物理化学上册的答案第五版上册.doc
天津大学_第五版_物理化学上习题答案
第一章 气体的pVT 关系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 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 CH ρ 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(天大第五版全册)课后习题答案
资料范本本资料为word版本，可以直接编辑和打印，感谢您的下载物理化学(天大第五版全册)课后习题答案地点：__________________时间：__________________说明：本资料适用于约定双方经过谈判，协商而共同承认，共同遵守的责任与义务，仅供参考，文档可直接下载或修改，不需要的部分可直接删除，使用时请详细阅读内容气体pVT性质1-1物质的体膨胀系数与等温压缩系数的定义如下：试导出理想气体的、与压力、温度的关系？解：对于理想气体，pV=nRT1-5 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。
若将其中一个球加热到100℃，另一个球则维持0℃，忽略连接管中气体体积，试求该容器内空气的压力。
解：方法一：在题目所给出的条件下，气体的量不变。
并且设玻璃泡的体积不随温度而变化，则始态为终态（f）时1-8 如图所示一带隔板的容器中，两侧分别有同温同压的氢气与氮气，二者均克视为理想气体。
（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）隔板抽去前后，H2及N2的摩尔体积是否相同？（3）隔板抽去后，混合气体中H2及N2的分压力之比以及它们的分体积各为若干？解：（1）抽隔板前两侧压力均为p，温度均为T。
（1）得：而抽去隔板后，体积为4dm3，温度为，所以压力为（2）比较式（1）、（2），可见抽去隔板后两种气体混合后的压力仍为p。
（2）抽隔板前，H2的摩尔体积为，N2的摩尔体积抽去隔板后所以有，可见，隔板抽去前后，H2及N2的摩尔体积相同。
（3）所以有*1-17 试由波义尔温度TB的定义式，试证范德华气体的TB可表示为TB=a/（bR）式中a、b为范德华常数。
解：先将范德华方程整理成将上式两边同乘以V得求导数当p→0时，于是有当p→0时V→∞，（V-nb）2≈V2，所以有 TB= a/（bR）第二章热力学第一定律2-1 1mol理想气体于恒定压力下升温1℃，试求过程中气体与环境交换的功W。
天津大学 第五版 物理化学上册习题答案
第一章 气体的pVT 关系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 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 CH ρ 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。
天津大学_第五版_物理化学上册习题答案
天津大学_第五版_物理化学上册习题答案(总94页)-本页仅作为预览文档封面，使用时请删除本页-目录第一章 气体的pVT 关系 .................................... 错误!未定义书签。
第二章 热力学第一定律 ..................................... 错误!未定义书签。
第三章 热力学第二定律 ..................................... 错误!未定义书签。
第四章 多组分系统热力学 ................................. 错误!未定义书签。
第五章 化学平衡 ................................................. 错误!未定义书签。
第六章 相平衡 .................................................... 错误!未定义书签。
第一章 气体的pVT 关系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 T VV p nR V T p nRT V T V V p p V α 1211 )/(11-=⋅=⋅=⎪⎪⎭⎫ ⎝⎛∂∂-=⎪⎪⎭⎫ ⎝⎛∂∂-=p p VV pnRT V p p nRT V pV V T T T κ 1-2 气柜内有、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 C n/v=（÷）=小时1-3 0℃、的条件常称为气体的标准状况。
(完整版)天津大学第五版物理化学上册习题答案
第一章 气体的pVT 关系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 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 CH ρ1-4 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。
