Ideal gas adiabatic/isovolumetric help

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SUMMARY

The discussion focuses on calculating the change in entropy for an ideal gas undergoing an adiabatic expansion followed by an isobaric compression. The gas, with specific heat capacities Cp = 20.78 J/mole·K and Cv = 12.47 J/mole·K, expands to four times its original volume, where the adiabatic condition (Q = 0) applies. The correct approach involves using the adiabatic equation PV^γ = constant to find the new pressure and temperature after expansion, and then applying the isobaric process equations to determine the change in entropy, ultimately leading to a calculated value of ΔS = 24.94 J/K.

PREREQUISITES
  • Understanding of ideal gas laws and equations (PV = nRT)
  • Knowledge of thermodynamic processes, specifically adiabatic and isobaric processes
  • Familiarity with specific heat capacities (Cp and Cv) and their applications
  • Ability to apply entropy change formulas in thermodynamics
NEXT STEPS
  • Study the derivation and application of the adiabatic process equation PV^γ = constant
  • Learn how to calculate temperature changes in isobaric processes using the ideal gas law
  • Research the natural logarithm formula for calculating entropy change in non-constant temperature scenarios
  • Explore the implications of heat transfer in thermodynamic processes, particularly in adiabatic conditions
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, particularly those tackling problems involving ideal gases, entropy calculations, and the principles of adiabatic and isobaric processes.

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Homework Statement


two moles of an ideal gas is initially at P = 2*20^5 Pa expands adiabaticaly to four times its original volume. it is then compressed at constant pressure to its original volume. what is the change in entropy of the gas

Cp = 20.78 joules/mole-deg
Cv = 12.47 joules/mole-deg
gamma = 5/3



Homework Equations



adiabatic expansion --> PV^gamma = constant, Q = 0
isovolumetric compression --> nCv(deltaT)
entropy deltaS= deltaQ/T

The Attempt at a Solution



i am not sure how to use the adiabatic equation because although P and V is given as well as gamma, i can solve for 'constant' but where does constant come into play?

since for an adiabat Q = 0, do i just take it as zero and move on or do i actually need to solve for something.

as for the isovolumetric compression, moles is given, Cv is given but not deltaT so i solved the Pv=nRT eq for T and subbed it in, i then assumed P and V to be constants as stated in the problem?? and solved for Q getting 1.5 joules

when i solved for entropy i did (compression - expansion, so isovolumetric minus adibat = i subbed in the equation from the isovolumetric equation in and canceled out the T's and was able to get deltaS = 24.94 joules per kelvin

did i take the correct approach? help appreciated...
 
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As you have recognized, the heat transferred in an adiabatic process is zero. Therefore, the ratio of Q/T is zero, and there is no change in entropy.

And the second process is at constant pressure (isobaric), not constant volume, so use Cp in the second equation.

You need to find the two temperatures, and for that you need to find the pressure after the adiabatic expansion. That's where the gamma equation comes in. (You said volume is given but you did not state what initial volume is). Once you have the pressure after the expansion, you can get the temperature simply using PV=nRT

Then find the delta T for the isobaric compression.

And to find the delta S, you need a slightly different formula, since T is not constant (look for a natural log formula).
 
hi thanks, just to clarify the problem does not specify an actual volume just the amount at which it decreases/increased
 

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