物理化学热力学第一定律习题答案概要

(3)绝热可逆膨胀到50kPa;

(4)绝热反抗50 kPa恒外压不可逆膨胀。

解:(1)理想气体恒温可逆膨胀到50 kPa:

?50?103?Wr?nRTln?p2/p1??10?8.3145?350ln?J??40.342kJ ?200?103???J??40342??(2)恒温反抗50 kPa恒外压不可逆膨胀:

W??pamb(V2?V1)??pamb?(nRT/pamb)?(nRT/p1)? ??21826J??21.826kJp2?(3)绝热可逆膨胀到50kPa: T2????p???1?R/Cp,m ?-nRT?1-(pamb/p1)???10?8.3145?350?1?(50/200)?J

R/(7R/2)?50?103??T1???200?103?????350K?235.53K

绝热,Q=0,

W??U??nCV,mdT?n?CV,m?(T2?T1)T1T2

?10?5?8.3145?(235.53?350)J??23793J??23.793kJ2(4)绝热反抗50 kPa恒外压不可逆膨胀 绝热,Q=0, W??U ?pabm(V2?V1)?nCV,m(T2?T1)?pamb?(nRT2/pamb)?(nRT1/p1)??n?(5/2)R(T2?T1)

上式两边消去nR并代入有关数据得

?T2?0.25?350K?2.5T2?2.5?350K

3.5T2=2.75×350K 故 T2=275K W??U??nCV,mdT?n?CV,m?(T2?T1)T1T2

?10?5?8.3145?(275?350)J??15590J??15.590kJ2

2-12 0.5 mol 双原子理想气体1mol 从始态300K,200 kPa,先恒温可逆膨胀到压力为50kPa,再绝热可逆压缩末态压力200 kPa。求末态温度T及整个过程的Q,W,△U及△H。

解:整个过程如下

T1?300KT2?300KT200kPa?p1?恒温可逆膨胀?????50kPa?p2?绝热可逆压缩?????200kPa?p3 0.5mol0.5mol0.5molp3?理想气体绝热可逆过程:T????p???2?R/Cp,m?200?103??T2???50?103????R/(7R/2)?400K?445.80K

恒温可逆膨胀过程:

?50?103? Wr?nRTln?p2/p1??0.5?8.3145?300ln??200?103???J??1729J??1.729kJ??因是理想气体,恒温,△U恒温=△H恒温=0 绝热可逆压缩:Q=0,故

5W绝??U绝?nCV,m(T?T1)?0.5?R(T?T2)2

5 ?0.5??8.314?(445.80?300)?J?1515.2J?1.515kJ27?H绝?nCp,m(T?T1)?5?R(T?T1)2

7 ?0.5??8.314?(445.80?300)?J?2121.3J?2.121kJ2故整个过程:

W=Wr+W绝= (-1.729+1.515)kJ=0.214 kJ △U=△Ur+△U绝=(0+1.515)=1.515kJ △H=△Hr+△H绝=(0+2.121)=2.121kJ

2-13 已知水(H2O,l)在100℃的饱和蒸气压p*=101.325 kPa,在此温度、压力下水的摩尔蒸发焓

?vapHm?40.668kJ?mol?1。求在100℃,101.325 kPa 下使1 mol

水蒸气全部凝结成液体水时的Q,W,△U及△H。设水蒸气适用理想气体状态方程。

解:凝结过程为 1molH2O(g),1000C,101.325kPa△H(凝结)= —△H(蒸发)

Q?Qp?n?(??vapHm)?1?(?40.668)kJ??40.668kJ??H W??pamb(Vl?Vg)?pVg?ngRT?(1?8.314?373.15)J?3.102kJ

