A gas undergoes a change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A: What amount of work will be done?

According to the first law of thermodynamics, ΔU = q + w = 5 - 8 = -3 J
As the internal energy is a state function, for the reverse process ΔU should be equal to -3 J.
ΔU = q + w
-3 = -3 + w
→ w = 0 J
However, this is incorrect because the given heat evolved is -3 J and we are given options. Let's recalculate with given heat and work done:
For the process A to B:
ΔU = q + w = 5 J - 8 J = -3 J
For the process B to A:
q = -3 J (heat evolved)
ΔU = -3 J (since internal energy is a state function)
ΔU = q + w
-3 J = -3 J + w
w = 0 J
This calculation suggests that no work is done in the reverse process. However, this is not consistent with the given options. There appears to be an error in the problem statement or the provided options. Let's assume there's a mistake and recalculate using the options to find which option satisfies the first law of thermodynamics.
Option A: 6 J work done by surroundings on gas (w = +6 J)
ΔU = q + w = -3 J + 6 J = 3 J (Incorrect)
Option B: 10 J work done by gas (w = -10 J)
ΔU = q + w = -3 J - 10 J = -13 J (Incorrect)
Option C: 6 J work done by gas (w = -6 J)
ΔU = q + w = -3 J - 6 J = -9 J (Incorrect)
Option D: 10 J work done by surroundings on gas (w = +10 J)
ΔU = q + w = -3 J + 10 J = 7 J (Incorrect)
None of the options are consistent with the first law of thermodynamics and the given data. There must be an error in the question or options provided.