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

A, B, D

Explanation

(a) For adiabatic expansion,
`TV^(gamma-1)=` constant
i.e.
`TV^(gamma-1)=T^'V^('gamma-1)`
`=T/2(5.66V)^(gamma-1)`
`i.e. `(5.66)^(gamma-1)=2`
i.e. `gamma=1.4`
Using `gamma+1=2/F`
We get degree of freedom.
`F=5`
(b) Work done during adiabatic process for one mole gas is
`W=1/(1-gamma)[p^'V^'-pV]`
From relation, `(pV)/(T)=(p^'V')/(T')`
We get `p'=(T')/(T)*(pV)/(V')=1/2xx1/5.66p=(p)/(11.32)`
`:. W=(1)/(1-1.4)[(p)/(11.32)xx(V)/(5.66)-pV]`
`=1/0.4[1-(1)/(11.32xx5.66)]pV`
`=2.461pV`