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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Mar 2008 18:05:00 IST
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Q1)
For process 1:
V is constant and P is doubled => Temp is doubled
So Q = U = nCvT
For process 2:
P is constant and V is doubled => Temp is again doubled
So Q = nCp2T ( As initially temp was 2T(after process 1) and so now it is 4T)
So net Q = nCvT + 2nCpT
And net change in Temp is 3T
nC3T = nCvT + 2nCpT
Cancelling and putting value of Cp and Cv for diatomic gas,we get
C = 19R/6 Option B
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