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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 − 21−γ = (γ − 1) ln2.
Solution
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 − 21−γ
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