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

`Delta Q_("iaf") = Delta U_("iaf")+W_("iaf")rArr 500 =Delta U_("iaf") + 200J`
`Delta U_("iaf") = 300 J" ":." "Delta U_("ibf") = 300 J`
` W_("ibf") = W_("ib") + W_("bf") = 50 + 100 = 150 j`
`Q_("ibf") = W_("ibf") + Delta U_("ibf") = 150 + 300 =450" ":." "Delta Q_("ibf") = 450 J = Q_("jb") = + Q_("bf")`
`U_("jb")=Delta U_("jb") + W_("jb")" ":." "Q_("jb") = 150 J " ":." "450 = Q_("jb") + Q_("bf")rArr 450 = 150 J + Q_("bf")`
`Q_("bf") =300 J rArr 450 = 150 J +Q_("bf")`