An ideal gas having initial pressure p, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66V, while its temperature falls to `T//2`.
(a) How many degrees of freedom do the gas molecules have?
(b) Obtain the work done by the gas during the expansion as a function of the initial pressure p and volume V.
Given that `(5.66)^0.4=2`

Explanation

a. For an adiabatic change
`(V.)/V^((gamma-1))=T/(T.)`
Given `V.=5.66V` and `T.=T/2`
`implies(5.66)^((gamma-1))=2`
`(gamma-1)log(5.66)=log(2)`
`impliesgamma=1+(log(2))/(log(5.66))=1+0.3010/0.7528=1+0.4=1.4`
Since `gamma=1.4` the gas is diatomic. For a diatomic gas, the number of degrees of freedom of the molecules =5
b. We knwo that the work done by the gas during adiabatic expansion is given by `w=1/((gamma-1))(PV-P.V.)` (1)
Where pressure P. after expansion is obtained from the relation
`(P.V.)/(T.)=(PV)T`
or `P.=PxxV/(V.)xx(T.)/T=PxxV/(5.66V)xx(T//2)/T=P/11.32`
Putting `gamma=1.4,v.=5.66V` and `P.=P/11.32` in Eq. (1) we have
`w=1/((1.4-1))(PV-P/11.32xx5.66V)`
`=1/0.4(PV-1/2PV)=1.25PV`