1molH2O(l),1000C,101.325kPa

?U?Q?W?(?40.668?3.102)kJ??37.566kJ

2-14 100 kPa 下,冰(H2O,s)的熔点为0℃,在此条件下冰的摩尔熔化焓

?fusHm?6.012kJ?mol?1。已知在-10℃~0℃范围内过泠水(H2O,l)和冰的摩尔定

?1?1?1?1压热容分别为Cp,(H2O,l)=76.28J?mol?K和Cp,(H2O,s)=37.20J?mol?K。mm

求在常压条件下–10℃时过泠水结冰的摩尔凝固焓。

解:

H2O(l),?100CHm?????H2O(s),?100C

△H1,m △H3,m

H2O(l), 00C2,m????H2O(s), 00C

?H?H2,m???fusHm??6.012kJ?mol?1

?Hm??H1,m??H2,m??H3,m ??273.15K263.15KCp,m(H2O,l)dT??H2,m??263.15K273.15KCp,m(H2O,s)dT ?Cp,m(H2O,l)?(273.15K?263.15K) ??H2,m?Cp,m(H2O,s)?(263.15K?273.15K) ?(76.28?10?6012?37.2?10)J?mol?1 ??5621J?mol?1??5.621kJ?mol?1

2-15 25℃下,密闭恒容的容器中有10g 固体萘C10H8(s)在过量的O2(g)中完全燃烧成CO2(g)和H2O(l)。过程放热401.727 kJ。求 (1)C10H8(s)?12O2(g)?10CO2(g)?4H2O(l)的反应进度;

???U?HCmCm(2)C10H8(s)的; (3)C10H8(s)的。

解:(1)M萘=128.173,

反应进度:???n/???n/1??n??(2)C10H8(s)的?CUm:

10?0.078019mol?78.019mmol

128.173Qv=△U= -401.727 kJ

每摩尔萘的恒容恒温燃烧热为

??cUm(298.15K)??rUm(298.15K)?128.173?(?401.727)kJ?mol?1(压力变化对燃烧热基本10 ??5149kJ?mol?1无影响,因为温度未变,dU=CvdT公式具有普适性,适用于气、液、固)

(3)所以本题所给反应的标准摩尔反应焓为

???rHm(298.15K)??rUm(298.15K)???B(g)?RT ?-5149000?(-2)?8.314?298.15 ?-5154kJ?mol-1???CHm??rHm?-5154kJ?mol-1

??Hcm2-16 已知25℃甲酸乙酯(HCOOCH3,l)的标准摩尔摩尔燃烧焓为-979.5

kJ?mol?1,甲酸(HCOOH,l)、甲醇(CH3OH,l)、水(H2O,l)及二氧化碳

(CO2,g)的标准摩尔生成焓数据

?1?1??fHm?1分别为-424.72kJ?mol,?1-238.66kJ?mol,-285.83kJ?mol及-393.509kJ?mol。应用这些数据求25℃时下列反应的标准摩尔反应焓。

HCOOH(l)?CH3OH(l) HCOOCH3(l)?H2O(l)

?解:(1)先求?fHm(HCOOCH3,l)。

燃烧反应 HCOOCH3(l)?2O2(g) 2CO2(g)?2H2O(l)

?????rHm?2??fHm(HCOOCH3,l) (CO2,g) + 2×?fHm(H2O,l)-?fHm?? ?rHm=?CHm(HCOOCH3,l)(反应热即为燃烧热)

所以有

????(HCOOCH3,l) (H2O,l)-?CHm?fHm(HCOOCH3,l)=2??fHm(CO2,g) + 2×?fHm ={2×(-393.509)+2×(-285.83)-(-979.5)}kJ·mol = - 379.178 kJ·mol (2)HCOOH(l)?CH3OH(l) HCOOCH3(l)?H2O(l)

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????rHm??fHm(HCOOCH3,l)+?fHm(HO2,l)

??-?fHm(HCOOH,l)-?fHm(CH3OH,l)

={(-379.178)+(-285.83)-(-424.72)-(-238.66)}kJ·mol = - 1.628 kJ·mol

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