I'm reading an introduction into heat pump cycles right now, and have question about the compression part:

The optimum compression would be an adiabatic one, after which you end at a certain temperature T(1) and pressure p(1). If the compression is non-ideal you still want to reach pressure p(1) but now the textbook says that we have a new temperature T(2) which is higher than T(1). I dont get how this can happen?

Following that logic a part of the compression work would lead to an isobaric compression, and increase the temperature only. But how can the pressure stay constant while I increase temperature? If I want to keep the pressure constant while decreasing the volume, I need an equal amount of heat exchange with the surroundings- again cooling the gas itself?

kind regards

  • $\begingroup$ Why don't you just derive the equations for the two situations (i.e., model them) and see how it plays out? $\endgroup$ Jun 22, 2020 at 15:20
  • $\begingroup$ i would if i knew how. not studying physics, Im just interested in the topic. I have calculated that for an adiabatic compression T(1) / T(0) = p(1) / p(0) but no idea how to calculate a combined sometimes isobaric- sometimes not isobaric equation $\endgroup$
    – racctor
    Jun 22, 2020 at 16:12
  • $\begingroup$ your equation is incorrect for either reversible or irreversible compressions $\endgroup$ Jun 23, 2020 at 0:37

1 Answer 1


OK. Here's how it plays out. In the non-ideal (irreversible) adiabatic compression, you start at with the gas at the same p(0) as for the ideal compression. There is an external pressure p(0) balancing this gas pressure so that the system is initially in equilibrium. Then, at time zero, you suddenly increase the external pressure from p(0) to p(1) and hold it at this new higher value for the entire compression. The higher externally applied pressure causes the gas to compress very rapidly until the volume decreases to a new volume v(2). This new value of the volume is a little higher than the final volume in the ideal compression (even though the final pressure is the same) because the final temperature is a little higher. Basically, in the non-ideal case, you need to do more work on the gas to compress it than in the ideal case. The imposed pressure being higher throughout the compression than in the ideal case offsets the lower volume change, and more work is done on the gas. So the final gas temperature is a little higher in the non-ideal case than in the ideal case.

Here is how the math plays out. For the irreversible adiabatic expansion or compression of an ideal gas at the suddenly imposed external pressure p(1) (with the gas having started out at p(0)), the first law of thermodynamics tells us that:

$$\Delta U=-W$$ or, equivalently, $$\Delta U=-p_1(V_1-V_0)$$where U is the internal energy of the gas. For the case of an ideal gas, this equation reduces to: $$nC_v(T_1-T_0)=p_1\left(\frac{nRT_1}{p_1}-\frac{nRT_0}{p_0}\right)$$where n is the number of moles of gas (which cancels from the equation). So the gas temperature changes as a result of it exchanging energy in the form of work, with the surroundings.

Solving this equation for the ratio of the final temperature to the initial temperature yields: $$\left(\frac{T_1}{T_0}\right)=\frac{1+(\gamma-1)(p_1/p_0)}{\gamma}$$

  • $\begingroup$ see the additional math development that I included. $\endgroup$ Jun 22, 2020 at 23:36
  • $\begingroup$ Thanks for your answer. But I still can't figure out where the increased temperature should come from. If we have 10 Joule of work put into the compression, and lets say 8 J are used ideally, while 2 J are dissipation loss to the surrounding. The temperature of the gas still doesnt change, only the surroundings heat up. Yes, I need some additional work to make up for those 2 Joules, but still no additional gas temperature gained. The only example of a pressure drop at constant T that comes to my mind, would be an expansion into vacuum, but I doubt that is the case here. $\endgroup$
    – racctor
    Jun 24, 2020 at 0:45

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