# What is the meaning of $PV^\gamma=$ constant in an adiabatic process?

In an adiabatic process, heat transfer doesn't occur and hence $$\Delta U=-W$$ (an increase in internal energy due to work done on the system or a decrease in internal energy due to work done on the surrounding).

My textbook then says (for monatomic gas) adiabatic process takes the form: $$PV^{5/3} =$$ constant

What does this equation means? From my understanding, "$$PV^{5/3} =$$ constant" pinpoints to a specific Pressure and Volume of the system and implies a specific Temperature of the system. This is pretty much equivalent to $$PV=nRT$$ where a specific Pressure and Volume implies a specific Temperature of the system.

But in an Adiabatic process the Temperature, Pressure and Volume can change (otherwise if nothing changes, what's the point of there being a process? and from $$\Delta U=-W$$ there should be something going on, the temperature should change which will cause a change in the volume of the system [but not a change in pressure since we're assuming the process is quasi-static and thus pressure is constant]) and thus what is the meaning of the equation? Why is this equation specific to adiabatic process?

The equation for adiabtic process you wrote does not fix both volume and pressure...you can pick any pressure and then and only then, the volume is fixed, and it is calculated via formula you wrote. I can give you this example: lets say that we observe isotermic process, that is, with constant temperature, and we know that for this process there is a formula PV=const. That does not mean that both P and V are fixed. Lets say, PV=2. Then, we can write: P=2/V. So, if V=8, for example, P= 2/8=0.25. Same thing for adiabtic process with the difference that the dependence goes with -5/3 and not -1 like here.

• Thanks for the input. But if P and V changes in the isotermic process, then that'd imply work is done on the system/surrounding? And thus ΔU is non-zero? And thus temperature isn't constant? Oct 19, 2018 at 18:46
• System can change its internal energy U through basicaly two processes: doing work and exchange of heat. However, temperature is not the same as heat. Isothermal means for example, if you do the work on the system, you give it enough time to adjust and dissipate heat so it would be a slow process. In this process, you disipate heat just to stay on the same temperature. Also, if the system does the work, then it is given enough time to absorb heat to stay on the same temperature. So, system stays on the same temp and dU is zero, but dW and dQ are not, they are actually the same: dW + dQ = 0 Oct 19, 2018 at 19:20
• Oh right sorry I mixed up your isothermal explanation with the adiabatic process. But I think I'm starting to know where my misunderstanding is, thanks for explaining. Oct 19, 2018 at 19:38
• I used isothermal because I dont know how to write exponents here :-) Oct 19, 2018 at 19:40

You are correct in saying that, in an adiabatic reversible process, the pressure, volume, and temperature are changing in tandem. However, the temperature is changing in such a way that, along the process path, the equation $$PV^{\gamma}=const$$ is always satisfied. And, in an adiabatic process, the parameter $$\gamma$$ is equal to the ratio of the specific heat at constant pressure to the specific heat at constant volume.

When you mentioned a quasi-static process path, I think you were thinking of a stair-step arrangement in which, over each step, the incremental change takes place at constant pressure. This is only an approximation to a perfectly reversible process in which the applied pressure is changing very gradually (with time). However, it is close enough to say that, if you determine the pressure and volume at the beginning and end of each incremental step, to an excellent approximation $$P_iV_i^{\gamma}=P_fV_f^{\gamma}$$.

• Thanks for the input. You mentioned the 'excellent approximation equation' which there's something I'm not so sure about. My textbook mentioned that relationship only applies to when there is a constant temperature (but that can never be the case for an adiabatic process I believe?) Oct 19, 2018 at 19:46
• If your textbook actually said that (which I doubt), then your textbook is wrong. Please provide an exact quote. Oct 19, 2018 at 19:49
• Right, it'd make sense if that's wrong. Maybe I did word it wrongly, my textbook said "at constant temperature" instead of what i said "only applies to constant temperature". I've attached the screenshot of my textbook's explanation on adiabatic process here (I highlighted the related part) imgur.com/a/UiLoDo3 But from my understanding, "at constant temperature" should also mean "only applies when there is a constant temperature"? Oct 19, 2018 at 19:58
• The last statement in the highlighted section is incorrect. They meant to say that the equation applies to an adiabatic change, not to constant temperature. Oct 19, 2018 at 20:03

It means that during an adiabatic process if you monitor both the pressure and the volume then when you calculate the quantity $$PV^\gamma$$ every point you will get the same result. For example, $$PV=nRT$$ can be stated as $$PV=$$ constant for an isothermal process. You are correct that temperature is also changing for an adiabatic process, so you can combine the ideal gas law with $$PV^\gamma=$$ constant to get $$PT^{\gamma/(1-\gamma)}=$$ constant or $$TV^{\gamma-1}=$$ constant.