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In above diagram it can be seen that the curve for adiabatic compression is more steeper than curve for adiabatic expansion. The difference between p1 and p4 is more than p2 and p3 - it means that the weight lifted from the piston during adiabatic expansion is less than the weight added to the piston in adiabatic compression. This implies that the change in internal energy in over all Carnot cycle is greater than zero since the internal energy which decreased in adiabatic expansion is less than the energy gained during adiabatic compression. Kindly tell me where I am wrong.

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    $\begingroup$ Why aren't you considering anything happening on the other two curves? Internal energy is a state variable, so going around the loop should not change the internal energy. $\endgroup$ Jul 26, 2018 at 12:06
  • $\begingroup$ The curve between 3 and 4 is qualitatively incorrect. There can't be a minimum in the pressure for the isothermal change. The curve is a hyperbola, PV = const. Maybe that's why you misjudged by eyeball the equal amounts of work for the adiabatic steps. $\endgroup$ Jul 26, 2018 at 16:12

1 Answer 1


The fact that the difference between $p_1$ and $p_4$ is greater than the difference between $p_2$ and $p_3$ does not actually imply that the increase of the internal energy during the adiabatic compression is greater than the decrease of the internal energy during the adiabatic expansion. They are the same, since the internal energy of an ideal gas is determined by its temperature and the temperature difference is the same in both cases.

Your conclusion is based on the assumption that the work is determined by the pressure only, but the work is also determined by the volume change, which is greater during the adiabatic expansion than during the adiabatic compression.

  • $\begingroup$ I understand now that change in internal energy will be same because change in temperature difference is same. During adiabatic compression the pressure difference is large but volume difference is smaller because at higher values large change in pressure causes small change in volume. $\endgroup$
    – hood
    Jul 27, 2018 at 5:50
  • $\begingroup$ @hood You've got it! $\endgroup$
    – V.F.
    Jul 27, 2018 at 10:11

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