天津大学-第五版-物理化学上习题答案
第一章 气体的pVT 关系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、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 C n/v=（÷）=小时1-3 0℃、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。
充以4℃水之后，总质量为。
若改用充以25℃、的某碳氢化合物气体，则总质量为。
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。
物理化学上册第五版天津大学出版社第二章 热力学第一定律习题答案
物理化学上册第五版天津大学出版社第二章 热力学第一定律习题答案2-1 1mol 理想气体于恒定压力下升温1℃，试求过程中气体与环境交换的功W 。
解：J T nR nRT nRT pV pV V V p W am b 314.8)(121212-=∆-=+-=+-=--=2-2 1mol 水蒸气（H 2O ，g ）在100℃，101.325 kPa 下全部凝结成液态水。
求过程的功。
解： )(g l am b V V p W --=≈kJ RT p nRT p V p gam b 102.315.3733145.8)/(=⨯===2-3 在25℃及恒定压力下，电解1mol 水（H 2O ，l ），求过程的体积功。
)(21)()(222g O g H l O H +=解：1mol 水（H 2O ，l ）完全电解为1mol H 2（g ）和0.50 mol O 2（g ），即气体混合物的总的物质的量为1.50 mol ，则有)()(2l O H g am b V V p W --=≈)/(p nRT p V p g am b -=-kJ nRT 718.315.2983145.850.1-=⨯⨯-=-= 2-4 系统由相同的始态经过不同途径达到相同的末态。
若途径a 的Q a =2.078kJ ，W a = -4.157kJ ；而途径b 的Q b = -0.692kJ 。
求W b 。
解：因两条途径的始末态相同，故有△U a =△U b ，则 b b a a W Q W Q +=+所以有，kJ Q W Q W b a a b 387.1692.0157.4078.2-=+-=-+=2-5 始态为25℃，200kPa 的5 mol 某理想气体，经a ，b 两不同途径到达相同的末态。
途径a 先经绝热膨胀到 – 28.57℃，100kPa ，步骤的功W a = - 5.57kJ ；在恒容加热到压力200 kPa 的末态，步骤的热Q a = 25.42kJ 。
物理化学上册的答案,第五版上册(完整资料).doc
此文档下载后即可编辑第一章 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 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。
物理化学答案_上册_高教_第五版[1]
![物理化学答案_上册_高教_第五版[1]](https://img.taocdn.com/s3/m/756e882a453610661ed9f454.png)
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[上册]完整习题的答案解析]第五版高等教育出版社
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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。
第五版物理化学第三章习题答案完整版.doc
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。
（2）系统先与55℃的热源接触至热平衡，再与100℃的热源接触。
物理化学第五版上册习题答案
物理化学第五版上册习题答案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。
43331432r133r2r13a1222r2－r12)(1?10)1?10?9?33－10－6)2?r2?r4?r＝2?58.91?100.1?10 ?6?32?58.91?100.1?10 ?6?34?58.91?100.1?10 ?6?3－－－－2?r＝2?22.3?105?10 ?5?3＝892 pa?3h＝2?cos?r?g＝2?22.3?105?10 ?5?1?789.4?9.8＝0.1153 m－时，计算(1) 开始形成水滴的半径；(2) 每个水滴中所含水分子的个数。
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。
【物理化学上册完整习题答案】第五版
物理化学上册完整习题答案第五版第一章热力学基础习题一(a)热容量定义为物体单位质量或单位物质量的温度升高所需的热量。
它的单位是J/(g·°C)。
