A monoatomic ideal gas undergoes a process `ABC`. The heat given to the gas is

Options

A. `7.5 PV`
B. `12.5 PV`
C. `16.5 PV`
D. `20.5 PV`

Answer

C

Explanation

If `T` is temperature of gas

at `A` then `T_(B) = 3T = T_(C)`
in process,`AB, DeltaW = (5PV)/(2)`
and `Delta U = 1/2nR(2T)=3xx2 PV`
so, `DeltaQ = DeltaW + Delta U =8.5PV`
in process `BC DeltaU = 0 as DeltaT = 0`
`DeltaW = 4V.2P = 8PV hence `DeltaQ_(2) = 8 PV`
so, `DeltaQ_(T) = 16.5 PV`.