## Problem 8: A system absorbs 1176 J of heat and at the same time does 352.8 J of external work. Find the change in internal energy of the system. Find the change in internal energy of the system when it absorbs 1050 J of heat while 84 J of work is done. What will be the change in internal energy of the gas if 210 J of heat is removed at constant volume?

Solution

Solution (a):       Heat absorbed, ΔQ = 1176 J         Work done, ΔW = 352.8 J

Change in internal energy, Δu =?

Apply first law of thermodynamics, ΔQ = Δu + ΔW, we have, 1176 = Δu + 352.8

⇒ Δu = 1176 – 352.8 ⇒ Δu = 823.2 J

Solution (b):      Heat absorbed, ΔQ = 1050 J         Work done, ΔW = 84 J

Change in internal energy, Δu =?

Apply first law of thermodynamics, ΔQ = Δu + ΔW, we have, 1050 = Δu + 84

⇒ Δu = 1050 – 84 ⇒ Δu = 966 J

Solution (c): Heat removed, ΔQ = -210 J                                Work done, ΔW = 0 (because volume is constant)

Change in internal energy, Δu =?

Apply first law of thermodynamics, ΔQ = Δu + ΔW, we have, -210 = 0 + Δu ⇒ Δu = -210 J