(b)热容量可以用以下公式来计算：Q = mcΔT，其中Q表示热量，m表示物体的质量，c表示物质的比热容，ΔT表示温度变化。
(c)水的比热容是4.18 J/(g·°C)。
(a)焓的定义是在恒定压力下，物体单位质量的温度升高所需的热量。
焓可以用以下公式来计算：H =Q/m，其中H表示焓，Q表示热量，m表示物体的质量。
(b)焓变是指在化学反应或物理过程中，物体单位质量的焓的变化。
1.辅助函数的作用是简化热力学计算过程。
它可以帮助我们计算焓、熵和自由能等。
习题二1.当物体从低温区域向高温区域传递热量时，热量会按照热流从高温区域传递到低温区域。
这个现象符合热力学第一定律，即能量守恒定律。
2.传送速率可以通过热传导定律来计算。
热传导定律可以用以下公式表示：Q = kA(ΔT/Δx)，其中Q是传输的热量，k是热传导系数，A是传输的面积，ΔT是温度差，Δx是传输的距离。
3.传热的三种途径是热传导、热对流和热辐射。
习题三1.升华是指固体直接从固体相转变为气体相的过程，而不经过液体相。
凝华是指气体直接从气体相转变为固体相的过程。
2.溶解度是指溶质在溶剂中溶解所达到的最大浓度。
溶解度可以受到温度、压力和溶剂性质等因素的影响。
3.相变是指物质在一定条件下从一个相态转变为另一个相态的过程。
常见的相变包括融化、凝固、汽化、液化、升华和凝华。
第二章热力学函数习题一(a)熵是描述系统无序程度的物理量。
它的单位是J/(mol·K)。
(b)熵的变化可以通过以下公式计算：ΔS = Q/T，其中ΔS表示熵的变化，Q表示热量，T表示温度。
(a)熵增定律是指在孤立系统中，系统熵总是增加或至少保持不变。
(b)熵增定律可以用以下公式表示：ΔStotal =ΔSsystem + ΔSsurroundings ≥ 0，其中ΔStotal表示系统和周围环境的总熵变化，ΔSsystem表示系统的熵变化，ΔSsurroundings表示周围环境的熵变化。
物理化学第五版全册课后习题答案
V
1 V 1 (nRT / p) 1 nR 1 V T 1 V T p V T p V p V T
T
1 V 1 (nRT / p) 1 nRT 1 V 2 p 1 V p T V p V p T V p
第七章 电化学
余训爽
1-8 如图所示一带隔板的容器中， 两侧分别有同温同压的氢气与氮气， 二者均克视为理 想气体。 H2 p 3dm T
3
N2 p
n n1,i n2,i 2 piV /( RTi )
第 1 页 共 128 页创建时间：
第七章 电化学
余训爽
终态（f）时
n n1, f n2, f
pf V V R T1, f T2, f
pfV R
T2, f T1, f T T 1, f 2, f
1-6 0℃时氯甲烷（CH3Cl）气体的密度ρ随压力的变化如下。试作ρ/p—p 图，用外推 法求氯甲烷的相对分子质量。 P/kPa ρ/（g·dm ）
-3
101.325 2.3074
67.550 1.5263 67.550 0.02260
50.663 1.1401 50.663 0.02250
33.775 0.75713 33.775 0.02242
25.331 0.56660 25.331 0.02237
解：将数据处理如下： P/kPa 101.325 -3 (ρ/p)/ （g· dm · kPa） 0.02277 作(ρ/p)对 p 图
0.0229 0.0228 0.0227 0.0226 0.0225 0.0224 0.0223 0.0222 0 20 40
物理化学上册第五版的答案(东南大学用)上册
第一章 气体pVT 性质希望同学们能够正确地运用答案，一定要自己先尝试着做一做再来参考答案。
声明：此答案如有错误，本人不承担任何责任。
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 p nRT V p p nRT V p V V TT 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。
物理化学-第五版-天津大学-上册习题答案
第一章 气体的pVT 关系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、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=（÷）=小时1-3 0℃、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 CH ρ 1-4 一抽成真空的球形容器，质量为。
充以4℃水之后，总质量为。
若改用充以25℃、的某碳氢化合物气体，则总质量为。
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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气体pVT 性质1. 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 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133153.144145.621090109032-⋅=⨯=⨯=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。
