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Question

Find the change in the internal energy of the system in each of situation in Column I.
Given J=4.184
Column IColumn IIi. A system absorbs 500 cal of heat anda.5020 Jat the same time does 400 J of workii. A system absorbs 300 cal and at theb.1692 Jsame time 420 J of work is done on itiii. Twelve hundred calories heat is removedc.1675 Jfrom a gas held at constant volume

A
i-b, ii-c, iii-a
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B
i-c, ii-b, iii-a
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C
i-b, ii-a, iii-a
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D
i-b, ii-a, iii-c
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Solution

The correct option is A i-b, ii-c, iii-a
ib;iic;iiia
i.dQ=dU+dw
Here
dQ=+500×4.184J=2092 J
dW=+400 J. Since the work is done by the system,
dU=dQdW=2092400=1692 J
ii.ΔU=ΔQΔW=(300)(4.184)(420)=1675 J (approx).
iii.ΔU=(ΔQΔW)=(1200)(4.184)0=5020 J
(Note that ΔQ is positive when heat is added to the system and ΔW is positive when the system does the work)

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