How Does Entropy Change in a Reversible Cooling Process?

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SUMMARY

The discussion focuses on the entropy change in a reversible cooling process involving 1000 mol of helium gas, initially at 2000 K and 1 MPa, cooled to 500 K under the constraint that PV3 = constant. The initial volume of the gas is determined, and the first law of thermodynamics is applied to derive the relationship between heat transfer and volume change. The entropy change ΔS is calculated as -1.152 x 104, prompting a discussion on the implications of negative entropy values in reversible processes, confirming that ΔS + ΔSsurrounding = 0 is a valid assumption.

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  • Understanding of the first law of thermodynamics
  • Familiarity with the ideal gas law (PV = nRT)
  • Knowledge of entropy and its calculation in thermodynamic processes
  • Concept of reversible processes in thermodynamics
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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 VA of the gas

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

dQin = (3/2)nRdT + (PAVA3dV)/V3

c) eleminate P from the two process equations PV = nRT and PV3 = PAVA3

and hence that

blah blah blah

d) Use dS/dQinrev/T to find the entropy change ΔS = SB - SA

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 ΔSuniverse = 0 such that ΔS + ΔSsurrounding = 0

would this be a good assumption?
 
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Liquidxlax said:
I thought to myself that the ΔSuniverse = 0 such that ΔS + ΔSsurrounding = 0

would this be a good assumption?

Yes; that's what "reversible" implies.
 
Mapes said:
Yes; that's what "reversible" implies.

derp :facepalm: thanks for the help
 

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