2nd Law of Thermo: Specific entropy

[imscavok]

Homework Statement

A parcel of dry air is lifted isothermally such that its pressure decreases from 1000 hPa to 850 hPa. Calculate the change in specific entropy of the parcel.

Homework Equations

ΔS=n*R*ln(p1/p2)

The Attempt at a Solution

Using R=8.314 J/mol K
p1=1000hPa
p2=850hPa

And assuming n=1

ΔS=1.35J/K

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The problem is that's just the change in entropy. I need the change in specific entropy, which units are in J/g*K.

The problem gives no information on the amount of mass within the parcel. Even if I had the mass, I'm not sure how I would find the specific entropy with it. What am I missing?

edit: Nevermind. R is specific for dry air which is R=287 J/kg*K

 

[anathema]

. So the specific entropy is just the change in entropy divided by the mass. In this case, ΔS=0.0047 J/g*K.

Thank you for your question. To find the change in specific entropy, you will need to use the specific gas constant, which is different for different gases. For dry air, the specific gas constant is 287 J/kg*K. This means that to find the specific entropy, you will need to divide the change in entropy (1.35 J/K) by the mass of the parcel. However, the problem does not give information on the mass of the parcel, so you will not be able to find the specific entropy without this information.

If you had information on the mass of the parcel, you would simply divide the change in entropy by the mass to find the specific entropy. In this case, the specific entropy would be 0.0047 J/g*K.

I hope this helps clarify your understanding of specific entropy. If you have any further questions, please feel free to ask.