An ideal gas having initial pressure P, volume V and temperature T is allowed to expands adiabatically until its volume becomes 5.66V while its temperature falls to T/2.
(i) How many degrees of freedom do the gas molecules have?
Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V.

Answer

A, B

Explanation

(i) Let `pressure =P, Volume=V and Temperature =T` be
the initial quantities and `Pressure=P,Volume=5.66V`
`Temperature =T//2` be the final quantities.
For adiabatic process
`TV^(gamma-1)=T/2(5.66V)^(gamma-1) rArr 2=(5.66)^(gamma-1)`
Taking log on both sides, `log2=(gamma-1)log5.66`
`rArrgamma=1.4`
But `gamma=1+2/f rArr 1.4=1+2/f`
`rArrf=2/0.4=5`
Thus degrees of freedom of gas molecules=5
(ii) For adiabatic process the pressure-volume relationship
is
`P_1V_1(gamma)=P_2V_2(gamma)`
`rArrP_2=P/((5.66)^(1.4)=P/11.32`
Work done for adiabatic process
`W=(P_1V_1-P_2V_2)/(gamma-1)=((PV-(P/11.32)(5.66V))/(1.4-1)=1.25PV`