Reversible Adiabatic Expansion for an Ideal Gas

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SUMMARY

The discussion focuses on the reversible adiabatic expansion of 1 mole of a monoatomic ideal gas, transitioning from 30 L and 400 K to 60 L. The molar heat capacity is established as (3/2)R, indicating that the process is adiabatic with no heat exchange (Q=0). The initial pressure (P1) is calculated to be 1.1 atm, leading to a final pressure (P2) of 0.55 atm, while the final temperature remains constant at 400 K. The confusion arises from the application of the isothermal equation P1V1 = P2V2, which is incorrect for adiabatic processes.

PREREQUISITES
  • Understanding of ideal gas laws, specifically PV=nRT.
  • Knowledge of adiabatic processes and their characteristics.
  • Familiarity with the concept of molar heat capacity, particularly for monoatomic gases.
  • Ability to apply the first law of thermodynamics in the context of adiabatic systems.
NEXT STEPS
  • Study the equations governing adiabatic processes, such as P1V1^γ = P2V2^γ, where γ is the heat capacity ratio.
  • Explore the derivation and implications of the first law of thermodynamics for adiabatic systems.
  • Learn about the differences between isothermal and adiabatic processes in thermodynamics.
  • Investigate the behavior of monoatomic gases under varying conditions using simulation tools like PhET Interactive Simulations.
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adiabaffled
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Homework Statement


1 mole of a monoatomic gas undergoes reversible expansion from 30 L and 400K to 60 L. The molar heat capacity in this situation is (3/2)R, independent of temperature. Calculate the final pressure and temperature of this process if it is adiabatic.

Homework Equations


Q= 0 if it is adiabatic
Q= nCv(dT)
PV=nRT
P1V1= P2V2
monoatomic Cv= (3/2)R <-- heat capacity at const. vol.

The Attempt at a Solution


Using Q=0, 0= nCv(dT)
0= (1 mol)(3/2)(R)(dT)
dT= 0
P1V1=nRT or (P1)(30L)=(1)(8.314 J/mol K)(400 K)
P1= 1.1 atm (using conversion factor)
Substituting into P1V1=P2V2, I get 0.55 atm.
Temperature is still 400 K at the end

I wanted to make sure if this was correct (the fact that the change in temp is zero while it was adiabatic confused me).
 
Last edited:
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adiabaffled said:
P1V1= P2V2
That's for an isothermal process, but this one is adiabatic. There is a different, but similar-looking, equation to use instead.
 
Last edited:

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