·例6.02的气态方程为：pV(1-Bp)=nRTB=-0.00094,求在 273K下，将0.5mol02从10个大气压降至1个大气压的△G? 。解：dT=0 △G=Vdp V=nRT/p(1-βp) △GnRT/p(1-βp)dp=nRT[1/p(l-Bp)ldp ● =nRT[1/p+β/(1-βpldp=nRT[dp/ptp/1-pp)dpl ● AG=nRT[In(p2/p)-In(1-Bp2)/(1-Bp1)] =0.5·8.314273[ln1/10-ln(1-βp2)/1-βp1)l =-2604J 此过程的△G=-2604J • 例6. O2的气态方程为: pV(1-p)=nRT =-0.00094, 求在 273K下,将0.5mol O2从10个大气压降至1个大气压的G? • 解: dT=0 G=∫Vdp • V=nRT/p(1-p) • G=∫nRT/p(1-p)dp=nRT∫[1/p(1-p)]dp • =nRT∫[1/p+/(1-p)]dp=nRT[∫dp/p+∫/(1-p)dp] • G=nRT[ln(p2 /p1 )-ln(1-p2 )/(1-p1 )] • =0.5·8.314·273·[ln1/10-ln(1-p2 )/(1-p1 )] • =-2604 J • 此过程的G=-2604J