A thermodynamic system is taken from an initial state i with internal energy Ui = 100 J 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 paths af, ib and bf are Waf = 200 J, Wib = 50 J and Wbf = 100 J respectively. The heat supplied to the system along the path iaf, ib and bf are Qiaf, Qib and Qbf respectively. If the internal energy of the system in the state b is Ub = 200 J and Qiaf = 500 J, the ratio Qbf / Qib is :

Engineering

Physics

Thermodynamics

Cyclic Processes

Question

A thermodynamic system is taken from an initial state i with internal energy U_i = 100 J 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 paths af, ib and bf are W_af = 200 J, W_ib = 50 J and W_bf = 100 J respectively. The heat supplied to the system along the path iaf, ib and bf are Q_iaf, Q_ib and Q_bf respectively. If the internal energy of the system in the state b is U_b = 200 J and Q_iaf = 500 J, the ratio Q_bf / Q_ib is :

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500 = U_f – 100 + 200 + 0

U_f = 400 J

Q_ib = W + $∆$ U = 50 + (200 – 100) = 150 J

Q_bf = W_bf + 400 – 200

= 100 + 200 = 300

$\frac{Q_{bf}}{Q_{ib}} = 2$