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

B

Explanation

Applying first law of thermodynamics to path iaf
`Q_(iaf)=DeltaU_(iaf)+W_(iaf)`
`500=DeltaU_(iaf)+200 :. DeltaU_(iaf)=300J`
Now, `Q_(ibf)=DeltaU_(ibf)+W_(ib)+W_(bf)`
`=300+50+100`
`Q_(ib)+Q_(bf)=450J….(1)`
Also `Q_(ib)=DeltaU_(ib)+DeltaW_(ib)`
`:. Q_(ib)=100+50=150J....(2)`
From (1) & (2) `(Q_(bf))/(Q_(ib))=300/150=2`