Two samples `A` and `B` of the same gas have equal volumes and pressure. The gas in sample `A` is expanded isothermally to four times of its initial volume and the gas in `B` is expanded adiabatically to double its volume. It work done in isothermal process is twice that of adiabatic process, then show that `gamma` satisfies the equation `1-2^(1-gamma) =(gamma-1)In2`.

Explanation

Work done by gas `A` in isothermal process
`W_(A) = P_(0) V_(0)ln ((4V_(0))/(V_(0))) = 2P_(0)V_(0)ln2`
Work done by gas `B`in adiabatic process
`W_(B) = (P_(i)V_(i)-P_(f)V_(f))/(gamma-1) = (P_(0)V_(0)-P_(2)(2)^(-gamma)2V_(0))/(gamma-1)`
`= (P_(0)V_(0)(1-2^(1-gamma)))/(gamma-1)`
According to question. `W_(A) = 2W_(B)`
`2P_(0)V_(0) ln2 = (2P_(0)V_(0)(1-2^(1-gamma)))/(gamma-1)`
`rArr 1-2^(1-gamma) = (gamma-1) ln2`