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Question

Two moles of an ideal gas is heated at constant pressure of one atmosphere from 27oC to 127oC. If Cv,m=20+102T 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, 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
We have to calculate q, heat given and work.
Work here will be w=nR T
putting the values we get
=2×8.314×100=1662.8 J

Also we know,
U=n400300Cv,mdT=2×400300(20+102T)dT
putting the values we get,
=2[20×100+1022(40023002)]=4700 J

from first law of thermodynamics we know
ΔU=q+w
Putting the values, we get ;
4700=q1662.8
q=6362.8 J.

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