How Is Entropy Calculated in an Irreversible Gas Expansion?

[MichaelWiteman]

Hi,

An irreversible gas expansion is often described in textbooks with a compressed gas in a cylinder pushing up a weight (with mass m) via a hypothetical friction-less and weightless piston. It is said the work done by the gas is equal to -mg × h and from this you can derive the work for a irreversible expansion (w = - P_ext×ΔV). This makes sense since work is F × displacement. However, wouldn't the force exerted by the gas on the weight be much larger than mg at the beginning and reach mg only at the end of the expansion? What am I missing here?

Cheers,
Mike

 

[Andrew Mason]

Hi,

An irreversible gas expansion is often described in textbooks with a compressed gas in a cylinder pushing up a weight (with mass m) via a hypothetical friction-less and weightless piston. It is said the work done by the gas is equal to -mg × h and from this you can derive the work for a irreversible expansion (w = - P_ext×ΔV). This makes sense since work is F × displacement. However, wouldn't the force exerted by the gas on the weight be much larger than mg at the beginning and reach mg only at the end of the expansion? What am I missing here?

The work done by the gas must be equal to the energy imparted to the weight plus work done on the atmosphere (Pext×ΔV). That consists of the change in potential energy plus the change in kinetic energy of the weight plus Pext×ΔV

The gas pressure will be higher than mg/A + Pext initially and the weight will accelerate upward until the pressure reaches mg/A +Pext. At that point the weight will deccelerate until the weight stops, at which point the pressure will be less than mg/A. The work done by the gas up to that point is mgh plus Pext×Ah since KE =0. So the work done by the gas is (mg + PextA)h where h is the maximum height reached.

AM

 

[MichaelWiteman]

The work done by the gas must be equal to the energy imparted to the weight plus work done on the atmosphere (Pext×ΔV). That consists of the change in potential energy plus the change in kinetic energy of the weight plus Pext×ΔV

The gas pressure will be higher than mg/A + Pext initially and the weight will accelerate upward until the pressure reaches mg/A +Pext. At that point the weight will deccelerate until the weight stops, at which point the pressure will be less than mg/A. The work done by the gas up to that point is mgh plus Pext×Ah since KE =0. So the work done by the gas is (mg + PextA)h where h is the maximum height reached.

AM

Thank you for your reply! I should had mentioned that this is in vacuum, but it doesn't invalidate your answer. If I understand correctly, you are using the fact the total total energy (where the system is the expanding gas, and surroundings is the moving weight) is equal to zero and that the final kinetic energy of the weight is zero. This makes a lot more sense now!

I still don't understand why textbooks show the P vs V graphs for the irreversible expansion with a instantaneous drop of P with no volume change, followed by volume expansion at a constant pressure. Other than that the area under the curve is equal to work, the graph doesn't represent that actual expansion.

Mike

 

[Andrew Mason]

Thank you for your reply! I should had mentioned that this is in vacuum, but it doesn't invalidate your answer. If I understand correctly, you are using the fact the total total energy (where the system is the expanding gas, and surroundings is the moving weight) is equal to zero and that the final kinetic energy of the weight is zero. This makes a lot more sense now!

The total energy is not zero. It starts with the internal energy of the gas. As the gas expands it does work and internal energy decreases since it is an adiabatic expansion (Q = 0).

$$W_{gas} = \int_{V_0}^{V_f} P_{gas}dV$$ where $W_{gas}$ is the work done BY the gas.

The problem is that this is a difficult integral to compute because we don't know what P is in this dynamic process. But we do know that the work done by the gas is not lost as heat due to friction so it must be imparted entirely to the weight ie. as kinetic and potential energy change of the weight.

I still don't understand why textbooks show the P vs V graphs for the irreversible expansion with a instantaneous drop of P with no volume change, followed by volume expansion at a constant pressure. Other than that the area under the curve is equal to work, the graph doesn't represent that actual expansion.

Perhaps you can show us such an example. If the gas expands against a constant external pressure, then the work done by the gas in the cylinder is $W_{gas} = -\int_{V_0}^{V_f} P_{ext}dV$. That may be what the PV graph is showing.

