高教第五版物化答案013_khdaw

[物理化学(上册)完整习题答案解析]第五版高等教育出版社

第一章气体pVT 性质1-1物质的体膨胀系数V α与等温压缩系数T κ的定义如下：1 1TT p V p V V T V V ⎪⎪⎭⎫⎝⎛∂∂-=⎪⎭⎫ ⎝⎛∂∂=κα 试导出理想气体的V α、T κ与压力、温度的关系？解：对于理想气体，pV=nRT111 )/(11-=⋅=⋅=⎪⎭⎫⎝⎛∂∂=⎪⎭⎫ ⎝⎛∂∂=T TVV p nR V T p nRT V T V V p p V α 1211 )/(11-=⋅=⋅=⎪⎪⎭⎫ ⎝⎛∂∂-=⎪⎪⎭⎫ ⎝⎛∂∂-=p p V V pnRT V p p nRT V p V V T T T κ 1-2 气柜内有121.6kPa 、27℃的氯乙烯（C 2H 3Cl ）气体300m 3，若以每小时90kg 的流量输往使用车间，试问贮存的气体能用多少小时？解：设氯乙烯为理想气体，气柜内氯乙烯的物质的量为mol RT pV n 623.1461815.300314.8300106.1213=⨯⨯⨯== 每小时90kg 的流量折合p 摩尔数为13353.144145.621090109032-⋅=⨯=⨯=h mol M v Cl H C n/v=（14618.623÷1441.153）=10.144小时1-3 0℃、101.325kPa 的条件常称为气体的标准状况。

试求甲烷在标准状况下的密度。

解：33714.015.273314.81016101325444--⋅=⨯⨯⨯=⋅=⋅=m kg M RT p M V n CH CH CHρ 1-4 一抽成真空的球形容器，质量为25.0000g 。

充以4℃水之后，总质量为125.0000g 。

若改用充以25℃、13.33kPa 的某碳氢化合物气体，则总质量为25.0163g 。

试估算该气体的摩尔质量。

解：先求容器的容积33)(0000.10010000.100000.250000.1252cm cm V l O H ==-=ρn=m/M=pV/RTmol g pV RTm M ⋅=⨯-⨯⨯==-31.301013330)0000.250163.25(15.298314.841-5 两个体积均为V 的玻璃球泡之间用细管连接，泡内密封着标准状况条件下的空气。

物理化学第五版课后习题答案解析

第五章化学平衡5－1．在某恒定的温度和压力下，取n 0﹦1mol 的A (g )进行如下化学反应：A (g )B (g )若0B μ﹦0A μ，试证明，当反应进度﹦0.5mol 时，系统的吉布斯函数G 值为最小，这时A ，B 间达到化学平衡。

解：设反应进度为变量A (g )B (g )t ﹦0 n A , 0﹦n 0 0 0﹦0t ﹦t 平 n A n B﹦BBn ν n B ﹦B，n A ﹦n 0－n B ﹦n 0－B，n ﹦n A ＋n B ﹦n 0气体的组成为：y A ﹦A n n ﹦00B n n νξ-﹦01n ξ-，y B ﹦B nn﹦0n ξ各气体的分压为：p A ﹦py A ﹦0(1)p n ξ-，p B ﹦py B ﹦p n ξ各气体的化学势与的关系为：0000ln ln (1)A A AA p p RT RT p p n ξμμμ=+=+- 0000lnln B B B B p p RT RT p p n ξμμμ=+=+⋅ 由 G ＝n AA＋n BB＝(n A 0A μ＋n B 0B μ)＋00ln(1)A p n RT p n ξ-＋00ln B p n RT p n ξ⋅ ＝[n 0－A μ＋0B μ]＋n 00lnpRT p ＋00()ln(1)n RT n ξξ--＋0ln RT n ξξ 因为 0B μ﹦0A μ，则G ＝n 0(0A μ＋0lnpRT p)＋00()ln(1)n RT n ξξ--＋0ln RT n ξξ ,0()ln T p G RT n ξξξ∂=∂- 20,20()()T p n RT Gn ξξξ∂=-∂-＜0 令 ,()0T p Gξ∂=∂011n ξξξξ==-- ﹦0.5 此时系统的G 值最小。

5－2．已知四氧化二氮的分解反应 N 2O 4 (g ) 2 NO 2(g )在298.15 K 时，0r m G ∆＝4.75kJ ·mol －1。

物理化学答案_上册_高教_第五版[1]