若将其中一个球加热到100℃，另一个球则维持0℃，忽略连接管中气体体积，试求该容器内空气的压力。
解：方法一：在题目所给出的条件下，气体的量不变。
并且设玻璃泡的体积不随温度而变化，则始态为 )/(2,2,1i i i i RT V p n n n =+=终态（f ）时⎪⎪⎭⎫⎝⎛+=⎪⎪⎭⎫ ⎝⎛+=+=f f ff f f f f f f T T T T R Vp T V T V R p n n n ,2,1,1,2,2,1,2,1 kPaT T T T T p T T T T VR n p f f f f i i ff f f f 00.117)15.27315.373(15.27315.27315.373325.1012 2,2,1,2,1,2,1,2,1=+⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛+=⎪⎪⎭⎫ ⎝⎛+=1-6 0℃时氯甲烷（CH 3Cl ）气体的密度ρ随压力的变化如下。
试作ρ/p —p 图，用外推法求氯甲烷的相对分子质量。
解：将数据处理如下：P/kPa 101.325 67.550 50.663 33.775 25.331(ρ/p)/（g ·dm -3·kPa ）0.02277 0.02260 0.02250 0.02242 0.02237作(ρ/p)对p 图当p →0时，(ρ/p)=0.02225，则氯甲烷的相对分子质量为()10529.5015.273314.802225.0/-→⋅=⨯⨯==mol g RT p M p ρ1-7 今有20℃的乙烷-丁烷混合气体，充入一抽真空的200 cm 3容器中，直至压力达101.325kPa ，测得容器中混合气体的质量为0.3879g 。
试求该混合气体中两种组分的摩尔分数及分压力。
解：设A 为乙烷，B 为丁烷。
mol RT pV n 008315.015.293314.8102001013256=⨯⨯⨯==- B A B B A A y y mol g M y M y n m M 123.580694.30 867.46008315.03897.01+=⋅==+==- （1） 1=+B A y y（2）联立方程（1）与（2）求解得401.0,599.0==B By ykPap y p kPa p y p B B A A 69.60325.101599.063.40325.101401.0=⨯===⨯==1-81-9 氯乙烯、氯化氢及乙烯构成的混合气体中，各组分的摩尔分数分别为0.89、0.09和0.02。
于恒定压力101.325kPa 条件下，用水吸收掉其中的氯化氢，所得混合气体中增加了分压力为2.670 kPa 的水蒸气。
试求洗涤后的混合气体中C 2H 3Cl 及C 2H 4的分压力。
解：洗涤后的总压为101.325kPa ，所以有kPa p p H C Cl H C 655.98670.2325.1014232=-=+02.0/89.0///423242324232===H C Cl H C H C Cl H C H C Cl H C n n y y p p 联立式（1）与式（2）求解得kPa p kPa p H C Cl H C 168.2 ;49.964232==1-10 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置换，步骤如下向釜内通氮直到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。
这种步骤共重复三次。
求釜内最后排气至年恢复常压时其中气体含氧的摩尔分数。
设空气中氧、氮摩尔分数之比为1∶4。
解: 高压釜内有常压的空气的压力为p 常，氧的分压为常p p O 2.02=每次通氮直到4倍于空气的压力，即总压为p=4p 常，第一次置换后釜内氧气的摩尔分数及分压为常常常常p y p p p p pp y O O O O ⨯=⨯=====05.005.042.042.01,1,1,2222第二次置换后釜内氧气的摩尔分数及分压为常常常常p y p p p p p p y O O O O ⨯=⨯====405.0405.0405.02,2,1,2,2222所以第三次置换后釜内氧气的摩尔分数%313.000313.01605.04)4/05.0(2,3,22=====常常p p pp y O O 1-11 25℃时饱和了水蒸汽的乙炔气体（即该混合气体中水蒸汽分压力为同温度下水的饱和蒸气压）总压力为138.7kPa ，于恒定总压下泠却到10℃，使部分水蒸气凝结成水。
试求每摩尔干乙炔气在该泠却过程中凝结出水的物质的量。
已知25℃及10℃时水的饱和蒸气压分别为3.17kPa 和1.23kPa 。
