Question

The figure shows a cycle consisting of five paths: AB is isothermal at 300 K, BC is adiabatic with work = 5.5 J, CD is at a constant pressure of 5 atm, DE is isothermal, and EA is adiabatic with a change in internal energy of 7.2 J. What is the change in internal energy of the gas along path CD?

Answer 1

Solution)

Hete, AB is isothermal

Total internal energy change = 0

for AB , change is 0 (isothermal)

BC, adiabatic

so ΔU = -5.5 J

Work is done by the system or it is negative. Work is done on the system as volume expands.

DE is isothermal so change in internal energy=0

EA is adiabatic so ΔU = 7.2J.

Here, work is done on the system as volume contracts.

7.2-5.5 + CD = 0

CD = -1.7J (Ans)

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Good luck!:)