A的为系统，求经可逆膨胀到解：过程图示如下
。始态温度时，系统的
，压力及过程的
。今以气体 B 。
将 A 和 B 共同看作系统，则该过程为绝热可逆过程。作以下假设（ 1）固体 B 的体积不随温度变化；（2）对固体 B ，则
从而
对于气体 B
2.26 已知水（H2O, l）在 100 °C 的饱和蒸气压尔蒸发焓结成液体水时的
（2）气体 B 得到的功（3）气体 A 的末态温度
。。。
（4）气体 A 从电热丝得到的热解：过程图示如下
由于加热缓慢，B 可看作经历了一个绝热可逆过程，因此
功用热力学第一定律求解
气体 A 的末态温度可用理想气体状态方程直接求解，
将 A 与 B 的看作整体，W = 0，因此
2.25 在带活塞的绝热容器中有 4.25 mol 的某固态物质 A 及 5 mol 某单原子理想气体 B，物质
因此，
1.9 如图所示，一带隔板的容器内，两侧分别有同温同压的氢气与氮气，二者均可视为理想气体。
（1）
保持容器内温度恒定时抽去隔板，且隔板本身的体积可忽略不计，试求两种气体混合后的压力。（2）隔板抽取前后，H2 及 N2 的摩尔体积是否相同？（3）隔板抽取后，混合气体中 H2 及 N2 的分压立之比以及它们的分体积各为若干？解：（1）等温混合后
要确定
，只需对第二步应用绝热状态方程
，
对
双
原
子
气
体
因此
由于理想气体的 U 和 H 只是温度的函数，
整个过程由于第二步为绝热，计算热是方便的。而第一步为恒温可逆
2.24 求证在理想气体 p-V 图上任一点处，绝热可逆线的斜率的绝对值大于恒温可逆线的绝对值。

《物理化学》第五版(天津大学物理化学教研室著)课后习题答案高等教育出版社

由于汽缸为绝热，因此
2.20 在一带活塞的绝热容器中有一固定的绝热隔板。隔板靠活塞一侧为 2 mol，0 C 的
单原子理想气体 A，压力与恒定的环境压力相等；隔板的另一侧为 6 mol，100 C 的双原子
理想气体 B，其体积恒定。今将绝热隔板的绝热层去掉使之变成导热板，求系统达平衡时的
T 及过程的
与温度的函数关系查本书附录，水
的比定压热容
。
解：300 kg 的水煤气中 CO(g)和 H2(g)的物质量分别为
300 kg 的水煤气由 1100 C 冷却到 100 C 所放热量
设生产热水的质量为 m，则
2.18 单原子理想气体 A 于双原子理想气体 B 的混合物共 5 mol，摩尔分数
，始态温
（1）
（2）
的；
（3）
的；
解：（1）C10H8 的分子量 M = 128.174，反应进程
。
（2）
。
（3） 2.34 应用附录中有关物资在 25 C 的标准摩尔生成焓的数据，计算下列反应在 25 C 时的及。
解：将气相看作理想气体，在 300 K 时空气的分压为
由于体积不变（忽略水的任何体积变化），373.15 K 时空气的分压为
由于容器中始终有水存在，在 373.15 K 时，水的饱和蒸气压为 101.325 kPa，系统中水蒸气的分压为 101.325 kPa，所以系统的总压
第二章热力学第一定律
解：该过程图示如下
设系统为理想气体混合物，则
1.17 一密闭刚性容器中充满了空气，并有少量的水。但容器于 300 K 条件下大平衡时，容器内压力为 101.325 kPa。若把该容器移至 373.15 K 的沸水中，试求容器中到达新的平衡时应有的压力。设容器中始终有水存在，且可忽略水的任何体积变化。300 K 时水的饱和蒸气压为 3.567 kPa。

物理化学部分答案_上册_高教_第五版

，始态膨胀到
平衡态。求末态温度
解：过程图示如下
分析：因为是绝热过程，过程热力学能的变化等于系统与环境间以功的形势所交换的能量。因此，
单原子分子
，双原子分子
由于对理想气体 U 和 H 均只是温度的函数，所以
2.19 在一带活塞的绝热容器中有一绝热隔板，隔板的两侧分别为 2 mol，0 °C 的单原子理想气体 A 及 5 mol，100 °C 的双原子理想气体 B，两气体的压力均为 100 kPa。活塞外的压力维持在 100 kPa 不变。今将容器内的隔板撤去，使两种气体混合达到平衡态。求末态的温度 T 及过程的。解：过程图示如下
解：过程图示如下
显然，在过程中 A 为恒压，而 B 为恒容，因此
同上题，先求功
同样，由于汽缸绝热，根据热力学第一定律
2.23 5 mol 双原子气体从始态 300 K，200 kPa，先恒温可逆膨胀到压力为 50 kPa，在绝热可逆压缩到末态压力 200 kPa。求末态温度 T 及整个过程的解：过程图示如下及。
，并将蒸汽看作理想气体，则过程的焓变为
（注：压力对凝聚相焓变的影响可忽略，而理想气体的焓变与压力无关）查表知
因此，
2.31 100 kPa 下，冰（ H2O, s ）的熔点为 0 °C 。在此条件下冰的摩尔融化
热
。已知在-10 °C ~ 0 °C 范围内过冷水（H2O, l）和冰的摩尔定压热容分别为。求在常压及-10 °C 下过冷水结冰的摩尔凝固焓。解：过程图示如下和
冰后，系统末态的温度。计算时不考虑容器的热容。解：经粗略估算可知，系统的末态温度 T 应该高于 0 °C, 因此