解：p y p B B =，故有)/(///B B A B A B A B p p p n n y y p p -=== 所以，每摩尔干乙炔气含有水蒸气的物质的量为 进口处：)(02339.017.37.13817.3222222mol p p n n H C O H H C O H =-=⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛进进 出口处：)(008947.01237.138123222222mol p p n n H C O H H C O H =-=⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫⎝⎛出出 每摩尔干乙炔气在该泠却过程中凝结出的水的物质的量为 0.02339-0.008974=0.01444（mol ）1-12 有某温度下的2dm 3湿空气，其压力为101.325kPa ，相对湿度为60％。
设空气中O 2和N 2的体积分数分别为0.21和0.79，求水蒸气、O 2和N 2的分体积。
已知该温度下水的饱和蒸气压为20.55kPa （相对湿度即该温度下水蒸气分压与水的饱和蒸气压之比）。
解：水蒸气分压＝水的饱和蒸气压×0.60＝20.55kPa ×0.60＝12.33 kPa O 2分压＝（101.325-12.33 ）×0.21＝18.69kPa N 2分压＝（101.325-12.33 ）×0.79＝70.31kPa33688.02325.10169.18222dm V p p V y V O O O =⨯=== 33878.12325.10131.70222dm V pp V y V N N N =⨯=== 32434.02325.10133.12222dm V pp V y V O H O H O H =⨯=== 1-13 一密闭刚性容器中充满了空气，并有少量的水，当容器于300K 条件下达到平衡时，器内压力为101.325kPa 。
若把该容器移至373.15K 的沸水中，试求容器中达到新的平衡时应有的压力。
设容器中始终有水存在，且可忽略水的体积变化。
300K 时水的饱和蒸气压为3.567kPa 。
解：300K 时容器中空气的分压为 kPa kPa kPa p 758.97567.3325.101=-='空373.15K 时容器中空气的分压为)(534.121758.9730015.37330015.373kPa p p =⨯='=空空 373.15K 时容器中水的分压为 =O H p 2101.325kPa所以373.15K 时容器内的总压为p=空p +=O H p 2121.534+101.325=222.859（kPa ）1-14 CO 2气体在40℃时的摩尔体积为0.381dm 3·mol -1。
设CO 2为范德华气体，试求其压力，并与实验值5066.3kPa 作比较。
解：查表附录七得CO 2气体的范德华常数为 a=0.3640Pa ·m 6·mol -2；b=0.4267×10-4m 3·mol -15187.7kPa5187675250756176952362507561100.338332603.5291)10381.0(3640.0104267.010381.015.313314.8)(3-23432==-=-⨯=⨯-⨯-⨯⨯=--=---PaV a b V RT p m m相对误差E=5187.7-5066.3/5066.3=2.4%1-15今有0℃、40530kPa 的氮气体，分别用理想气体状态方程及范德华方程计算其摩尔体积。
其实验值为70.3cm 3·mol -1。
解：用理想气体状态方程计算如下：1313031.56000056031.0 4053000015.273314.8/--⋅=⋅=÷⨯==molcm molm p RT V m将范德华方程整理成0/)/()/(23=-++-p ab V p a V p RT b V m m m(a)查附录七，得a=1.408×10-1Pa ·m 6·mol -2，b=0.3913×10-4m 3·mol -1 这些数据代入式（a ），可整理得100.1)}/({100.3 )}/({109516.0)}/({131392134133=⨯-⋅⨯+⋅⨯-⋅------mol m V mol m V mol m V m m m解此三次方程得 V m =73.1 cm 3·mol -11-16 函数1/（1-x ）在-1＜x ＜1区间内可用下述幂级数表示：1/（1-x ）=1+x+x 2+x 3+…先将范德华方程整理成2/11mmm V aV b V RT p -⎪⎪⎭⎫ ⎝⎛-=再用述幂级数展开式来求证范德华气体的第二、第三维里系数分别为B （T ）=b-a （RT ） C=（T ）=b 2解：1/（1-b/ V m ）=1+ b/ V m +（b/ V m ）2+… 将上式取前三项代入范德华方程得3222221m m m m m m m V RTb V a RTb V RT V a V b V b V RT p +-+=-⎪⎪⎭⎫ ⎝⎛++=而维里方程（1.4.4）也可以整理成32mm m V RTCV RTB V RT p ++=根据左边压力相等，右边对应项也相等，得 B （T ）=b – a/（RT ） C （T ）=b 2*1-17 试由波义尔温度T B 的定义式，试证范德华气体的T B 可表示为T B =a/（bR ）式中a 、b 为范德华常数。