A perfect gas goes from state A to another state B by absorbing 8 × 10^5 J of heat and doing 6.5 × 10^5 J of external work. It is now transferred between the same two states in another process in which it absorbs 10^5 J of heat. Then in the second process

Q: A perfect gas goes from state A to another state B by absorbing 8 × 105 J of heat and doing 6.5 × 105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. Then in the second process

(a) Work done on the gas is 0.5 × 105 J

(b) Work done by gas is 0.5 × 105 J

(c) Work done on gas is 105 J

(d) Work done by gas is 105 J

Ans: (a)

Sol: In Process First:

Applying First Law of Thermodynamics ,

∆Q = ∆U + ∆W

8 × 105 = ∆U + 6.5 × 105

∆U = 1.5 × 105 J

In Process Second :

Applying First Law of Thermodynamics ,

∆Q = ∆U + ∆W

10^5 = 1.5 × ∆W

⇒ ∆W = – 0.5 × 105 j