Two moles of an ideal gas is heated at constant pressure of one atmosphere from 27oCto127oC. If Cv,m=20+10−2TJK−1mol−1, 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=−1662.8J
Also we know, △U=n∫400300Cv,mdT=2×∫400300(20+10−2T)dT
putting the values we get, =2[20×100+10−22(4002−3002)]=4700J
from first law of thermodynamics we know ΔU=q+w
Putting the values, we get ; 4700=q−1662.8 ∴q=6362.8J.