How to know whether a given process is reversible?

[Pushoam]

Homework Statement

How do we know that an adiabatic process is reversible , while isothermal process is not?
Do we have to calculate the entropy change of the process or we have to check experimentally whether the system is in eqbm at infinitesimal time interval( but this seems impossible)?

Homework Equations

The Attempt at a Solution

 

[DrClaude]

How do we know that an adiabatic process is reversible , while isothermal process is not?

Why do you say that? The Carnot cycle is built up of adiabatic and isothermal steps, all reversible.

Do we have to calculate the entropy change of the process or we have to check experimentally whether the system is in eqbm at infinitesimal time interval( but this seems impossible)?

Reversibility is an idealized concept. In reality, no thermodynamic process is truly reversible, but reversibility is a good approximation for many cases. As far as I know, reversibility is most often stated, rather than inferred. In other words, we want a change to be reversible, and imagine a physical situation where that would be the case. Some processes, of course, are only irreversible.

 

[Chestermiller]

Homework Statement

How do we know that an adiabatic process is reversible , while isothermal process is not?
Do we have to calculate the entropy change of the process or we have to check experimentally whether the system is in eqbm at infinitesimal time interval( but this seems impossible)?

Homework Equations

The Attempt at a Solution

I think your real question is "what are the physical characteristics of an irreversible process compared to a reversible process?" Is that what you are asking?

 

[Pushoam]

I think your real question is "what are the physical characteristics of an irreversible process compared to a reversible process?" Is that what you are asking?

Yes.
If I have been given a process, then how will I decide whether it is irreversible or not?
One method will be to check entropy of the isolated system.
Is there any other method?

 

[Chestermiller]

Yes.
If I have been given a process, then how will I decide whether it is irreversible or not?
One method will be to check entropy of the isolated system.
Is there any other method?

Ok. I will be back in a while to help you get a feel for the characteristic differences between reversible and irreversible processes. Are you familiar with spring-dashpot mechanical systems?

 

[Pushoam]

Are you familiar with spring-dashpot mechanical systems?

I haven't heard it before.
Will you please post a picture of the system?
I may recognize it.

 

[Chestermiller]

I haven't heard it before.
Will you please post a picture of the system?
I may recognize it.

Instead of using this approach to help you understand irreversible processes, I think I'll try a different approach. Are you familiar with the transport (non-equilibrium) physical properties of gases and liquids known as viscosity and thermal conductivity?

 

[Pushoam]

Sorrry for replying late, please forgive me for this.

Thank you for reminding me this.

Are you familiar with the transport (non-equilibrium) physical properties of gases and liquids known as viscosity and thermal conductivity?

I am a little bit familiar with viscosity. We can start with this.

 

[Chestermiller]

When a gas is experiencing an irreversible expansion or compression, the rate of deformation of the gas is very rapid. When this happens (i.e., the process is irreversible), the behavior of the gas does not obey the ideal gas law, which is a thermodynamic equilibrium relationship. Instead, the behavior of the gas depends on the rate at which the gas is deforming. To give you an idea how this works, a very crude approximation to the behavior of an ideal gas in a cylinder under irreversible expansion or compression conditions can be written as:
$$\frac{nRT}{V}-2\frac{\eta}{V}\frac{dV}{dt}=P_{ext}\tag{1}$$where $\eta$ is the viscosity of the gas and $P_{ext}$ is the external force per unit area applied to the gas. Note the time derivative of the gas volume in the viscosity term in the equation.

Here is an example of how this plays out. Suppose you have an adiabatic expansion or compression, so that the amount of heat added or removed is zero. From the first law of thermodynamics applied to this system, we have (in differential form):
$$nC_v\frac{dT}{dt}=-P_{ext}\frac{dV}{dt}\tag{2}$$If we combine these two equations, we obtain:
$$nC_v\frac{dT}{dt}=-nRT\frac{d\ln{V}}{dt}+2\eta V\left(\frac{d\ln{V}}{dt}\right)^2\tag{3}$$If we divide this equation by T, we obtain:
$$nC_v\frac{d\ln{T}}{dt}+nR\frac{d\ln{V}}{dt}=2\frac{\eta}{T} V\left(\frac{d\ln{V}}{dt}\right)^2\tag{4}$$
But, for an ideal gas, the left hand side of this equation is the rate of change of entropy:
$$\frac{dS}{dt}=2\frac{\eta}{T} V\left(\frac{d\ln{V}}{dt}\right)^2\tag{5}$$Note that, because the rate of change of volume is squared on the right hand side of this equation, the right hand side of this equation is positive definite (and thus does not depend on whether the gas is expanding or compressing). So, according to this equation, in an adiabatic irreversible expansion or compression of an ideal gas in a cylinder, the rate of generation of entropy is always positive, and proportional to the square of the rate of change of volume.

Like I said, this development only very crude and approximate, but it contains the key physical mechanism (viscous dissipation) responsible for the generation of entropy.

 

[Pushoam]

But, for an ideal gas, the left hand side of this equation is the rate of change of entropy:

Looking at eqn. 2, I think that LHS of this eqn. is $\frac{dQ}{dT}$.
Then, only the first part of the LHS of eqn (4) is $\frac {dS}{dT}$. Isn't it so?

 

[Chestermiller]

Looking at eqn. 2, I think that LHS of this eqn. is $\frac{dQ}{dT}$.

For someone presently involved in trying to understand the 2nd law of thermodynamics, you seem awfully confused about how even to apply the 1st law of thermodynamics to a problem. In terms of Q, what do the words "adiabatic expansion or compression" in my post #9 mean to you? Eqn. 2 is the result of applying the 1st law of thermodynamics to our problem of adiabatic expansion or compression for an ideal gas. The left hand side represents the change in internal energy, and the right hand side represents the work done by the surroundings on the gas.

Then, only the first part of the LHS of eqn (4) is $\frac {dS}{dT}$. Isn't it so?

No. You are aware that, in addition to temperature, entropy also varies with volume, correct?