Two moles of a gas is expanded to double its volume by two different processes. One is isobaric and the other is isothermal. If `W_(1)` and `W_(2)` are the works done respectively them

Options

A. `W_(2) = (W_(1))/(loge^(2))`
B. `W_(2) = W_(1)`
C. `W_(2) = w_(1) loge^(2)`
D. `W_(2) = W_(1) loge (2)`

Answer

d

Explanation

Given
In the first condition, work done = `W_(1)`
In the second condition, work done = `W_(2)`
Now accroding to the question,
In the first condition,
`W_(1) = p(2V - v)`
or , `W_(1) pV`
In the second condition,
`W_(2) = hRT log_(e) ((V_(2))/(V_(1)))`
or `W_(2) = pV log_(e) ((2V)/(V))` `[:' pV = nRT]`
From Eq. (i),
`W_(2) = W_(1) log_(4) (2)`