**JEE Advanced 2014 Paper 1, Question 20**

A thermodynamic system is taken from an initial state with internal energy to the final state along two different paths and , as schematically shown in the figure. The work done by the system along the paths and are and respectively. The heat supplied to the system along the path and are and respectively. If the internal energy of the system in the state is and , the ratio is

**Solution**

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)

where is the work done *by* the system. We apply this first to the path . Note that no work is done along since volume is constant, therefore . However, the internal energy at can be different from that at (for example, for an ideal gas undergoing an isochoric/constant-volume process, which means its internal energy changes). Thus,

(2)

Similarly, along the path ,

(3)

Finally, along

(4)

Therefore, the ratio .