Understanding the PV^(gamma) = C Equation in Adiabatic Processes and Ideal Gases

[mune]

Hi guys,I have some questions about this equation:

P.V^(gamma) = constant, gamma = C_p / C_v

Is it only valid for a adiabatic process, plus ideal gas? I thought it was at the first place, as I saw its derivation uses adiabatic properties(dQ=0) and assumes the gas is ideal (PV=mRT).
But when I read my textbook, it doesn't mention anything about adiabatic. It says those processes follow:

P.V ^ n = constant (n is another constant)

are called 'polytropic'. If the gas is ideal, then the process is isothermal.

I become more confuse when I saw 'isothermal', as I thought it should be adiabatic.

And, is the equation only valid for a closed system?

Last question, is 'TV^(gamma - 1) = constant' valid only for adiabatic process and ideal gas?

I like thermodynamics, but I find some parts of it quite confusing, hope someone can clear my doubts.

thank you.

 

[siddharth]

Hi guys,I have some questions about this equation:

P.V^(gamma) = constant, gamma = C_p / C_v

Is it only valid for a adiabatic process, plus ideal gas? I thought it was at the first place, as I saw its derivation uses adiabatic properties(dQ=0) and assumes the gas is ideal (PV=mRT).
But when I read my textbook, it doesn't mention anything about adiabatic. It says those processes follow:

P.V ^ n = constant (n is another constant)

are called 'polytropic'.

These are two different process. In an adiabatic reversible process for an ideal gas, you'll get $$PV^{\gamma}=c$$ where $$\gamma$$ is the ratio of C_p and C_v.

A polytropic process is generally different from an adiabatic process, where n can take any value depending on the system.

If the gas is ideal, then the process is isothermal.

I become more confuse when I saw 'isothermal', as I thought it should be adiabatic.

I think that's wrong. When the value of n is 1, only then will the reversible polytropic process be isothermal.

And, is the equation only valid for a closed system?

Yes, most of the equations you use are derived for a closed system. For an open system, you'd use the general form of the first law,
$$dU = TdS - PdV + \mu dn$$
where $$\mu$$ is the chemical potential.

Last question, is 'TV^(gamma - 1) = constant' valid only for adiabatic process and ideal gas?

Yes. Can you derive this from PV^gamma is constant? Hint: Use the ideal gas law to substitute for P.

I like thermodynamics, but I find some parts of it quite confusing, hope someone can clear my doubts.
thank you.

Hope this helped. If you have any further questions, feel free to ask. Thermodynamics can indeed be quite interesting

 

[Andrew Mason]

I think that's wrong. When the value of n is 1, only then will the reversible polytropic process be isothermal.

It need not be reversible. Not every isothermal process is reversible but, for an ideal gas, PV=constant where T is constant.

AM

 

[siddharth]

It need not be reversible. Not every isothermal process is reversible but, for an ideal gas, PV=constant where T is constant.

Yeah, that's true. However, for a process which is irreversible and not in equilibrium, the idea of a macroscopic pressure and temperature don't hold, do they?

 

[olgranpappy]

Yeah, that's true. However, for a process which is irreversible and not in equilibrium, the idea of a macroscopic pressure and temperature don't hold, do they?

As long as the initial state and the final state are macroscopic equilibrium states then it can certainly be said that they possesses an internal Energy (U) and, for an ideal gas this obeys U=3NT/2 where T is the temperature and N is the number of particles. So that the initial and final states both have some well-defined temperature... even thought there might not be any meaningful temperature for the in-between non-equilibrium.

 

[rbj]

As long as the initial state and the final state are macroscopic equilibrium states then it can certainly be said that they possesses an internal Energy (U) and, for an ideal gas this obeys U=3NT/2 where T is the temperature and N is the number of particles.

that ideal gas equation (we are assuming that units are chosen so that Boltzmann = 1) is also only valid for monatomic particles (like inert gasses) with 3 degrees of freedom (the x, y, z axis). if you have diatomic particles (like N₂ and O₂, together comprising 99% of air), the idealized equation for mean energy is 5/2 T per particle. this difference is important in getting the theoretical speed of sound in normal dry air to agree with experiment.

 

[olgranpappy]

that ideal gas equation (we are assuming that units are chosen so that Boltzmann = 1) is also only valid for monatomic particles (like inert gasses) with 3 degrees of freedom (the x, y, z axis). if you have diatomic particles (like N₂ and O₂, together comprising 99% of air), the idealized equation for mean energy is 5/2 T per particle. this difference is important in getting the theoretical speed of sound in normal dry air to agree with experiment.

true.

 

[mune]

thanks siddharth. You are right, for the process to be isothermal, n has to be equal to 1. I checked my textbook again, and I realized I missed that part last time.

Today I just learned that PV^n = constant is also called path equation, all process in thermodynamics can be represent by this equation. (Correct me if I am wrong.)

Thanks to others also, although I don't understand what some of you said. lol.

 

[Andrew Mason]

Yeah, that's true. However, for a process which is irreversible and not in equilibrium, the idea of a macroscopic pressure and temperature don't hold, do they?

A reversible process is, by definition, one in which the system is in constant equilibrium (or, to be more precise, one in which equilibrium can be restored by an infinitessimal change in conditions). Temperature is only defined for a system in equilibrium. So, you make a good point. A true isothermal process, in which temperature is defined and constant at all times during the process, must be reversible.

AM

 

[bananenplukker]

v^(gamma)P=c

heey, can you explain how to come from Vftf^f/2 = ViTi^f/2 to VT^f/2 and then to

PV^gamma = c

thanks