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

Options

A. 1
B. 2
C. 3
D. 4

Answer

B

Explanation

As`Q_(iaf)= 500J and 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= 400J`
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= 300J`
`(Q_(bf))/(Q_(ib))=(300)/(150)=2`