1.

1 mole of CO_(2) gas at 300 K is expanded under adiabatic conditions such that its volume becomes 27 times . What is work done ?(gamma = 1.33 and C_(v) = 6 cal mol^(-1) for CO_(2))

Answer»

900 cal
1000 cal
1200 cal
1400 cal

Solution :for ADIABATIC condition `(T_(2))/(T_(1))=((V_(1))/(v_(2)))^(gamma-1)=(1/27)^(1//3)=1/3`
`T_(2)=300xx1/3 =100K`
thus `T_(2) LT T_(1)` HENCE cooling takes place due to expansion under adiabatic condition.
`W=DeltaE=-C_(v)(T_(2)-T_(1))`
`[DeltaE=(-)` ve , expansion]
`-6(100-300)=1200 cal`


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