The volume of one mole of an ideal gas with adiabatic exponent `gamma` is varied according to `V=b/T`, where b is a constant. Find the amount of Heat is absorbed by the gas in this process temperature raised by `DeltaT`

Options

A. `((1-gamma)/(gamma+1))RDeltaT`
B. `R/(gamma-1)DeltaT`
C. `((2-gamma)/(gamma-1))RDeltaT`
D. `(RDeltaT)/(gamma-1)`

Answer

C

Explanation

`V=b/T rArr T=b/V`…(1)
`PV=muRT`
`=(muRb)/V`
`therefore PV^2=muRb`=constant
Now for adiabatic change `PV^x` =constant by comparing it x=2
Now, `c=R/(gamma-1)+R/(1-x)`
`=R/(gamma-1)+R/(1-2)`
`=R/(gamma-1)-R`
`=R[(1-gamma+1)/(gamma-1)]=R[(2-gamma)/(gamma-1)]`
Heat absorbed `DeltaQ=muCDeltaT` here `mu=1` mole
`therefore DeltaQ=RDeltaT((2-gamma)/(gamma-1))`