Two samples A and B of the same gas have equal volumes and pressures . The gas is sample A is expanded isothermally to double its volume and the the gas in B is expanded adibatically to double its volume . If the work done by the gas is the same for the two cases, show that gamma satisfies the equation`(1-2 ^(1-gamma)= (gamma - 1 ) 1n2`.

Explanation

`P_1` = Initial pressure,
`V_1` = Initial volume ,
`P_2` =Final pressure,
`V_2`= Final volume
Given ,`V_2 = 2V_1`
Isothermal work done
`=nRT_1l n (V_2)/(V_1)`
Adiabatic work done
`=(P_1V_1-P_2V_2)/(gamma-1)`
Given that work done at both clases are same.
Hence,
`nRT_1l n = V_2/V_1`
`=(P_1V_1-P_2V_2)/(gamma-1)` ...(1)
At adiabatic process,
`P_2 = P_1(V_2/V_1)^gamma=P_1(1/2)^gamma`
From the equation (1)
`nRT_1l n2=(P_1V_1(1-1/2^(gamma)2))/(gamma-1)`
and `nRT_1 = P_1V_1`
So `l n2 = (1-1/2^(gamma)2)/(gamma-1)`
or `(gamma-1)l n2=1-2^(1-gamma)`