When a system is taken from state I to state f along thepath iaf, it is found that Q = 50 cal and W = 20 cal. Along the path ibf, Q = 36 cal. W along the path ibf is

When a system is taken from state I to state f along thepath iaf, it i

1.	When a system is taken from state I to state f along thepath iaf, it is found that Q = 50 cal and W = 20 cal. Along the path ibf, Q = 36 cal. W along the path ibf is
Answer» 14 CAL 6 cal 16 cal 66 cal Solution :According to first law of thermodynamics, For the path iaf, `Q_(iaf) = DeltaU_(iaf) + W_(iaf)` or `Delta U_(iaf) = Q_(iaf) - W_(iaf) = 50 - 20 = 30 cal` For the path IBF, `Q_(ibf) = Delta U_(ibf) + W_(ibf)` As change in INTERNAL energy is path independent, so `Delta U_(iaf) = Delta U_(ibf) :. Q_(ibf) = Delta U_(iaf) + W_(ibf)` or `W_(ibf) = Q_(ibf) - Delta U_(iaf) = 36-30 = 6 cal`