Q.
6 moles of oxygen, 4 moles of Ar. and 2 moles of water vapour form an ideal gas mixture. The mixture is made to expand adiabatically from an initial temperature 127°C to a final temperature 27°C, decrease of internal energy of the mixture is: (take R=2 cal/mol-K)
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a
3600 calorie
b
4800 calorie
c
5400 calorie
d
6600 calorie
answer is C.
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Detailed Solution
Decrease in internal energy =-∆U=(n1CV1+n2CV2+n3CV3)(T1-T2) =6×52R+4×32R+2×3R100=2700 R=5400 cal
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