## 4F.9 7th Edition

$\Delta S = \frac{q_{rev}}{T}$

klarratt2
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### 4F.9 7th Edition

"Calculate the change in entropy when the pressure of 1.50 mol Ne(g) is decreased isothermally from 15.0 atm to 0.500 atm. Assume ideal behavior."

I got the correct numerical answer for this, but why is the answer not negative? In the back of the book, it says that 42.4 J/K is the answer but I got -42.4 J/K. Wouldn't the answer be negative because pressure is decreasing?

Chem_Mod
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### Re: 4F.9 7th Edition

When the volume of a system decreases, then yes the $\Delta S$ would be negative. However, we are dealing with pressure, not volume. Since P and V are on the same side of ideal gas law, when one increases the other must decrease. So if we are decreasing the pressure from 15 atm to 0.5 atm then its the same as increasing the volume. The equation for $\Delta S$ when pressure is changing is $\Delta S=nRln(P_{1}/P_{2})$. So you'll notice that in this equation the initial pressure is in the numerator, and the final pressure is in the denominator. This is reverse of what the equation is for volume (V2 is in the numerator and V1 is in the denominator), so that is why $\Delta S$ is positive when the pressure is decreasing.