Question by Cam N: What is the entropy change for Sulfur Dioxide acting as an ideal gas?
I am trying to calculate the entropy for sulfur dioxide, while it acts like an ideal gas. It is of 273.15 K and 1.5929 bar and 450 K and 35 bar go. I get a negative value, this is even possible and if so, what would be the entropy change?

Best answer:

Answer by Lee Wang Tui
T1 = 273.15 K
P1 = 1.5929 bar
T2 = 450 K
P2 = 35 bar
From source, Cp = 0.64 kJ/kgK and R = R’/M = 0.13 kJ/kgK
Using Tds equation. Tds = dh – vdp, assume constant cp, ideal gas, and integrate from T1 to T2
∆s = cp ln T2/T1 – R ln P2/P1 = 0.64 [kJ/kgK] ln 450/273.15 – 0.13 [kJ/kgK] ln 35/1.5929
∆s = 0.3195 – 0.4017 = – 0.0822 kJ/kgK

Yes, you are right. Entropy change is negative.
High temperature means entropy system higher (+) than previous state, but high pressure indicate entropy system lower (-). Tds equation counting both change.

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