Question

A system of idea gas is changed from state A to state B by the following sequence of steps:
Step 1: The system absorbs 5.25 kJ of heat and expands against a constant external pressure of 1.50 atm from 25 L to 125L.
Step 2: The system loses 2.10 kJ of heat to the surroundings at constant volume.
Step 3: The surroundings do 0.75 kJ of work on the system and this energy is lost as heat.
What is the change in Internal Energy, for each step and for the overall change from State A to State B?

Answer 1

Change in Internal Energy(delta U) = Q(heat energy) + W(work)
Work = - P x delta V (change in volume)

Step 1. Work = - 1.50 atm · (125-25) = -150 L·atm
Since 1 J = 0.00987 L·atm
then Work = -150/0.00987 = - 15198 J
Change in Internal Energy(delta U) = 5250 + (-15198)=-9948 J

Step 2. Change in Internal Energy(delta U) = -750 + 0=-750 J

Step 3. Change in Internal Energy(delta U) = -750 + 750= 0 J

Overall: -9948 + (-750) + 0 = -10698 J