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+1 vote
1.4k views
in Chemical thermodynamics by (64.7k points)

For the reaction, 2CO + O2 → 2CO2 ; ∆H= -560 kJ. 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 deviate appreciably from ideal behaviour. If the pressure in the vessel changes 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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2 Answers

+2 votes
by (58.3k points)

∆H =∆U +∆U +∆(pV) = ∆U +V∆p

⇒∆U =∆H − ∆Vp 

= –560 – 1× 30 × 0.1 = –563 kJ

0 votes
by (90 points)

∆H =∆U +∆U +∆(pV) = ∆U +V∆p

⇒∆U =∆H − ∆Vp = –560 – 1× 30 × 0.1

= –563 kJ This is the required answer.

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