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' :

For every process relating the same gas, the first law of thermodynamics is written (using the chemistry sign convention):
$$\Delta U = q - w$$
where
  • $$q$$ is the heat absorbed by the gas (positive when absorbed).
  • $$w$$ is the work done by the gas on the surroundings (positive when the gas expands).

Process X → Y

Given: $$q_1 = +10\; \text{J}$$ (heat absorbed),
       $$w_1 = +18\; \text{J}$$ (work done by the gas).
Thus
$$\Delta U_1 = q_1 - w_1 = 10 - 18 = -8\; \text{J} \; -(1)$$

Process Y → X

The gas returns to the initial state by a different path. In this reverse path
       heat evolved (released) $$= 6\; \text{J} \implies q_2 = -6\; \text{J}$$

Because internal energy $$U$$ depends only on state, the total change around the closed cycle X → Y → X must be zero:
$$\Delta U_1 + \Delta U_2 = 0 \Longrightarrow \Delta U_2 = -\Delta U_1 = +8\; \text{J} \; -(2)$$

Apply the first-law equation to the reverse path:

$$\Delta U_2 = q_2 - w_2$$
Substitute the known values:
$$8 = (-6) - w_2$$
$$-w_2 = 14$$
$$w_2 = -14\; \text{J}$$

Negative $$w_2$$ means the work is done on the gas by the surroundings. Its magnitude is $$14\; \text{J}$$.

Hence, in the reverse process Y → X, 14 J of work is done on gas A by the surroundings.

Option D which is: 14 J of the work is done on the gas 'A' by the surrounding.