One mole of an ideal gas is allowed to expand reversible and adiabatically from a temperatureof`27^(@)C)`if the work done during the process is`3`kJ,the final temperature will be equal to`(C_(v)=20JK^(-1))`

Options

A. 100K
B. 150K
C. 195K
D. 255K

Answer

B

Explanation

Given number of moles = 1
Initial temperature = `27^(@)C = 300K`
Work done by the system = 3 kJ = 3000J
It will be (-) because work is done by the system. Heat capacity at constant volume `(C_(V)) = 20` J/K
We know that work done
`W=-nC_(v) (T_(2)- T_(1)), 3000= -1 xx 20 (T_(2)- 300)`
`3000= - 20T_(2)+ 6000`
`20T_(2)= 3000, T_(2)= (3000)/(20)= 150K`