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1、物理化学答案_上册_高教_第五版（Answer _ _ on physical chemistry of higher education _ Fifth Edition）Chapter 1 gasPVTNature1.1 material expansion coefficientIsothermal compression rateThe definition is as followsTest the introduction of ideal gas,Relationship with pressure and temperature.According to the ideal ga
2、s equation1.5 two glass balls with V volume are connected by a small tube, and the bubbles are sealed with standard air. If soA ball is heated to 100 degrees C, while the other ball is maintained at 0 degrees C. The volume of the gas in the pipe is neglected and the container is soughtInternal air p
3、ressure.Solution: from the question to the condition, (1) the total mass of the system is constant; (2) the pressure remains the same in the two ball.Standard state:Therefore,As figure 1.9 shows, the container area partition, both sides are at the same pressure of hydrogen and nitrogen, two are rega
4、rded as idealGas.Page(1) when the temperature in the container is kept constant, the separator is removed and the volume of the barrier itself is negligibleSeek the pressure of two gases mixed.(2) is the molar volume of H2 and N2 equal before and after the separation of the separator?(3) after the s
5、eparation of the separator, the partial pressure of H2 and N2 in the mixture gas and the volume of their respective parts are severalSolution: (1) after isothermal mixingThat is, under the above conditions, the mixing pressure of the system is considered.(2) how is the molar volume of a component in
6、 a mixed gas defined?(3) according to the definition of volumeFor partial pressure1.11 atmospheric pressure air in a autoclave at room temperature for safety during the experiment and the use of pure nitrogen at the same temperatureThe steps are as follows: nitrogen is sent to the kettle until it is
7、 4 times as high as the air, and then the mixture gas is discharged from the tank until it is recoveredCompound atmospheric pressure. Repeat three times. The mole fraction of oxygen contained in the gas is obtained when the final exhaust gas reaches the normal atmospheric pressure.Analysis: every ti
8、me after the nitrogen gas, the gas returned to the atmospheric pressure P, and the mole fraction of the mixed gas remained unchanged.Before the first charge of nitrogen, the mole fraction of oxygen in the system isAfter nitrogen, the mole fraction of oxygen in the systembyThen,. Repeat the process a
9、bove, nAfter secondary nitrogen, the mole fraction of the system is,therefore.1.13 there are 0 degrees C, 40.530 kPa N2 gas, using ideal gas equation of state and van, der, Waals equation calculationIts molar volume. The experimental value is.Solution: the equation of state of ideal gas is usedPageU
10、sing van, der, and Waals calculations, look-up tables are known for N2 gas (Appendix seven)MatLab fzero function is used to obtain the solution of the equationDirect iteration can also be used,Take initial valueIterate ten timesAt 1.1625 degrees C, saturated acetylene gas saturated with water (i.e.,
11、 the partial pressure of the steam in the mixture is saturated at the same temperature)And vapor pressure) the total pressure is 138.7 kPa and is cooled to 10 DEG C at constant total pressure, which condenses some of the water vapor into water. Try to seek forThe amount of material that condenses in
12、 the cooling process of each mole of dry acetylene gas. Saturated vapor of water at 25 C and 10 degrees CThe pressures are 3.17 kPa and 1.23 kPa, respectively.Solution: the process is illustrated as followsIf the system is an ideal gas mixture, then1.17 a tightly packed rigid container filled with a
13、ir and with a small amount of water. But containers are in great balance at 300 K conditions, containersThe internal pressure is 101.325 kPa. If the container is moved to boiling water of 373.15 K, try to achieve a new equilibrium in the containerSome pressure. There is always water in the container
14、, and any volume change of water can be neglected. 300 K the saturated vapor pressure of waterFor 3.567 kPa.Solution: when gas phase is regarded as ideal gas, at 300 K, the partial pressure of air isPageAs the volume is constant (ignoring any change in volume of water), the partial pressure of the a
15、ir at 373.15 K isSince there is always water present in the container, the saturated vapor pressure of water at 373.15 K is 101.325 kPaThe partial pressure of water vapor is 101.325 kPa, so the total pressure of the systemPageThe second chapter is the first law of thermodynamics2.5 the initial state
16、 is 25 C, 200 kPa, 5 mol of an ideal gas, by way of a, B two different ways to the same final.By way of a, the adiabatic expansion to -28.47 degrees C, 100 kPa, the steps of the workThe constant volume heats up to the pressure200 kPa the final step of heat. Pathway B is a constant pressure heating p
17、rocess. Seeking ways of Band.Solution: first determine the initial and final systemFor the way B, his work isAccording to the first law of thermodynamics2.64 mol of an ideal gas, the temperature increased by 20 degrees CValue.Solution: according to the definition of enthalpy2.102 mol an ideal gas,.T
18、he initial state is 100 kPa, 50 DMThreeThe constant volume heating increases the volume of pressureAs large as 150 DMThreeConstant pressure cooling reduced the volume to 25 DMThree. Seeking the whole process.Solution: the process is illustrated as followsPageBecauseThen,Yes, ideal gasandIts just a f
19、unction of temperatureThe approach involves only constant volume and constant pressure processes, so it is convenient to calculate the workAccording to the first law of thermodynamicsThe 2.13 known 20 DEG C liquid ethanol (C2H5OH, l) of the expansion coefficientIsothermal compression rateDensityMola
20、r heat capacity at constant pressure. Seek 20 DEG C, liquid ethanol.Solution: the second law of thermodynamics can prove that the relationship between constant pressure molar heat capacity and constant volume molar heat capacity is as follows2.14 volume is 27 mThreeA small heating element is arrange
21、d in the insulating container, and a small hole is communicated with the atmosphere of the 100 kPa,To maintain constant air pressure in the vessel. The heating device is used to heat the air in the apparatus from 0 DEG C to 20 DEG CHow much heat is given to the air in the container?. Known air.If th
22、e air is an ideal gas, the temperature of the air in the container will be uniform during heating.Solution: in this problem, the pressure of the air in the container is constant, but the mass of the material varies with the temperaturePageNote: cannot be applied in the above questionsAlthough the vo
23、lume of the container is constant. This is because, fromThe air out of the hole will do the work to the environment. The work done is as follows:When the temperature is T, the system temperature is increased by dT, and the mass of the air discharged from the container isWork doneThats exactly what i
24、ts worthandThe difference in heat calculated.2.15 volume is 0.1 MThreeAn insulating partition board is arranged in the constant volume closed container, wherein the sides are respectively 0 DEG C, 4 mol Ar (g) and150 degrees C, 2 mol Cu (s). The partition removed, the system reaches heat balance, te
25、mperature and process for final t.Known: molar heat capacity at constant pressure of Ar (g) and Cu (s)Respectively asandAnd assume that they do not change with temperature.Here is the diagram belowIt is assumed that the insulation wall is in close contact with the copper block, and the volume of cop
26、per block varies with temperatureThe process can be viewed as a constant volume process, thereforePageIt is assumed that gas can be regarded as an ideal gas,Then,The temperature of the water gas at the outlet of the 2.16 water gas producer is 1100 degrees C, and the mole fractions of CO (g) and H2 (
27、g) are 0.5.If 300 kg of water gas is cooled from 1100 C to 100 C per hour, and heated by the recovered heat, the water temperature is caused by25 degrees C rise to 75 degrees C. Seek the quality of hot water produced per hour. Molar constant heat capacity of CO (g) and H2 (g)TemperatureFunction rela
28、tions, book, appendix, waterSpecific heat capacity at constant pressure.Interpretation: the mass fraction of CO (g) and H2 (g) in 300 kg water gas is respectively300 kg of water gas from 1100 degrees C to 100 degrees C cooling heatThe quality of producing hot water is m2.18 monatomic ideal gas A in
29、a mixture of diatomic ideal gas B, a total of 5 mol, mole fractionInitial statetemperaturePressure. The mixture is insulated against constant external pressureExpand toEquilibrium state. For the final temperatureAnd process.Solution: the process is illustrated as followsAnalysis: because of the adia
30、batic process, the change of thermodynamic energy in process is equal to the form of work between the system and environmentThe energy exchanged by the potential. Therefore,PageMonatomic moleculeDiatomic moleculeSince the ideal gases, U and H, are only functions of temperature, so2.19 there is an in
31、sulating partition in the insulated container of the piston in the vicinity. The sides of the bulkhead are 2 mol and 0 C respectivelyGas A and 5 mol, 100 degrees C diatomic ideal gas B, two gas pressure is 100 kPa. Pressure dimension outside the pistonRemain unchanged at 100 kPa. This will be remove
