Question

The above pV diagram shows the expansion of 5.0 moles of a monatomic ideal gas from state a to state b. As shown in the diagram, Pa = Pb = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3. The pressure is then reduced to 200 Pa without changing the volume, as the gas is taken from state b to state c.

c. Determine Q for the process bc.

d. Determine the change in thermal energy of the gas for the two processes ab and bc together; that is, Eth,c - Eth,a.

Answer 1

from a to b is isobaric process

change in internal energy Ub- Ua = n*Cv*dT

dU = n*Cv*dT

dU = (3/2)*n*R*(Pb*Vb - Pa*va)/nR

dU = (3/2)*(Pb*Vb - Pa*va)

dU = (3/2)*((600*9) -(600*3)) = 5400

Work = Pa*(Vb - va) = 600*(9-3) = 3600 J

from Ist law of thermodynamics

Q = dU + W = 9000 J   <<<<=========ANSWER

part (b)

Work = Pa*(Vb - va) = 600*(9-3) = 3600 J <<<<=========ANSWER

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part(c)

change in internal energy Ub- Ua = n*Cv*dT

dU = n*Cv*dT

dU = (3/2)*n*R*(Pc*Vc - Pb*vb)/nR

dU = (3/2)*(Pc*Vc - Pb*vb)

dU = (3/2)*((200*9) -(600*9)) = -5400

dU = -5400 J

work = 0

Q = dU = -5400 J   <<<<=========ANSWER

part(d) Uc - Ua = Uc - Ub + (Ub-Ua) = 0 <<<<=========ANSWER