How Is Entropy Calculated in an Isothermal Expansion of CH4 in a Carnot Engine?

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SUMMARY

The discussion focuses on calculating the entropy increase during the isothermal expansion of 1 kg of methane (CH4) in a Carnot engine, where the efficiency is 25% and the volume ratio is 4. The initial calculations suggested an incorrect entropy change of 720 J/K, while the correct value is established as 294 J/K. Key equations utilized include the ideal gas law and the relationship between efficiency and temperature in a Carnot cycle, specifically η = 1 - (T1/T2). The confusion arose from miscalculating the number of moles and misunderstanding the volume ratios at different stages of the cycle.

PREREQUISITES
  • Understanding of the Carnot cycle and its efficiency calculations.
  • Familiarity with the ideal gas law and entropy equations.
  • Knowledge of the relationship between pressure, volume, and temperature in thermodynamics.
  • Basic skills in sketching thermodynamic cycles for visual analysis.
NEXT STEPS
  • Study the derivation of the Carnot cycle efficiency formula, η = 1 - (T1/T2).
  • Learn how to apply the ideal gas law in isothermal and adiabatic processes.
  • Explore the concept of entropy in thermodynamic systems, focusing on isothermal expansions.
  • Practice sketching and analyzing thermodynamic cycles to clarify volume relationships.
USEFUL FOR

This discussion is beneficial for students of thermodynamics, particularly those studying heat engines, as well as engineers and physicists involved in energy systems and efficiency analysis.

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Homework Statement



A Carnot engine operates on 1 Kg of CH4, which we shall consider an ideal gas. Take \gamma=1.35. The ratio of max to min volume is 4 and the cycle efficency is 25%. Find the entropy increase during the isothermal expansion.
[Classical and Statistical Thermodynamics, Ashley H. Carter, pg.104]

Homework Equations



dU = TdS-PdV

The Attempt at a Solution



Ideal gas, isothermal expansion,\Rightarrow dU = 0

\Rightarrow dS=\frac{P}{T}dV=nR\frac{dV}{V}

\Rightarrow \DeltaS=R(62.5)ln(4)=720J/K


The correct answer is 294J/K

I know my problem has to be with the number of moles, probably accounting for the efficiency. working backwards I see there should be 25.5 moles of methane for n.
How do I account for the efficiency and get this answer?
 
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In a Carnot cycle, the efficiency is connected to the temperatures of various processes. Does this help? Have you sketched out the cycle?

(Also, 1 kg methane is not 25.5 moles.)
 
Mapes said:
In a Carnot cycle, the efficiency is connected to the temperatures of various processes. Does this help? Have you sketched out the cycle?

\eta=1-\frac{T_{1}}{T_{2}}

Is that what you mean? How does that help for an isotherm? I'm not sure what information to use regarding the Carnot cycle itself considering the question only asks about the isothermal expansion.

For just the expansion with the ratio of volumes given my answer should be correct, there is some relevant information I am not seeing.


(Also, 1 kg methane is not 25.5 moles.)

I know, its 62.5 moles. I got 25.5 from solving
nRln(\frac{V2}{V1})=294J/K
for n.
 
Sketch the cycle; are the volumes at the beginning and end of the isothermal process the maximum and minimum volumes in the cycle?
 
Thank you Mapes.

Of course those are not the max/min volumes. I was able to obtain the correct volume ratio buy utilizing PV^{\gamma}=constant for an an adiabat,\eta=1-\frac{T_1}{t_2}, and the given information.
 
I'm doing this question right now, and I am wondering how to get 294 J/K as an answer. I don't understand why the final volume doesn't relate to the max volume, and how to use PV^gamma to get the correct ratio.

Thanks.
 
Seem you guys forget that the initial volumn is not at the isothermal expansion, you have to trace it back to the start of adiabatic expansion then the isothermal expansion. (draw the diagram and you will see)

Ratio Max/Min volume is the volume at the tip of adiabatic expansion.

You'll need to use both PVg = PV/T = constant

Good brush up exercise though.
 

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