Practice portal › Thermodynamics › Cyclic Processes

A sample of an ideal gas is taken through the cyclic process abca shown in the figure. The change in the internal energy of the gas along the path ca is -180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is

Asked in JEE Main 12th April 1st Shift 2019 · Energy Balance round a Cycle

Figure: Energy Balance round a Cycle
Answer: (2) 130 J

Step-by-step solution

Round the closed cycle Δ U=0, so

Δ U_ab+Δ U_bc=-Δ U_ca=+180 J.

Heat in along abc: Q_abc=250+60=310 J.

First law on abc: W_abc=Q_abc-Δ U_abc=310-180=130 J.

Why the other options are wrong

More Cyclic Processes questionsAll Cyclic Processes questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer