1.

In adiabatic conditions 1 mole of CO_(2) gas at 300 K is expanded such that its volume becomes 27 times. Calculate the work done. (C_(V)=6" cal "mol^(-1) and gamma=1.33 is given)

Answer»

1400 cal
1000 cal
900 cal
1200 cal

Solution :In adiabatic conditions `(T_(2))/(T_(1))=((V_(1))/(V_(2)))^(gamma-1)`
`(T_(2))/(T_(1))=((1)/(27))^(1.33-1)=((1)/(27))^(0.33)=((1)/(27))^(1//3)=(1)/(3)`
`T_(2)=300xx(1)/(3)=100K`
Thus, `T_(2) lt T_(1)` hence cooling TAKES place due to EXPANSION under adiabatic CONDITION
`DeltaE=q+W`
`DeltaEneW`(q=0 for adiabatic change)
`DeltaE=-ve` because gas exapands.
`W=-DeltaE=-C_(V)(T_(2)-T_(1))`
`=-6(100-300)=1200" cal "mol^(-1)`


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