物理化学(第五版)》第十三章复习题答案

九、Freundlich 等温式
q kp
1/n
复习题3
• 3.用学到的关于界面现角的知识解释以下几种做法或现象的基体原理：①人工降雨； ②有机蒸馏中加沸石；③多孔固体吸附蒸气时的毛细凝聚； ④过饱和溶液，过饱和蒸气，过冷液体等过饱和现象； ⑤重量分析中的“陈化” 过程；⑥喷洒农药时，为何常常在农药中加入少量表面活性剂。
复习题14
14.接触角的定义是什么？它的大小受哪些因素影响？如何用接触角的大小来判断液对固体的润湿情况？
• 接触角为在气液固三相交接处，气-液界面与固-液界面之间的夹角，其大小由三种界面张力的相对大小决定； • 接触角为零，液体在固体表面铺展， • 小于 90°液体能润湿固体， • 大于 90°，不能润湿。
小结——第十三章
总结
基本公式
一、比表面能二、表面热力学
G A T , P
def
dG SdT V dp dA
( G ) dA T ,p
S A TA T ,p ,p
( G ) SdT dA p
复习题8
• b 管垂直高度为 h，
• c 管调整表面曲率半径但不溢出，
• d 管不溢出，
• e 管高度为到扩大部分为止，但上去后不会下降仍
然为 h
• f 管将下降为凸液面，
• g 管为 1/n倍h。
复习题9
9.把大小不等的液滴（或萘粒）密封在一玻璃罩内，隔相当长时间后，估计会出现什么现象？ • 小液滴消失，大液滴更大。弯曲液面所产生的附加压力将使液体在以小液滴形式分散存在时比大量聚集存在时具有更大的饱和蒸气压。
的附加压力大，化学势高。

[物理化学[上册]完整习题的答案解析]第五版高等教育出版社

第一章气体pVT 性质1-1物质的体膨胀系数V α与等温压缩系数T κ的定义如下：1 1TT p V p V V T V V ⎪⎪⎭⎫⎝⎛∂∂-=⎪⎭⎫ ⎝⎛∂∂=κα 试导出理想气体的V α、T κ与压力、温度的关系？解：对于理想气体，pV=nRT111 )/(11-=⋅=⋅=⎪⎭⎫⎝⎛∂∂=⎪⎭⎫ ⎝⎛∂∂=T TVV p nR V T p nRT V T V V p p V α 1211 )/(11-=⋅=⋅=⎪⎪⎭⎫ ⎝⎛∂∂-=⎪⎪⎭⎫ ⎝⎛∂∂-=p p V V pnRT V p p nRT V p V V T T T κ 1-2 气柜内有121.6kPa 、27℃的氯乙烯（C 2H 3Cl ）气体300m 3，若以每小时90kg 的流量输往使用车间，试问贮存的气体能用多少小时？解：设氯乙烯为理想气体，气柜内氯乙烯的物质的量为mol RT pV n 623.1461815.300314.8300106.1213=⨯⨯⨯== 每小时90kg 的流量折合p 摩尔数为13353.144145.621090109032-⋅=⨯=⨯=h mol M v Cl H C n/v=（14618.623÷1441.153）=10.144小时1-3 0℃、101.325kPa 的条件常称为气体的标准状况。

试求甲烷在标准状况下的密度。

解：33714.015.273314.81016101325444--⋅=⨯⨯⨯=⋅=⋅=m kg M RT p M V n CH CH CHρ 1-4 一抽成真空的球形容器，质量为25.0000g 。

充以4℃水之后，总质量为125.0000g 。

若改用充以25℃、13.33kPa 的某碳氢化合物气体，则总质量为25.0163g 。

试估算该气体的摩尔质量。

解：先求容器的容积33)(0000.10010000.100000.250000.1252cm cm V l O H ==-=ρn=m/M=pV/RTmol g pV RTm M ⋅=⨯-⨯⨯==-31.301013330)0000.250163.25(15.298314.841-5 两个体积均为V 的玻璃球泡之间用细管连接，泡内密封着标准状况条件下的空气。

