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))`

Answer

A, B, C

Explanation

dW=dU
`nC_V DeltaT=3xx10^3`
`1xx20xxDeltaT=3xx10^3`
`DeltaT=3000/20=300-T_2` (Since after expansion the temperature will be lowered)
`therefore T_2=-300/2 + 300=150K=-123^@C`