In a thermodynamic process helium gas obeys the law `TP^(2//5)` = constant,If temperature of `2` moles of the gas is raised from `T` to `3T`, then

Options

A. heat given to the gas is `9RT`
B. heat given to the is zero
C. increase in internal energy is `6RT`
D. work done by the gas is `-6RT`.

Answer

B, C, D

Explanation

In adiabatic process `DeltaQ=0`
`DeltaU=nc_(v)DeltaT=2((3)/(2)R)(3T-T)=6RT` ltBRgt `DeltaW=-DeltaU=-6RT , DeltaW=-DeltaU=-6RT`