物理化学第五版课后习题答案

第十章界面现象10－1 请回答下列问题：(1) 常见的亚稳定状态有哪些？为什么产生亚稳态？如何防止亚稳态的产生？(2) 在一个封闭的钟罩内，有大小不等的两个球形液滴，问长时间放置后，会出现什么现象？(3) 下雨时，液滴落在水面上形成一个大气泡，试说明气泡的形状和理由？ (4) 物理吸附与化学吸附最本质的区别是什么？(5) 在一定温度、压力下，为什么物理吸附都是放热过程？答： (1) 常见的亚稳态有：过饱和蒸汽、过热液体、过冷液体、过饱和溶液。

产生这些状态的原因就是新相难以生成，要想防止这些亚稳状态的产生，只需向体系中预先加入新相的种子。

(2) 一断时间后，大液滴会越来越大，小液滴会越来越小，最终大液滴将小液滴“吃掉”，根据开尔文公式，对于半径大于零的小液滴而言，半径愈小，相对应的饱和蒸汽压愈大，反之亦然，所以当大液滴蒸发达到饱和时，小液滴仍未达到饱和，继续蒸发，所以液滴会愈来愈小，而蒸汽会在大液滴上凝结，最终出现“大的愈大，小的愈小”的情况。

(3) 气泡为半球形，因为雨滴在降落的过程中，可以看作是恒温恒压过程，为了达到稳定状态而存在，小气泡就会使表面吉布斯函数处于最低，而此时只有通过减小表面积达到，球形的表面积最小，所以最终呈现为球形。

(4) 最本质区别是分子之间的作用力不同。

物理吸附是固体表面分子与气体分子间的作用力为范德华力，而化学吸附是固体表面分子与气体分子的作用力为化学键。

(5) 由于物理吸附过程是自发进行的，所以ΔG ＜0，而ΔS ＜0，由ΔG =ΔH －T ΔS ，得 ΔH ＜0，即反应为放热反应。

10－2 在293.15K 及101.325kPa 下，把半径为1×10－3m 的汞滴分散成半径为1×10－9m 的汞滴，试求此过程系统表面吉布斯函数变(ΔG )为多少？已知293.15K 时汞的表面张力为0.4865 N ·m －1。

解： 3143r π＝N ×3243r π N ＝3132r rΔG ＝21A A dA γ⎰＝γ(A 2－A 1)＝4πγ·( N 22r －21r )＝4πγ·(312r r －21r )＝4π×0.47×(339(110)110--⨯⨯－10－6)＝5.9062 J10－3 计算时373.15K 时，下列情况下弯曲液面承受的附加压力。

第五版物理化学课后习题答案

物化第二章热力学第一定律2－1． 1mol 理想气体在恒定压力下温度升高1℃，求过程中系统与环境交换的功。

解： n ＝ 1molp 1, V 1, T 1−−−→−恒压升温p 2, V 2, T 2 W ＝－p a m b ΔV ＝－p (V 2－V 1) ＝－nR (T 2－T 1) ＝－8.314J2－2． 1mol 水蒸气(H 2O ,g )在100℃，101.325kPa 下全部凝结成液态水。

求过程的功。

假设：相对于水蒸气的体积，液态水的体积可以忽略不计。

解： n ＝ 1molH 2O (g )−−−−−→−kPa101.325100℃，H 2O (l ) W ＝－p a m b ΔV ＝－p (V l －V g ) ≈ pVg ＝ nRT ＝ 3.102kJ2－3．在25℃及恒定压力下，电解1mol 水(H 2O ,l )，求过程的体积功。

H 2O (l ) H 2(g ) + 12O 2(g )解： n ＝ 1molH 2O (l )−−−−−→−kPa 101.325100℃，H 2(g ) + O 2(g )n 1＝1mol 1mol + 0.5mol ＝ n 2 V 1 ＝ V l V (H 2) + V (O 2) ＝ V 2W ＝－p amb ΔV ＝－(p 2V 2－p 1V 1)≈－p 2V 2 ＝－n 2RT ＝－1.5×R ×298.15＝－3.718kJ2－4．系统由相同的始态经过不同的途径达到相同的末态。

若途径a 的Q a ＝2.078 kJ ，W a ＝－4.157 kJ ，而途径b 的Q b ＝－0.692kJ 。

求W b 。

解：Q a ＋W a ＝Q b ＋W bW b ＝Q a ＋W a －Q b ＝2.078－4.157＋0.692＝－2.079＋0.692＝－1.387kJ2－5．始态为25℃，200 kPa 的5 mol 某理想气体，经途径a ，b 两不同途径到达相同的末态。

物理化学答案_上册_高教_第五版(Answer _ _ on physical chemistry of higher education _ Fifth Edition)

