Question

A gas does 0.320 kJ of work on its surroundings and absorbs 120 J of heat from the surroundings. Hence, ΔU is
(a) 440 kJ
(b) 200 J
(c) 120.32 J
(d) -200 J

Answer 1

Explanation:

q =+120 J, w = 320 J

∆u = q- w

= 120 +320 = -200 J

Answer 2

Answer:

(d) -200 J

Explanation:

Given,

work done by the system = 0.320 KJ

                                           =320 J

Since, the work is done by the system so work done will negative i.e. -320 J

heat absorb by the system = 120 J

Now, the change in internal energy = work done on the system + heat added to the system

                                                        = -320 J + 120 J

                                                        = -200 J

Hence, the change in internal energy of the system is - 200 J which means 200 J of energy is extracted from the system.