H}{ ∂S})_{P,N} \tag {2.160}$$
$$V =(\frac{∂H}{ ∂P}) _{S,N} , \tag{2.161}$$
$$µ =(\frac{∂H}{ ∂N})_{S,P} . \tag{2.162}$$
解：由 $$H=E+PV\tag{2.29}$$ ，对其两边求全微分得： $$dH=dE+PdV+VdP\\=(TdS-PdV+µdN)+PdV+VdP\\=TdS+VdP+µdN$$
又由全微分定义可得： $$dH=T =(\frac{∂H}{ ∂S})_{P,N}dS+(\frac{∂H}{ ∂P}) _{S,N}dP+(\frac{∂H}{ ∂N})_{S,P}dN$$
对比以上两式系数，即可得到（2.160），（2.161），（2.162）

(b) Show that H is a minimum for an equilibrium system at fixed entropy.
解：由

$$\Delta A=\Delta E+P\Delta V-T\Delta S=\Delta G\le0\tag{2.152}$$
当固定熵时$\Delta S=0$,则有 $$\Delta A=\Delta E+P\Delta V=\Delta G=\Delta H\le0$$
所以平衡系统的焓为其最小值。

Problem 2.26. Maximum useful work and free energy changes
(a) Show that if the change in volume of the system is zero, ∆V = 0, and the initial and final temperatures are that of the heat bath, then the maximum useful work is −∆F.
解：由

$$W_{useful}\le-(\Delta E+P_{bath}\Delta V-T_{bath}\Delta S)=-\Delta A\\ $$ $\tag{2.171}$
当$\Delta V=0,T=T_{bath}$时，上式化为 $$W_{useful}\le-(\Delta E-T\Delta S)=-\Delta F$$
(b) Show that if the initial and final temperature and pressure are that of the bath, then the maximum useful work is −∆G.
解：此时由 $$W_{useful}\le-(\Delta E+P_{bath}\Delta V-T_{bath}\Delta S)=-\Delta A\\ $$ $\tag{2.171}$
代入$P=P_{bath},T=T_{bath}$有
$$W_{useful}\le-(\Delta E+P_\Delta V-T\Delta S)=-\Delta G.$$

Problem 2.27. More Maxwell relations
From the differentials of the thermodynamic potentials

$$dF = −SdT −PdV \tag {2.188}$$
$$dG = −SdT + VdP \tag{2.189}$$
$$dH = TdS + VdP \tag{2.190}$$
derive the Maxwell relations
$$(\frac{∂S}{ ∂V} )_T =(\frac{∂P}{ ∂T})_V \tag {2.191}$$ $$(\frac{∂S}{ ∂P})_T = −(\frac{∂V}{ ∂T})_P \tag {2.192}$$ $$(\frac{∂T}{ ∂P})_S =(\frac{∂V}{ ∂S})_P . \tag{2.193}$$

Also consider a variable number of particles to derive the Maxwell relations

$$(\frac{∂V}{ ∂N})_P =(\frac{∂µ}{ ∂P})_N , \tag{2.194}$$ and $$(\frac{∂µ}{ ∂V}_N =−(\frac{∂P}{ ∂N})_V .\tag {2.195}$$
解： $$dF = −SdT −PdV \tag {2.188}\\\implies S=-(\frac{\partial F}{\partial T})_V,P=-(\frac{\partial F}{\partial V})_T$$ 由于二阶偏导数与求导顺序无关，所以有
$$\frac{\partial^2 F}{\partial T\partial V}=\frac{\partial^2 F}{\partial V\partial T}\\\\\implies(\frac{∂S}{ ∂V} )_T =(\frac{∂P}{ ∂T})_V$$

$$dG = −SdT + VdP \tag{2.189}\\\implies S=-(\frac{\partial G}{\partial T})_P,V=(\frac{\partial F}{\partial P})_T$$ 由于二阶偏导数与求导顺序无关，所以有
$$\frac{\partial^2 G}{\partial T\partial P}=\frac{\partial^2 G}{\partial P\partial T}\\\\\implies(\frac{∂S}{ ∂P})_T = −(\frac{∂V}{ ∂T})_P$$

$$dH = TdS + VdP \tag{2.190}\\\implies T=(\frac{\partial H}{\partial S})_P,V=(\frac{\partial H}{\partial P})_S$$
由于二阶偏导数与求导顺序无关，所以有
$$\frac{\partial^2 H}{\partial S\partial P}=\frac{\partial^2 H}{\partial P\partial S}\\\\\implies(\frac{∂T}{ ∂P})_S =(\frac{∂V}{ ∂S})_P .$$

假设一个等温过程，dT=0，则有：

$$dG = \mu dN + VdP\\\\\implies \mu =(\frac{\partial G}{\partial N})_P,V=(\frac{\partial G}{\partial P})_N$$
由于二阶偏导数与求导顺序无关，所以有
$$\frac{\partial^2 G}{\partial N\partial P}=\frac{\partial^2 G}{\partial P\partial N}\\\\\implies(\frac{∂V}{ ∂N})_P =(\frac{∂µ}{ ∂P})_N$$