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 gas 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 pressure.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 regarded 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 separation 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 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 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 time 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 above, 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 usedPageUsing 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., 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 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 air and witha 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, 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 air 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 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 process. 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,.The initial state is 100 kPa, 50 DMThreeThe constant volume heating increases the volume of pressure As 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 gasandIt's just a function 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 rateDensityMolar 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 arranged 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 the 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 volume 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 doneThat's exactly what it's 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, temperature 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 withthe copper block, and the volume of copper 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 (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 relations, 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 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 adiabatic process, the change ofthermodynamic 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 insulating 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 removed 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 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 mol 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 the 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 diatomic 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 sureThe adiabatic equation of state can be applied to second steps onlyPairs of diatomic gasesthereforeSince the U and H of ideal gases are only functions oftemperature,PageThe whole process is convenient because the second step is adiabatic. The first step is constant temperature reversible2.24 it is proved 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 wireThe 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 piston 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, after 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.Solution: 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 insulated 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 systems, 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 water 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.Average 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 point the specific enthalpy of heat of ice is hotJ - G-1.Water and iceAverage heat capacity at constant pressureRespectively asand. Present 1 kg 50 in an adiabatic vesselWater C degrees C degrees kg degrees -20 ice 0.8 degrees. O:(1) the final temperature.(2) the quality of the final water and ice.Solution: 1 kg, 50 degrees C water cooling, 0 degrees C exothermic0.8 kg -20 degrees C ice warming caused by 0 degrees C endothermicHeat is needed when melted completelyPageTherefore, only part of the ice melted. So the final temperature is 0 ~ C. AIce melting of g,Is thereThe quality of the system ice and water, respectively2.30 steam boilers are continuously injected with water of 20 degrees C, heated and evaporated to 180 degrees C, saturated steam pressure as1.003 MPa of water vapor. The amount of heat needed to produce 1 kg of water vapor.Known: waterMolar enthalpy of vaporization at 100 CWater levelMean molar heat capacity at constant pressureSteamMolar constant heat capacity and temperatureSee Appendix for function.Solution: the process is considered as a constant pressure processThe initial and final system), respectively.and. Insertion of equilibrium phase change pointWhen the steam is regarded as an ideal gas, the enthalpy of the process becomes10 DEG C by the following three differentThe process heats up to 100 degrees C of water, asking for a process.(1) the system contacts with the heat source of 100 C.(2) the system contacts the heat source of the 55 degree C to the heat balance, and then contacts with the heat source of 100 C.(3) the system contacts the heat source of 40 degrees C and 70 degrees C to the heat balance, and then contacts with the heat source of 100 C.Solution: the entropy is the state function, and the entropy of the system changes the same in three casesIn the process, the heat obtained by the system is heat released by a heat source, so that3.8 known nitrogen (N2gThe molar heat at constant pressure is a function of temperatureThe initial state is 300 K, 100 kPa, the 1 mol N2 (g) is placed in the heat source of 1000 K, and the following is requiredPageProcess (1) by constant pressure process; (2) when the constant volume process reaches equilibrium state.Solution: in the case of constant pressureIn constant volume, nitrogen (N2gAs ideal gasbodytakeInstead of all of the aboveYou can obtain the required quantitiesThe 3.9 initial state is,A diatomic ideal gas, 1 mol, changed in the following waysTo,The final. Seek steps and ways.