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

`Q_(iaf) = W_(iaf) + DeltaU_(iaf)`
`500 = 200 + DeltaU_(iaf)`
`U_(iaf) = 300`
`U_(f) = 400`
` Q_(fb) = W_(fb) + DeltaU_(fb)`
=`50 + 100`
`Q_(ib) = 150`
`Q_(bf) = W_(bf) + DeltaU_(bf)`
=`100 + 200`
`Q_(bf) = 300`
`(Q_(bf))/(Q_(fb)) = (300)/(150) = 2` ans