Reversible Adiabatic Expansion for an Ideal Gas

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In a reversible adiabatic expansion of 1 mole of a monoatomic gas from 30 L at 400 K to 60 L, the final pressure is calculated to be 0.55 atm, while the temperature remains at 400 K. The confusion arises from the application of equations typically used for isothermal processes, such as P1V1 = P2V2. For adiabatic processes, a different equation is required, as the heat transfer (Q) is zero. The molar heat capacity used in the calculations is (3/2)R, consistent with the characteristics of a monoatomic gas. The discussion emphasizes the importance of using the correct equations for different thermodynamic processes.
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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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