(1) reversible expansion at constant temperature;(2) cool down the constant volume until the pressure drops to 100 kPa, and then heat up to the constant pressure;(3) expand first by adiabatic expansion until the pressure drops to 100 kPa, and then heat up to constant pressure.Solution: (1) reversible expansion of an ideal gas at constant temperature, DU = 0, thereforePage(2) calculate constant volume cooling until the pressure drops to 100 kPa, the temperature of the system is T:(3) similarly, the heat expands reversibly to a temperature of T: when the pressure drops to 100 kPaAccording to the equation of state of ideal gas adiabatic process,The thermodynamic quantities are calculated as follows2.122 mol diatomic ideal gas from the initial state 300 K, 50 DMThreeHeat the constant volume to 400 K, then heat it up to a constant pressureVolume increased to 100 DMThreeAsking for the whole process.Solution: the process is illustrated as followsPageFirst calculate the final temperatureTherefore,Two important formulasIdeal gas3.17 consists ofThe monatomic gas A is a mixture of 10 mol with ideal gas mixtures of diatomic gas B,Beginning stateAdiabatic reversible compression toEquilibrium state. Process seeking.Solution: the process is illustrated as followsPageThe adiabatic reversible equation of state for a mixed ideal gas is derived as followsEasy to getPage3.18 ideal mixture of monatomic gas A and diatomic gas B, 8 mol, consisting ofBeginning.state. Irreversible expansion to the final volume of this insulation against the constant pressureTheEquilibrium state. Process seeking.Solution: the process is illustrated as followsTo determine the final temperature, adiabatic processTherefore3.19 atmospheric pressure, 100 g, 27 C of water and 200 g, 72 degrees C of water in an insulated container mixed, the final water temperature TEntropy change of process. Specific heat at constant pressure of known water.Solution: the procedure is illustrated as followsPageIn the 321 adiabatic constant volume vessel, there is an insulating pressure bulkhead, with 200 K of 2 mol and 50 DM on the side of the partition boardThreeMonatomicThink gas A, and the other side is 3 mol, 400 K, 100 DMThreeDiatomic ideal gas B. Insulating partitions in a container todayThe way, A and B mixed gas gas balance. Process seeking.Solution: the process is illustrated as followsThe final temperature of the T system can be solved as followsEntropy change of the systemPageNote: for ideal gases, the presence of one component does not affect the other components. The final A and BVolume is the volume of the container.In the 322 adiabatic constant volume vessel, there is an insulating pressure bulkhead, with N2 (g) on both sides of the bulkhead. One side volume was 50 DMThreeThere is200 K N2 (g) 2 mol; the other side volume was 75 DMThreeWithin 500 K of N2 (g) 4 mol; N2 (g) can be considered as idealgas.This will be removed from the adiabatic partition in the container, allowing the system to achieve the equilibrium state. Process seeking.Solution: the process is illustrated as followsThe same problem, the final temperature of T is determined as followsAfter the first step, the volume of the two part isThat is, in addition to the partition, the 2 and final state of the same, soPagePay attention to the comparison between the 21 and the 22 questions.3.23 at atmospheric pressure, the melting point of ice is 0 degrees C, than the enthalpy of fusionConstant pressure hot melt of water. In an insulated vessel, there is 1 kg, 25 degrees C water, and 0.5 kg, 0 DEG C are added to the vesselThis is the beginning of the system. Seeking the equilibrium of a system; a process.Solution: the process is illustrated as followsThe process is considered as a constant pressure adiabatic process. As a result of 1 kg, water of 25 degrees C is cooled to 0 degrees CCan only lead toThe ice melts, therefore3.27 it is known that the melting point of ice at normal pressure is 0 degrees C, and the molar enthalpy of meltingBenzeneMelting point is 5.51 degrees C, molar melting enthalpy. Molar determination of liquid water and solid benzeneThe heat capacities areand. Today there are two containers surrounded by insulation, 0 C in a containerPageThe 8 mol H2O (s) is balanced with 2 mol H2O (L) and another container is 5.51 mol C6H6 (L) and 5 mol C6H6 (s) into 5 CBalance. The two containers are contacted to remove an insulating layer between the two containers so that the two container reaches a new equilibrium state. Process seeking.Solution: a rough estimate shows that 5 mol C6H6 (L) is completely solidified and will cause the 8 mol H2O (s) to melt completelyThe process is illustrated as followsThe total process is constant pressure adiabatic process,Therefore3.28 put a small glass bottle containing 0.1 mol ether (C2H5) 2O (L) into the volume of 10 DMThreeVacuum container with constant volume and airtightPageConstant temperature in the constant temperature trough at35.51 C. 35.51 C is the boiling point of ether at 101.325 kPa. Given this conditionMolar enthalpy of vaporization of diethyl ether. Now break the small glass bottle and the ether evaporates to equilibrium.seek(1) the pressure of ether vapor;(2) process.Solution: ether vapor is regarded as ideal gas due to constant temperatureThe changes in the state functions are calculated as followsIgnore the volume of the liquid etherThe 3.30. volume was 20 DMThreeThe closed vessel has 2 mol H2O gas liquid equilibrium. 80 degrees C, 100 degrees C saturation of the waterAnd vapor pressure, respectivelyandMolar enthalpy of vaporization of water at 25 CThe average pressure and molar heat capacities of water and water vapor at 25 ~ 100 degrees C are respectivelyand. Modern systemThe constant volume heating from 80 C to 100 C. Process seeking.Solution: first estimate 100 degrees C, whether there is liquid water in the system. There is only water vapor in the final stateThe quality is nbeObviously, only a portion of the water evaporation, the final is still the gas-liquid equilibrium. Therefore, the following process:Set the following channelsPageThe first and fourth steps are reversible phase change, the second step is constant temperature change of liquid water, and the third step is constant of liquid waterPressure changing temperature. The molar evaporation heat of water at 80 degrees C and 100 degrees C is first obtained3.31. O2 (g) is a function of the molar heat at constant pressure with temperaturePageThe standard molar entropy of O2 (g) under 25 degrees C is known. Seek O2 (g)Moore's entropy at 100 C and 50 kPa.Solution: by formulaknow3.32. if the molar heat capacity of each substance in the chemical reaction is expressed asTry to deduce the chemical reactionStandard molar reaction entropyAnd temperature T。

