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',
- 18 J of the work is done by the gas 'A'.
- 2 J of the work is done by the gas 'A'.
- 12 J of the work is done on the gas 'A' by the surrounding.
- 14 J of the work is done on the gas 'A' by the surrounding.
Answer
(4) 14 J of the work is done on the gas 'A' by the surrounding.
Solution
First law (NCERT sign convention): U = q + w, where w is the work done on the system.
X → Y: q = +10 J, w = −18 J (work done by the gas)
⇒ UX Y = 10 - 18 = -8 J
∵ U is a state function ⇒ UY X = +8 J
Y → X: q = −6 J (heat evolved)
⇒ w = U - q = 8 - (-6) = +14 J
∴ 14 J of work is done on the gas by the surroundings.
Answer: (4)