How Does Entropy Change in a Reversible Cooling Process?

[Liquidxlax]

Homework Statement

A cylinder contains 1000mol o He gas at an initial temp o 2000k and initial pressure of 1MPa. The He gas is now cooled to a final temp o 500k in a reversible process in which the volume and pressure are constrained to vary as PV³ = constant. Assume that the He is a monatomic ideal gas. Denote the initial and final states of the gas by A and B, respectively.

a) Find the initial volume V_A of the gas

b) use the 1st law o thermodynamics to show that

dQ_in = (3/2)nRdT + (P_AV_A³dV)/V³

c) eleminate P from the two process equations PV = nRT and PV³ = P_AV_A³

and hence that

blah blah blah

d) Use dS/dQ^inrev/T to find the entropy change ΔS = S_B - S_A

Homework Equations

all given in the question

The Attempt at a Solution

I solved a) b) c) with no trouble, but I'm just uncertain about d)

doing the integral i found that the change in entropy ΔS = -1.152x10⁴ and i was wondering i this is reasonable, because i thought entropy is supposed to be ≥0.

I thought to myself that the ΔS_universe = 0 such that ΔS + ΔS_surrounding = 0

would this be a good assumption?

 

[Mapes]

I thought to myself that the ΔS_universe = 0 such that ΔS + ΔS_surrounding = 0

would this be a good assumption?

Yes; that's what "reversible" implies.

 

[Liquidxlax]

Yes; that's what "reversible" implies.

derp :facepalm: thanks for the help