又有：

$$dF =−PdV +\mu dN\\\\\implies \mu =(\frac{\partial F}{\partial N})_V,P=-(\frac{\partial G}{\partial P})_N$$
由于二阶偏导数与求导顺序无关，所以有
$$\frac{\partial^2 F}{\partial N\partial V}=\frac{\partial^2 F}{\partial V\partial N}\\\\\implies(\frac{∂µ}{ ∂V}_N =−(\frac{∂P}{ ∂N})_V .$$

Problem 2.28. Show that the enthalpy of an ideal gas is a function of T only.
解：把焓视为温度和压强的函数$H=H(T,P)$，由

$$dH = TdS + VdP$$
所以有 $$(\frac{\partial H}{\partial P})_T=T(\frac{\partial S}{\partial P})_T+V\tag{*}$$
又有 $$(\frac{∂S}{ ∂P})_T = −(\frac{∂V}{ ∂T})_P$$
且 $$PV=NkT\implies (\frac{∂V}{ ∂T})_P=\frac VT$$
所以（*）式右边为0，即H仅为T的函数。

Problem 2.29. Free expansion of a van der Waals gas
Calculate (∂T/∂V )E for the van der Waals energy equation of state (2.24) and show that a free expansion results in cooling.
解：考虑气体从V1自由膨胀到V2，dQ=0,dW=0,由

$$E={3\over2}NkT-N{N\over V}a\tag{2.24}$$
$$(\frac{\partial T}{\partial V})_E= -\frac{(\partial E/ \partial V)_T}{(\partial E/ \partial T)_V}\\=-\frac{2Na}{3kV^2}$$

$$\Delta T=T_2-T_1=\int_{V_1}^{V_2}-\frac{2Na}{3kV^2}dV\\=\frac{2Na}{3k}( \frac 1{V_2}-\frac 1{V_1})$$
可见气体温度下降。

Problem 2.30. Low density limit of the thermal expansion coffeient
For simplicity, consider low densities and show that α in (2.223) is given by

$$α = \frac{k(1−bρ)}{ kT −2aρ(1−bρ)^2} \tag{2.224a}\\$$ $$≈ \frac{1}{T} [1−ρ(b− \frac{2a}{ kT})]. \tag{2.224b}$$
Use the approximation$(1 + x)^n ≈ 1 + nx$for small x to obtain (2.224b). Then show that in this approximation $$Tα−1 = ρ[(2a/kT)−b]. \tag{2.225}$$
解：由$(1 + x)^n ≈ 1 + nx$
$$α = \frac{k(1−bρ)}{ kT −2aρ(1−bρ)^2}$$ $$=\frac{k}{ \frac{kT}{1−bρ} −2aρ(1−bρ)}$$ $$≈\frac{k}{ kT(1+bρ) −2aρ}$$
上式略去了平方项，再由$(1 + x)^n ≈ 1 + nx$得
$$α ≈\frac{1}{T} [1−ρ(b− \frac{2a}{ kT})].$$
即 $$Tα−1 = ρ[(2a/kT)−b].$$

Problem 2.31.
Consider the function

$$z(x,y) = x^2y + 2x^4y^6. \tag{2.239}$$
Calculate $∂z/∂x, ∂z/∂y, ∂^2z/∂x∂y,$and$∂^2z/∂y∂x$ and show that$∂^2z/∂x∂y = ∂^2z/∂y∂x.$
解： $$∂z/∂x=2xy+8x^3y^6$$
$$∂z/∂y=x^2+12x^4y^5$$
$$∂^2z/∂x∂y=2x+48x^3y^5$$
$$∂^2z/∂y∂x=2x+48x^3y^5$$
可以看到$∂^2z/∂x∂y = ∂^2z/∂y∂x.$

Problem 2.32. Simple Legendre transforms
(a) Calculate the Legendre transform of $f(x) = x^3.$
解：由

$$g=f(x)-xm\tag{2.242}$$
其中$m=f^{'}(x)=3x^2\implies x=\sqrt{m\over3}$,所以$f(x) = x^3$的勒让德变换为：
$$g(m)=x^3-x(3x^2)=-2(\frac m3)^{\frac 32}$$
(b) Calculate the Legendre transforms of the functions f(x) = x and f(x) = sinx if they exist.
解：由 $$g=f(x)-xm\tag{2.242}$$
（1）当f(x)=x时，则有$f(x)=x\implies m=f^{'}(x)=1$，所以 $$g(m)=x-x=0$$
（2）$f(x)=sin x\implies m=f^{'}(x)=cos x$,$f^{'}(x)$不是x的单调函数，故勒让德变换不存在。

Problem 2.33. The Helmholtz free energy as a Legendre transform Start from the function E(S,V,N) and use the Legendre transform to find the function F(T,V,N).
解：

$$E(S,V,N)$$ ,由热力学温度的定义 $$\frac 1T=(\frac{\partial S}{\partial E})_{V,N}\tag{2.65}$$
所以 $$T=(\frac{\partial E}{\partial S})_{V,N}$$
$$F=E-TS$$
$$dF=dE-TdS-SdT$$ $$=TdS-PdV+\mu dN-TdS-SdT$$ $$=-SdT-PdV+\mu dN$$
可以看到亥姆霍兹自由能是F是内能E的勒让德变换。