The figure here shows five paths traversed by a gas on a P-V diagram. Rank the paths according to the change in internal energy of the gas, greatest first.

5 > 4 = 3 > 2 = 1

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5 > 4 = 3 > 1 = 2

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4 = 3 > 5 > 1 = 2

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5 > 4 = 3 = 2 = 1

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Solution

The correct option is D
5 > 4 = 3 = 2 = 1

For any ideal gas, the internal energy can be related to the temperature as -

$E_{i n t} = n C_{v} T$

$\Rightarrow Δ E_{i n t} = n C_{v} Δ T$ (for a change of state).

Thus the change in internal energy is solely dictated by the change in temperature.Since, all the processes 1, 2, 3, and 4 begin and end at the same temperatures, their $Δ T^{'} s (a n d, Δ E_{i n t} = n C_{v} (T_{2} - T_{1}))$ are the same. Process 5, on the other hand goes through a larger $Δ T (T_{3} - T_{1})$ , hence gains the highest internal energy.

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