作者oaoa0123 (OldFlame)
看板Chemistry
标题Re: [学科] 热力学问题
时间Mon Nov 27 00:40:55 2017
※ 引述《a1616001 (奈米栗子)》之铭言：
: http://i.imgur.com/FvrYmC9.jpg
: 小弟想问第三题
: 不知道要怎麽下手
: 想问各位大大的想法
: -----
: Sent from JPTT on my Asus ASUS_Z00LD.
q_P(gas->liq at 373K)=ΔH(gas->liq at 373K)=nΔH_m(vap)
=-(21.05/18)mol*40.65kJ/mol=47.54kJ.
q_P(liq at 298K->373K)=ΔH(liq at 298K->373K)=nCp_mΔT
=(415/18)mol*75.3J/K/mol*(373-298)K=130.2kJ.
So the final state of the system is liquid water with temperature T Kelvin.
Over the process, q_P=ΔH=0, which yields
ΔH(gas at 373K->liq at 373K)+ΔH(liq at 373K->TK)+ΔH(liq at 298K->TK)=0,
-47.54kJ+(21.05/18)mol*75.3J/K/mol*(T-373)K+(415/18)mol*75.3J/K/mol*(T-298)K=0,
T=325K.
ΔS(total)=ΔS(gas->liq at 373K)+ΔS(liq at 373K->325K)+ΔS(liq at 298K->325K)
^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
∫dq_P/T=∫dH/T=ΔH/T ∫dq_P/T=∫dH/T=∫nCp_mdT/T=nCp_mln(Tf/Ti)
=-47.54/373kJ/K+(21.05/18)*75.3*ln(325/373)J/K+(415/18)*75.3*ln(325/298)J/K
=-127-12+150J/K
=11J/K.
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