Gas 'A' undergoes change from state 'X' to state 'Y'. In this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state 'X' by another process during which 6 J of heat is evolved. In the reverse process of 'Y' to 'X',
D. 14 J of the work is done on the gas 'A' by the surrounding.
Using the first law of thermodynamics, U = q + w. For the process from state X to state Y: Heat absorbed by the gas, q_ X Y = +10 J Work done by the gas, w_ X Y = -18 J Change in internal energy, U_ X Y = q_ X Y + w_ X Y = 10 - 18 = -8 J For the reverse process from state Y to state X: Change in internal energy, U_ Y X = - U_ X Y = +8 J Heat evolved by the gas, q_ Y X = -6 J Let the work done on the gas be w_ Y X . U_ Y X = q_ Y X + w_ Y X 8 = -6 + w_ Y X w_ Y X = +14 J Since the work done is positive, 14 J of work is done on the gas by the surrounding. Answer: 14 J of the work is done on the gas 'A' by the surrounding.
Related: Chemistry — Thermodynamics (C) · All PYQ Banks