32、d from the container plate, the two kinds of gas mixture to reach an equilibrium state. The temperature for the final and TProcedural.Solution: the process is illustrated as followsIt is assumed that the insulation partition plate is replaced by a heat conduction baffle plate, then the heat balance
33、is removed, and then the clapboard is removed to make the mixture mixedPageThe work is convenient because of the constant external pressureSince the cylinder is insulated,2.20 there is a fixed insulated partition in the insulated container of the piston in the area. The bulkhead is single C with 2 m
34、ol and 0 DEG C on the piston sideThe sub ideal gas A is equal to the constant ambient pressure; the other side of the barrier is 6 mol, 100 C diatomicThink of gas B, whose volume is constant. The insulating layer of the insulating partition is removed so that it becomes a heat conducting plate and t
35、he system is T when it reaches equilibriumAnd process.Solution: the process is illustrated as followsObviously, in the process, A is constant voltage, and B is constant volume, thereforePageIbid., first seeking meritSimilarly, due to cylinder insulation, the first law of thermodynamics2.235 mol diat
36、omic gas from the initial state of 300 K, 200 kPa, first reversibly reversibly expanded to a pressure of 50 kPa, at adiabaticThe inverse compression pressure of 200 kPa to the final state. For the final temperature and the whole process of Tand.Solution: the process is illustrated as followsBe sureT
37、he adiabatic equation of state can be applied to second steps onlyPairs of diatomic gasesthereforeSince the U and H of ideal gases are only functions of temperature,PageThe whole process is convenient because the second step is adiabatic. The first step is constant temperature reversible2.24 it is p
38、roved that the absolute value of the slope of the adiabatic reversible line is greater than that of the constant temperature reversible line at the point where the ideal gas p-V diagram takes placeRight value.According to the adiabatic equation of ideal gas,have toTherefore. Therefore, the hot wireT
39、he slope isConstant temperature lineThe slope is. BecauseTherefore, adiabatic reversibleThe absolute value of the slope of the line is greater than the absolute value of the constant temperature reversible line.2.25, the insulated and constant volume cylinder is provided with an adiabatic ideal pist
40、on without friction, and the left and the right sides of the piston are respectively50 DMThreeThe monatomic ideal gases A and 50 DMThreeDiatomic ideal gas B. The two gases are 0 C, 100 kPa. A gasAn electric heating wire with negligible volume and heat capacity is arranged in the inner part. Now, aft
41、er energized, slowly heating the left gas A, so that the push pistonCompress the right gas B to the final pressure to 200 kPa. O:(1) the final temperature of gas B.Page(2) the work obtained by gas B;.(3) the final temperature of gas A.(4) the heat obtained by the gas A from the electric wire.Solutio
42、n: the process is illustrated as followsBecause of the slow heating, B can be regarded as undergoing an adiabatic reversible processSolution of the first law of thermodynamicsThe final A gas temperature is the ideal gas state equation solving,Treat A and B as whole, W = 0, therefore2.25 in an insula
43、ted vessel with a piston, there is a solid substance of 4.25 mol and 5 mol, a monatomic ideal gas, B, substance APageA. Initial temperaturePressure. This is gas BFor systems, reversible expansion toSystematicAnd process.Solution: the process is illustrated as followsWhen A and B are regarded as syst
44、ems, the process is an adiabatic reversible process. Assume the following (1) solidsThe volume of B does not vary with temperature; (2) for solid BThen,thusFor gas BPage2.26 known water (H2OlSaturation vapor pressure at 100 CAt this temperature and pressureEnthalpy of vaporization. Make the 1 kg wat
45、er vapor coagulate at 100 degrees C and 101.325 kPaForming liquid water. Set up ideal gas equation of state.Solution: the process is reversible phase change2.28 it is known that the melting point of ice under 100 kPa is 0 degrees C, at this point the specific enthalpy of heat of ice is hotJ - G-1.Av
46、erage waterheat capacity at constant pressure. 0.1 kg, 0 degrees C in 1 kg, 50 degrees C, in an adiabatic vesselThe ice after the final temperature system. The heat capacity of the vessel is not considered in the calculation.Solution: the rough estimate shows that the final temperature of T system should be higher than 0 DEG Ctherefore2.29 it is known that the melting point of ice under 100 kPa is 0 degrees C, at this poi