A thermodynamic system is taken from an initial state I with internal energy `U_i=-100J` 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 the pat af, ib and bf are `W_(af)=200J, W_(ib)=50J and W_(bf)=100J` respectively. The heat supplied to the system along the path iaf, ib and bf are `Q_(iaf), Q_(ib),Q_(bf)` respectively. If the internal energy of the system in the state b is `U_b=200J and Q_(iaf)=500J`, The ratio `(Q_(bf))/(Q_(ib))` 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`