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?
1
Expert's answer
2016-08-29T12:07:03-0400
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
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