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

Here, `Q_(iaf)= 500 Jand W_(af)= W_(iaf)= 200J`
`:. U_(iaf)= U_(f)-U_(i)= Q_(iaf)-W_(iaf)`
`= 500-200= 300J`
`:. U_(f)= 300+U_(i)= 300+100= 400 J`
Now, `Q_(ib)= U_(ib)+W_(ib)= (U_(b)-U_(i))+W_(ib)`
`=(200-100)+50= 150J`
and `Q_(bf)=U_(bf)+W_(bf)=(U_(f)-U_(b))+W_(bf)`
`=(400-200)+100= 300 J`
`:. (Q_(bf))/(Q_(ib))=(300)/(150)=2`