Usually, when I see a graph of reversible adiabatic and isothermal processes it's something like this.adiab isotherm curves Which makes sense mathematically since $yx^{\gamma} = k_1$ and $yx = k_2$ should intersect only once.

However, let's say if we have a system at $(P_1, V_1)$, then surely we can take it to $(P_2, V_2)$ adiabatically let's say in a thermoflask. We should also be able to take it from $(P_1, V_1)$ to $(P_2, V_2)$ isothermally.

So, my questions are:

  • How does that work mathematically? If the two graphs never intersect how can they reach the same position?
  • How would the $P-V$ graph look?
  • Does this have something to do with the temperature acting as a third dimension?
  • $\begingroup$ To get from the same initial state to the same final state, one of the process paths cannot be just isothermal or just adiabatic. E.g. for the curves on the chart,both curves may rejoin by isochoric heating or cooling. $\endgroup$
    – Poutnik
    Apr 27 at 5:57
  • $\begingroup$ The state doesn't need to be the same. Only the pressure and volume need to be the same. Temperature can vary, well it can vary for adiabatic not for isothermal of course. Would it be possible then that it happens in one single process? $\endgroup$
    – dotmashrc
    Apr 27 at 6:24
  • 1
    $\begingroup$ If P and V are the same, then T is the same. Your very basic premise is false: you can't get from P1,V1 to P2,V2 adiabatically. Instead, you get to P2 and some V that you don't get to choose. The word "adiabatic" sets the direction of the process, much like a highway. So does "isothermal", and that's a different highway. Interstate 10 from Houston will take you to San Antonio and El Paso. Can you get to Dallas? No. Or rather, yes, but not by using highway 10. Same thing here. $\endgroup$ Apr 27 at 6:35
  • $\begingroup$ Who says you can take it back from P2,V2 to P1,V1 isothermally? $\endgroup$ Apr 27 at 11:07

1 Answer 1


However, let's say if we have a system at $(P_1,V_1)$, then surely we can take it to $(P_2,V_2)$ adiabatically let's say in a thermoflask. We should also be able to take it from $(P_1,V_1)$ to $(P_2,V_2)$ isothermally.

The first sentence is correct. However, the second sentence violates the 2nd law! The latter is also in contradiction with your previous statement:

mathematically since $yx^γ=k_1$ and $yx=k_2$ should intersect only once.

This is more than mathematics, it has to do with a principal law of thermodynamics. Lets suppose that you are able to do it, and find that it contradicts the law. I illustrate the situation in the following figure, where the isothermal evolution crosses with the adiabatic evolution twice, and the adiabatic coefficient is $\gamma = 1.4$ (note that the isothermal evolution suffers an unrealistic twist near State 2, but this is because I am trying to nevertheless unite the paths):

enter image description here

  1. Adiabatic evolution: in absence of heat transfer between the system and the surroundings, and given that due to the expansion the temperature is decreasing, we have $$ Q_{12} = 0 \hspace{1 cm} W_{12} = \Delta U < 0 \tag{1,2} $$
  2. Isothermal evolution: if we assume an ideal gas behavior, then the change of internal energy is zero since $U=U(T)$. The gas is being compressed, so there is work done by the surroundings to the system, and the same amount of heat transfer from the system to the surroundings $$ W_{23} = -\int_{V_2}^{V_1} \; P \; dV > 0 \hspace{1 cm} Q_{23} = -W_{23} < 0 \tag{3,4} $$

By the $PV$ diagram, the work is the area below the curve of the particular evolution. If we sum them we clearly have that $W_{12} + W_{23} > 0$. However, there is only one heat exchange, and it happens in the isothermal process with $Q_{23} < 0$.

Conclusion: this is a cyclic device, that is capable of converting the heat rejected by the system completely into work done by the surroundings. This the inverse of the Kelvin-Planck's statement of the 2nd law. The process is impossible.

How would the P−V graph look? Does this have something to do with the temperature acting as a third dimension?

We have demonstrated that it impossible to go back by an isothermal path. We need something more, and there are many ways to accomplish this. I will use the proposal of Poutnik: (1) adiabatic cooling, (2) isochoric heating, (3) isothermal compression. We illustrate as follows:

enter image description here

  • Now the isothermal evolution behaves correctly, look how it cannot collide with the adiabatic evolution.
  • The work is the same as before, since an isochoric evolution has no work transfer. However, now we are absorbing heat from the surroundings in the isochoric evolution, and thereby we are not violating the 2nd law.
  • $\begingroup$ I have another question kind of related to this which is also quite silly. So, if we go from one state to another adiabatically while pressure is constant then we know $TV^{\gamma - 1} = const.$ but also since pressure is constant then $\frac{V}{T} = const.$ Am I missing something here? Surely they can't be related by both of those equations? $\endgroup$
    – dotmashrc
    Apr 27 at 20:22
  • $\begingroup$ @dotmashrc Hello dotmashrc. Are you asking if you can go from state $1$ to state $2$ adiabatically and at constant pressure, at the same time? $\endgroup$ Apr 27 at 20:26
  • $\begingroup$ Yeah pretty much. Or if a process is isobaric can it also be adiabatic? $\endgroup$
    – dotmashrc
    Apr 28 at 6:15
  • $\begingroup$ @dotmashrc As a rule, when you demand two constraints in ideal gases regarding this type of processes, you don't "move". If it is adiabatically reversible $P_2^{(1 - \gamma )/\gamma}T_2 = P_1^{(1 - \gamma )/\gamma}T_1$, but if $P_2=P_1$, then $T_2 = T_1$. Also, this means that $V_2=V_1$. Thereby if you ask for those two things, there is no evolution. $\endgroup$ Apr 28 at 9:53

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