Q 11-11-157JEE MainJEE Main 2019 (12 Apr, Shift 1)Easy
A sample of an ideal gas is taken through the cyclic process abca as 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:
Answer: (A) $130$ J
Over the cycle $\Delta U = 0$, so $\Delta U_{abc} = -\Delta U_{ca} = 180$ J.
$$W_{abc} = Q_{abc} - \Delta U_{abc} = (250 + 60) - 180 = 130\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics