One liter of a monoatomic ideal gas at 3 atm is subject to reversible adiabatic expansion...
One liter of a monoatomic ideal gas at 3 atm is subject to reversible adiabatic expansion to .5 atm to a final volume of 2.9 L. Calculate the work, the heat, and the enthalpy change in this process.
Homework Answers
Work done in adiabatic process =(P1V1 - P2V2) / (y-1)
Putting the values
W = (1 ×3 - 2.9×0.5) /(5/3 -1) = 1.55× 1.5 × 101.325 (converting to J = 235.58 J
As it is an adiabatic process, heat transfer is zero
∆H = ∆E + nR∆T
= 3/2 nR∆T + nR ∆T ( Cv for monoatomic = 3/2)
5/2 nR (∆T) = 392.64J
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