Design reversible process

H2O(g，383.2K， 1. 1p)——H2O(g，373.2K， 1. 1p)——H2O(g，373.2K，p)——H2O(l，373.2K，p)——H2O(l，383.2K，p)——H2O(l，383.2K， 1. 1p)

Entropy change of computing system;

S 1=Cp(H2Og)/T integral from T=383.2K to 373.2k.

S2 = Rial (1. 1p/p)

S3=-Hvap/T

S4=Cp(H2Ol)/T Integral from T=373.2K to 383.2k.

S5=0, because it is a condensed phase.

S=S 1+S2+S3+S4+S5

Computing environmental entropy change:

S`=-Q/TQ is the heat change in the actual process, which can be obtained from Q=U-W, and T is the ambient temperature of 373.2K K.

Calculate S+S'. If it is greater than 0, it is a spontaneous process. There is no calculator at hand, so I can only make suggestions. . . If you need it, I can help you in a few days.

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