Consider 0.05 mol of diatomic Oxygen in a container with initial volume 1 L and initial...

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Consider 0.05 mol of diatomic Oxygen in a container with initial volume 1 L and initial pressure 1 atm. The Oxygen undergoes

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Answer 1

Answer #1

(1) adiabatic constant $\gamma$ = 1.4

Final volume V2 = (P1/P2)V1 =( (1/3)1^1.4 )^1/1.4= 0.46 L

(2) work done w= (P1V1-P2V2)/( $\gamma$ -1) = (1.01325×10⁵×0.001- 1.01325× 3×10⁵×0.00046)/0.4= -97.96 Joule

(3) change in temperature T1-T2= w( $\gamma$ -1)/nR= (-97.96×0.4)/(0.05×8.314) = -94.27K

(4) cometing the process very fastly in isolated chamber.

Answer 2

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