物理化学第五版第三章答案

第三章热力学第二定律3.1 卡诺热机在的高温热源和的低温热源间工作。

求（1）热机效率；（2）当向环境作功时，系统从高温热源吸收的热及向低温热源放出的热。

解：卡诺热机的效率为根据定义3．2 卡诺热机在的高温热源和的低温热源间工作，求：（1）热机效率；（2）当从高温热源吸热时，系统对环境作的功及向低温热源放出的热解：(1) 由卡诺循环的热机效率得出（2）3．3 卡诺热机在的高温热源和的低温热源间工作，求（1）热机效率；（2）当向低温热源放热时，系统从高温热源吸热及对环境所作的功。

解：（1）(2)3．4 试说明：在高温热源和低温热源间工作的不可逆热机与卡诺机联合操作时，若令卡诺热机得到的功r W 等于不可逆热机作出的功－W 。

假设不可逆热机的热机效率大于卡诺热机效率，其结果必然是有热量从低温热源流向高温热源，而违反势热力学第二定律的克劳修斯说法。

证：（反证法）设 r ir ηη>不可逆热机从高温热源吸热，向低温热源放热，对环境作功则逆向卡诺热机从环境得功从低温热源吸热向高温热源放热则若使逆向卡诺热机向高温热源放出的热不可逆热机从高温热源吸收的热相等，即总的结果是：得自单一低温热源的热，变成了环境作功，违背了热力学第二定律的开尔文说法，同样也就违背了克劳修斯说法。

3．5 高温热源温度，低温热源温度，今有120KJ的热直接从高温热源传给低温热源，求此过程。

解：将热源看作无限大，因此，传热过程对热源来说是可逆过程3.6 不同的热机中作于的高温热源及的低温热源之间。

求下列三种情况下，当热机从高温热源吸热时，两热源的总熵变。

（1）可逆热机效率。

（2）不可逆热机效率。

（3）不可逆热机效率。

解：设热机向低温热源放热，根据热机效率的定义因此，上面三种过程的总熵变分别为。

3.7 已知水的比定压热容。

今有1 kg，10℃的水经下列三种不同过程加热成100 ℃的水，求过程的。

（1）系统与100℃的热源接触。

物理化学第五版课后习题答案

第十章界面现象10－1 请回答下列问题：(1) 常见的亚稳定状态有哪些？为什么产生亚稳态？如何防止亚稳态的产生？(2) 在一个封闭的钟罩内，有大小不等的两个球形液滴，问长时间放置后，会出现什么现象？(3) 下雨时，液滴落在水面上形成一个大气泡，试说明气泡的形状和理由？ (4) 物理吸附与化学吸附最本质的区别是什么？(5) 在一定温度、压力下，为什么物理吸附都是放热过程？答： (1) 常见的亚稳态有：过饱和蒸汽、过热液体、过冷液体、过饱和溶液。

产生这些状态的原因就是新相难以生成，要想防止这些亚稳状态的产生，只需向体系中预先加入新相的种子。

(2) 一断时间后，大液滴会越来越大，小液滴会越来越小，最终大液滴将小液滴“吃掉”，根据开尔文公式，对于半径大于零的小液滴而言，半径愈小，相对应的饱和蒸汽压愈大，反之亦然，所以当大液滴蒸发达到饱和时，小液滴仍未达到饱和，继续蒸发，所以液滴会愈来愈小，而蒸汽会在大液滴上凝结，最终出现“大的愈大，小的愈小”的情况。

(3) 气泡为半球形，因为雨滴在降落的过程中，可以看作是恒温恒压过程，为了达到稳定状态而存在，小气泡就会使表面吉布斯函数处于最低，而此时只有通过减小表面积达到，球形的表面积最小，所以最终呈现为球形。

(4) 最本质区别是分子之间的作用力不同。

物理吸附是固体表面分子与气体分子间的作用力为范德华力，而化学吸附是固体表面分子与气体分子的作用力为化学键。

(5) 由于物理吸附过程是自发进行的，所以ΔG ＜0，而ΔS ＜0，由ΔG =ΔH －T ΔS ，得 ΔH ＜0，即反应为放热反应。

10－2 在及下，把半径为1×10－3m 的汞滴分散成半径为1×10－9m 的汞滴，试求此过程系统表面吉布斯函数变(ΔG )为多少？已知时汞的表面张力为 N ·m －1。

解： 3143r π＝N×3243r π N ＝3132r rΔG ＝21A A dA γ⎰＝?(A 2－A 1)＝4??·( N 22r －21r )＝4??·(312r r －21r )＝4?×????×(339(110)110--⨯⨯－10－6) ＝ J10－3 计算时时，下列情况下弯曲液面承受的附加压力。

