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`

Answer

`(i) f= 5 (ii) W = 12.3 PV`

Explanation

For polytropic process `T_(1)V_(1)^(n-1) = T_(2)V_(2)^(n-1) implies TV^(n-1) = (T)/(2)(5.66V)^(n-1) implies 2 = 5.66^(n-1)`
Taking log both sides `ln2 = (n-1)ln 5.66`
`implies n= 1.4 = 1 + (2)/(f) implies f=5`
(i) `therefore` Degrees of freedom =`5`
(ii) Work done by gas =`(P_(1)V_(1) - P_(2)V_(2))/(gamma-1) = (PV-P_(2)(5.66V))/(1.4-1) = 12.3PV`
Where `P_(2)V_(2)^(gamma) = P_(1)V_(1)^(gamma) implies P_(2) = P((1)/(5.66))^(1//4)`