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Reversible Adiabatic Expansion for an Ideal Gas

1. 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.


2. 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.

3. 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:

Redbelly98

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Welcome to Physics Forums.

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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