第十三章物理化学(第五版、胡英)课后答案

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课后答案网
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第 13 章相倚子系统的统计热力学
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【优质】物理化学答案-word范文模板 (9页)

本文部分内容来自网络整理，本司不为其真实性负责，如有异议或侵权请及时联系，本司将立即删除！== 本文为word格式，下载后可方便编辑和修改！ ==物理化学答案篇一：物理化学第五版课后习题答案大学物理化学课后答案详解第一章气体的pVT性质1.1 物质的体膨胀系数与等温压缩率的定义如下试推出理想气体的，与压力、温度的关系。

解：根据理想气体方程1.5两个容积均为V的玻璃球泡之间用细管连结，泡内密封着标准状态下的空气。

若将其中的一个球加热到 100 ?C，另一个球则维持 0 ?C，忽略连接细管中气体体积，试求该容器内空气的压力。

解：由题给条件知，（1）系统物质总量恒定；（2）两球中压力维持相同。

标准状态：因此，1.9 如图所示，一带隔板的容器内，两侧分别有同温同压的氢气与氮气，二者均可视为理想气体。

（1）保持容器内温度恒定时抽去隔板，且隔板本身的体积可忽略不计，试求两种气体混合后的压力。

（2）隔板抽取前后，H2及N2的摩尔体积是否相同？（3）隔板抽取后，混合气体中H2及N2的分压立之比以及它们的分体积各为若干？解：（1）等温混合后即在上述条件下混合，系统的压力认为。

（2）混合气体中某组分的摩尔体积怎样定义？（3）根据分体积的定义对于分压1.11 室温下一高压釜内有常压的空气，为进行实验时确保安全，采用同样温度的纯氮进行臵换，步骤如下：向釜内通氮气直到4倍于空气的压力，尔后将釜内混合气体排出直至恢复常压。

重复三次。

求釜内最后排气至恢复常压时其中气体含氧的摩尔分数。

解：分析：每次通氮气后至排气恢复至常压p，混合气体的摩尔分数不变。

设第一次充氮气前，系统中氧的摩尔分数为，则，统的摩尔分数为，充氮气后，系统中氧的摩尔分数为。

重复上面的过程，第n次充氮气后，系因此，。

1.13 今有0 ?C，40.530 kPa的N2气体，分别用理想气体状态方程及van der Waals方程计算其摩尔体积。

实验值为解：用理想气体状态方程计算。

【物理化学上册完整习题答案】第五版-高等教育出版社

第一章气体pVT 性质1-1物质的体膨胀系数V α与等温压缩系数T κ的定义如下：1 1T T pV p V V T V V⎪⎪⎭⎫ ⎝⎛∂∂-=⎪⎭⎫⎝⎛∂∂=κα 试导出理想气体的V α、T κ与压力、温度的关系？解：对于理想气体，pV=nRT111 )/(11-=⋅=⋅=⎪⎭⎫⎝⎛∂∂=⎪⎭⎫ ⎝⎛∂∂=T TVV p nR V T p nRT V T V V p p V α 1211 )/(11-=⋅=⋅=⎪⎪⎭⎫ ⎝⎛∂∂-=⎪⎪⎭⎫ ⎝⎛∂∂-=p p V V pnRT V p p nRT V p V V T T T κ 1-2 气柜内有121.6kPa 、27℃的氯乙烯（C 2H 3Cl ）气体300m 3，若以每小时90kg 的流量输往使用车间，试问贮存的气体能用多少小时？解：设氯乙烯为理想气体，气柜内氯乙烯的物质的量为mol RT pV n 623.1461815.300314.8300106.1213=⨯⨯⨯== 每小时90kg 的流量折合p摩尔数为 133153.144145.621090109032-⋅=⨯=⨯=h mol M v Cl H Cn/v=（14618.623÷1441.153）=10.144小时1-3 0℃、101.325kPa 的条件常称为气体的标准状况。

试求甲烷在标准状况下的密度。

解：33714.015.273314.81016101325444--⋅=⨯⨯⨯=⋅=⋅=m kg M RT p M V n CH CH CHρ 1-4 一抽成真空的球形容器，质量为25.0000g 。

充以4℃水之后，总质量为125.0000g 。

若改用充以25℃、13.33kPa 的某碳氢化合物气体，则总质量为25.0163g 。

试估算该气体的摩尔质量。

解：先求容器的容积33)(0000.10010000.100000.250000.1252cm cm V l O H ==-=ρn=m/M=pV/RTmol g pV RTm M ⋅=⨯-⨯⨯==-31.301013330)0000.250163.25(15.298314.841-5 两个体积均为V 的玻璃球泡之间用细管连接，泡内密封着标准状况条件下的空气。

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