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Question

Two moles of an ideal gas heated at constant pressure of one atmosphere from 27oC to 127oC. If Cv,m=(20+102 T) JK1 mol1, then q and U for the process are respectively:


A
6362.8 J, 4700 J
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B
3037.2 J, 4700 J
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C
7062.8 J, 5400 J
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D
3181.4 J, 2350 J
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Solution

The correct option is A 6362.8 J, 4700 J
U (internal energy) is given as:
dU=nCvdT
integrating both sides we get
U=n400300(20+102T)dT=2[20×100+1022(40023002)]=4700J

we know at constant pressure heat absorbed is given as:
dq=nCpdT
and Cp=R+Cv
so, dq=n(Cv+R)dT

integrating both sides we get :
q=n400300(R+20+102T)dT=2[(R×100)+(20×100)+1022(40023002)]=6362.8J

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