Question

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)
A. 1400 cal
B. 1000 cal
C. 900 cal
D. 1200 cal

Answer 1

Correct Answer - D
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)`