大学物理答案 第11章 北京邮电大学出版社 主编：罗益民 余燕

第11章气体动理论11-1（1）由RTMmpV把p=10atm,T=(47+273)K=320K.m=0.1kg,M=32×10-3kgR=8.31J·mol-1·K-1代入.证V=8.31×10-3m3(2)设漏气后，容器中的质量为m′，则TRMmVp3201.08530030085RMRMmRMmpV)kg(151m漏去的氧气为kg103.3kg301kg)1511.0(2mmm11-2太阳内氢原子数HSmMN故氢原子数密度为3827303)1096.6(341067.11099.134sHSRmMVNn)(105.8329m由P=nkT知)(1015.11038.1105.81035.17232914KnkpT11-3如图混合前：2221112222111OHeTMmTMmRTMmpVRTMmpV气有对气有对①总内能222111212523RTMmRTMmEEE前②①代入②证1114RTMmE前混合后：设共同温度为T题11-2图RTMmTTEFRTMmMmE21210221125231,2523式得又由后③又后前EE，故由（2）（3）知)/53(8211TTTT11-4(1)000002020)(vvvvvavvvvavf（2）由归一化条件01d)(vvf得0020032123dd000vaavvavvvavvv（3）4dd)(00002/02/NvvvaNvvNfNvvvv（4）从图中可看出最可几速率为v0~2v0各速率.（5）0002/000ddd)(vvvvvavvvavvvvfv020911611vav（6）02/02/097ddd)(d)(00002121vvvvavvavvvvfvvvfvvvvvvvvv11-5氧气未用时，氧气瓶中TTpLVV111,atm130,32VRTMpVRTMpm11111①氧气输出压强降到atm102p时VRTMpVRTMpm22222②氧气每天用的质量000VRTMPm③L400,atm100VP设氧气用的天数为x,则021210mmmxmmxm由(1)(2)(3)知0021021)(VpVppmmmx)(6.932400110130天11-6（1）)(m1041.23001038.110325235KTpn（2）(kg)103.51002.61032262330NM（3）)kg/m(3.1103.51041.232625n（4）(m)1046.31041.21193253nl（5）认为氧气分子速率服从麦克斯韦布，故)(ms1046.4103230031.86.16.11-23MRTv（6）122ms1083.43MRTv（7）(J)1004.13001038.12522023KTi11-73112310m1006.12371038.1104kTpnnkTp)(cm1006.135故1cm3中有51006.1个氮气分子.m101.21006.111d43113n11-8由课本例11-4的结论知)ln(0ppMgRTh(m)1096.1)8.01ln(8.9102930031.83311-9(1)(J)1021.63001038.123232123KTt（2）看作理想气体，则3132310101030028.16.16.1KTv12ms1003.111-10(J)1074.330031.823233RTE平动(J)1049.230031.83RTE转动内能(J)1023.630031.825253RTE11-11（1）由KTpnnKTp∵是等温等压∴1:1:21nn（2）MRTv6.1是等温，∴4:1322::1221MMvv11-12317233102.33001038.11033.1mKTPnm)(8.71033.110923001038.1d2320232pKT11-13(1)8000021042.56.1d2zMRTvKTpnvnz（2）由公式MTRKpMRTKTpvnz222d26.1d2d2知z与T和P有关，由于T不变，故z只与P有关.则1854000071.01042.510013.11033.1::szppzppzz11-14（1）如图MRTv32∴AcAcTTvv::22又CB等温过程，故CBTT.由BAABVVPPRTMmpV2则ABTT2∴1:2:22AcVV（2）AAccAcPTPTpKT::d22CB等温过程ACAAACBBCCppVpVpVpVp221:2:AC11-15（1）MRTv73.12)(ms100.7102400031.873.1133（2）m10210)31(2122101021ddd（3）325202210710401042d2vnz110s10511-16（1）题11-14图MTRkpzKTpnMRTvvnz8d28d222①又由mREMTRTMmRTMmE3326②把②代入①知EmkMpKNEmkMpRz3d43d4022EmMpN3d402（2）MRTvP2把②代入得mEmREMMRVP3232（3）平均平动动能0232323mNEMmREMkkTt