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aG =-S [3D.8]; hence =-S, and =-S, (G,-G) AS = S, - 5, = ) +*). -(-73.13+ 42.8Jx =-42.8JK

aG =-S [3D.8]; hence =-S, and =-S, (G,-G) AS = S, - 5, = ) +*). -(-73.13+ 42.8Jx =-42.8JK

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The problem goes like this: The change in the Gibbs energy of a certain constant-pressure process was found to fit the expression ΔG/J = −73.1 + 42.8(T/K). Calculate the value of ΔS for the process. How did they come up with this answer?

aG =-S [3D.8]; hence =-S,, and =-S, (G,-G) aS = S, - S, = ) +). (-73.1J+ 42.8Jx) =-42.8JK
Transcribed Image Text:aG =-S [3D.8]; hence =-S,, and =-S, (G,-G) aS = S, - S, = ) +). (-73.1J+ 42.8Jx) =-42.8JK
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