Heat transferred along two paths

JEE Advanced 2014 Paper 1, Question 20

A thermodynamic system is taken from an initial state i with internal energy U_{i} = 100 \, {\rm J} to the final state f along two different paths i a f and i b f, as schematically shown in the figure. The work done by the system along the paths a f, i b and b f are W_{a f} = 200 \, {\rm J}, W_{i b} = 50 \, {\rm J} and W_{b f} = 100 \, {\rm J} respectively. The heat supplied to the system along the path i a f, i b and b f are Q_{i a f}, Q_{i b} and Q_{b f} respectively. If the internal energy of the system in the state b is U_{b} = 200 \, {\rm J} and Q_{i a f} = 500 \, {\rm J}, the ratio Q_{b f}/Q_{i b} is

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The problem gives us partial information about internal energies, heat transferred, and work done at various points in the PV diagram. We can complete the picture by using the first law of thermodynamics (see note)

(1)   \begin{equation*}   \Delta U = \Delta Q - W , \end{equation*}

where W is the work done by the system. We apply this first to the path iaf. Note that no work is done along ia since volume is constant, therefore W_{iaf} \equiv W_{af}. However, the internal energy at a can be different from that at b (for example, T \propto P for an ideal gas undergoing an isochoric/constant-volume process, which means its internal energy U(T) changes). Thus,

(2)   \begin{equation*}   U_f - U_i = Q_{iaf} - W_{af}   \implies   % U_f = U_i + Q_{iaf} - W_{af}   % U_f = 100 \text{ J} + 500 \text{ J} - 200 \text{ J}   U_f = 400 \text{ J} . \end{equation*}

Similarly, along the path bf,

(3)   \begin{equation*}   U_f - U_b = Q_{bf} - W_{bf}   \implies   Q_{bf} = 300 \text{ J} . \end{equation*}

Finally, along ib

(4)   \begin{equation*}   U_b - U_i = Q_{ib} - W_{ib}   \implies   Q_{ib} = 150 \text{ J} . \end{equation*}

Therefore, the ratio Q_{bf}/Q_{ib} = 2.

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