A thermodynamic system is taken from an initial state i with internal energy `U_i = 100 J` to the final state f along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along in the paths af, ib and bf are `W_(af) = 200 J, W _(jb) = 50 J` and `W_(bf) = 100 J` respectively. The heat supplied to the system along the path jaf, jb and are `Q_(jaf), Q_(jb)` and `Q_(bf)` respectively. If the internal energy of the system in the state b is `U_b = 200 J` and `Q_(Ja f)= 500 J`, the ratio `Q_(bf)//Q_(jb) ` is.

Answer

2

Explanation

`w_(ibf) = 150J`

`w_(iaf) = 200 J`
`Q_(iaf) = 500 J So U_(iaf) =300 J`
So `U_(f) = 400 J`
`U_(ib) = 100 J`
`Q_(ib) = 100 +50 = 150 J`
`Q_(ibf) = 300 +150 = 450 J`
So the required ratio `(Q_(ibf))/(Q_(ib)) - (450-150)/(150) = 2`