Question

Two samples A and B, of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 − 2^1−γ = (γ − 1) ln2.

Answer 1

Let,
Initial pressure of the gas = P₁
Initial volume of the gas = V₁ 
Final pressure of the gas= P₂ 
Final volume of the gas = V₂ 
Given, V₂ = 2 V₁, for each case.
In an isothermal expansion process,
work done = `"n""R""T"_1   "l""n"  "V"_2/"V"_1" `

Adiabatic work done,

`"W" = ("P"_1"V"_1 - "P"_2"V"_2)/ (gamma -1 )`

It is given that same work is done in both cases.
So, 

`"n""R""T"_1   "l""n" ("V"_2/"V"_1) =( "P"_1 "V"_1 - "P"_2"V"_2)/ (gamma -1)` ..(1)

In an adiabatic process,

`"P"_2 = "P"_1 ("V"_1/"V"_2)^gamma ="P"_1(1/2)^gamma`

From eq (1),

`"n""R""T"_1  "l""n" 2 = ("P"_1"V"_1(1-1/2^gamma xx 2)) /(gamma -1)`

and nRT₁ = P₁V₁

So, ln  2 =` (1 - 1/(2^gamma) . 2)/ (gamma -1)`

Or (γ − 1) ln 2 = 1 − 2^1−γ