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2CO+O22CO2;H=560kJ
Two moles of CO and one mole of O2 are taken in a container of volume 1 L. They completely form two moles of CO2, the gases deviated appreciably from ideal behavior. If the pressure in the vessel change from 70 to 40 atm, find the magnitude (absolute value) of U at 500 K. (1 L atm = 0.1 kJ)

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Solution

q=ΔH=560KJ
w=ΔPV
=(7040)×1
=30L atm=30×01KJ
=3KJ
ΔU=q+w
=(560+3)KJ
=563KJ

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