高教热统答案第三章

第三章单元系的相变习题3.2试由0vC及0)(TVp证明0pC及0)(SVp。证：由式(2.2.1)TCCVpVTppTVPCpTH=pTST;VCVTUVTSTdpdVVpTdTTpVdpdVVpSdSSpVdVVpSVSpdTTSdVVSVTTVpVSpTVS+SVp(1)VTpVSpTTS(2)由麦氏关系(2.2.3)代入(1)式中SVT-VSpTVpSVpSVTTVSSVpSVST,,TVTS,,SVpTVST,,SVTV,,TVST,,SVpSVTV,,2,,TVSTSVpVST2,,TVST由式(2.2.5)VCVTST；即0VVCTST.于是：0TVpSVp正数于是：SVp0PCPTSTpTpST,,VSpST,,pTVS,,SVpTpTVS,,SVpTVTVS,,pTVT,,SVpTVTSTpVSVpVTCpV0VC;因而0PC习题3.4求证：（1）nVT,VTnS,；（2）nTp,pTnV,证：(1)开系吉布斯自由能dnVdpSdTdG,),(TVppdndTTpdVVpVSdTdGVTdndVVPVdTTPVSnTnVVSTGnV,VTp①VVGnT,TVp②VTnG,③由式①nVnVTGTpVS,VTnS,VTnVnTG,,VnTG2VTnG2VTnS,nVT,第(1)式得证。(2)由式(3.2.6)得：pTnV,TnpG2TpnG2nTp,习题3.7试证明在相变中物质摩尔内能的变化为：dpdTTpLu1如果一相是气相，可看作理想气体，另一相是凝聚相，试将公式化简。解：由式(3.2.7)得：VpSTU；又由式(3.4.6)得：VTLdTdp；STL；dpdTTpLLUdpdTTpL1习题3.8在三相点附近，固态氨的蒸气压(单位为aP)方程为：Tp375492.27ln液态氨的蒸气压方程为：Tp306338.24ln，试求氨三相点的温度和压强，氨的汽化热、升华热及在三相点的熔解热。解：(1)固态氨的饱和蒸气压方程决定了固态-气态的相平衡曲线；液态氨的饱和蒸气压方程决定了氨的液态-气态的相平衡曲线。三相点是两曲线的交点，故三相点温度3T满足方程：TT306338.24375492.27；由此方程可解出3T，计算略；(2)相变潜热可由RTLApln与前面实验公式相比较得到：3754RLS，从而求出SL；类似可求出QL；计算略；(3)在三相点，有rQSLLL，可求得rL，计算略。习题3.10试证明，相变潜热随温度的变化率为：pcdTdL-pcTLvvLTvTvpp如果相是气相，相是凝聚相，试证明上式可简化为：ppccdTdL证：显然属于一级相变;)(SSTL;其中)(,TpTSS，在p～T相平衡曲线上。dTdppSTTSTSSdTdL其中：TSPTSPTSdTdppS[PTSPTS]dTdp又有：TCPPTS；)(SSTL由麦氏关系(2.2.4)：TpSPTV上几式联立(并将一级相变的克拉伯珑方程代入)得：pcdTdL-pcTLvvLTvTvpp若相是气相，相是凝聚相；V～0；pTV～0；相按理想气体处理，pV=RT，ppccdTdL习题3.11根据式(3.4.7)，利用上题的结果及潜热L是温度的函数。但假设温度的变化范围不大，定压热容量可以看作常数，证明蒸汽压方程可以表为：TCTBAplnln解：蒸汽压方程：21RTLdTdpp2TLdTpdp利用ex.3.10结果。pCdLdTCLTT0;温度变化的范围不大；设常数）(CCCCpPp20TLCRLdLpdpdLTLTTLTLCRp200200)(1ln0001ln1TLTCRTLCRL+T0=T；plnKTTCRTCR01ln1习题3.12蒸汽与液相达到平衡。以dTdv表在维持两相平衡的条件下，蒸汽体积随温度的变化率。试证明蒸汽的两相平衡膨胀系数为RTLTdTdvv111。解：由式(3.4.6)克拉珀珑方程。并注意到V～0.方程近似为：TVLTp，V—气相摩尔比容。VpTLTVV11①气相作理想气体，pV=RT②TRVppV③联立①②③式，并消去△p、P得：TLTVVVPTRTLVTVVVRTTRLTVVRTTR221RTLRTTVV；RTLTRTTTVVP111112习题3.13将范氏气体在不同的温度下的等温线的极大点N与极小点J联起来，可以得到一条曲线NCJ,如图3.17所示。试证明这条曲线的方程为：)2(3bvapv并说明这条曲线分出来的三条区域ⅠⅡⅢ的含义。解：范氏气体：RTbvvap2；2vabvRTp等温线上极值点，极值点组成的曲线：322)(vabvRT；由2vapbvRT)2(3bvapv习题3.14证明半径为r的肥皂泡的内压与外压之差为r4。(略解)：连续应用式(3.6.6)及(3.6.16)。习题3.16证明爱伦费斯公式：1212kkdTdp；)(1212TvccdTdppp证：对二级相变0)(dS；即2dS-1dS=00)(dV；即2dV-1dV=02dSdTTS2dppS1；1dSdTTS1dppS1)(0dS2dS-1dSdTTSTS12dppSpS12pSpSTSTSdTdp1212;将ppTSTC代入得。pSpSCCTdTdppP12121①由式(3.2.6)得：VpTpS2；结合式(3.7.2)即为：pS2121VpS；代入①得：1212TVCCdTdppP类似地，利用0)(dV可证第二式。（略）