AM

 

[Chestermiller]

Hi,

An irreversible gas expansion is often described in textbooks with a compressed gas in a cylinder pushing up a weight (with mass m) via a hypothetical friction-less and weightless piston. It is said the work done by the gas is equal to -mg × h and from this you can derive the work for a irreversible expansion (w = - P_ext×ΔV). This makes sense since work is F × displacement. However, wouldn't the force exerted by the gas on the weight be much larger than mg at the beginning and reach mg only at the end of the expansion? What am I missing here?

Cheers,
Mike

Using Newton's 2nd law of motion, what is the force balance on the combination of mass plus piston (letting F(t) represent the force that the gas exerts on the piston at time t)?

 

[MichaelWiteman]

The total energy is not zero. It starts with the internal energy of the gas(...)

Sorry, I meant the total energy change is zero.

--Mike

 

[Chestermiller]

Sorry, I meant the total energy change is zero.

--Mike

I'm still waiting for you to respond to my leading question in post #7. Do you think I asked it for my health? Very disrespectful.

 

[MichaelWiteman]

Using Newton's 2nd law of motion, what is the force balance on the combination of mass plus piston (letting F(t) represent the force that the gas exerts on the piston at time t)?

I'm not a physicist so I apologize if I write something that you might find ignorant. The gas has pressure and the piston has area, hence there is a force acting on it equal to F(t)=P(t)*A. The piston with the weight exerts a force on the gas equal to mg. So the net force is equal to F(t)-mg. This unbalanced force results in the piston initial acceleration. The problem with my thinking here is that if the weight was suddenly removed from the picture, there is still a force (from gas pressure), and if there is a unbalanced force I would think there is acceleration. If there is no weight on the weightless piston, but the gas still exerts a force and there is a displacement is work done? I know the answer is no, but this is not quite obvious for me. If the piston is weightless does the acceleration approach infinity?

I guess the moment the gas molecules make contact with the weightless piston the piston is displaced infinitely fast meaning that the gas never "has time" to exert force on it, therefore there can be no gas pressure with a weightless piston, just movement. Sorry for this jibber-jabber.

--Mike

 

[Chestermiller]

I'm not a physicist so I apologize if I write something that you might find ignorant. The gas has pressure and the piston has area, hence there is a force acting on it equal to F(t)=P(t)*A. The piston with the weight exerts a force on the gas equal to mg. So the net force is equal to F(t)-mg. This unbalanced force results in the piston initial acceleration. The problem with my thinking here is that if the weight was suddenly removed from the picture, there is still a force (from gas pressure), and if there is a unbalanced force I would think there is acceleration.

Correct. Assuming that there vacuum on the other side of the piston (i.e., no outside gas pressure pushing in), the force balance on the piston reads: $$F(t)-mg=ma=m\frac{dv}{dt}$$. If we multiply this equation by the velocity of the piston v = dx/dt, and integrate with respect to time, we get $$W_g(t)=\int{F\frac{dx}{dt}dt}=mg\Delta x+m\frac{v^2(t)}{2}$$where $W_g(t)$ is the work done by the gas on the piston up to time t. The second term on the rhs of this equation is the kinetic energy of the piston/mass. The piston (assumed frictionless) will oscillate about the final equilibrium position, analogous to a spring/mass system. Do you think that the piston will continue to oscillate forever, or do you think that some other damping mechanism (other than piston friction) will cause the piston to eventually settle at its final equilibrium position?

If there is no weight on the weightless piston, but the gas still exerts a force and there is a displacement is work done? I know the answer is no, but this is not quite obvious for me. If the piston is weightless does the acceleration approach infinity?

I guess the moment the gas molecules make contact with the weightless piston the piston is displaced infinitely fast meaning that the gas never "has time" to exert force on it, therefore there can be no gas pressure with a weightless piston, just movement. Sorry for this jibber-jabber.

--Mike

I will help you resolve this second issue later.

 

[MichaelWiteman]

(...)Do you think that the piston will continue to oscillate forever, or do you think that some other damping mechanism (other than piston friction) will cause the piston to eventually settle at its final equilibrium position?(...)

In a case with an ideal gas I believe there will be no damping and the piston with the weight will oscillate forever. However, in a real situation the gas will have some viscosity and the damping mechanism would be via internal friction.

 

[Chestermiller]

In a case with an ideal gas I believe there will be no damping and the piston with the weight will oscillate forever. However, in a real situation the gas will have some viscosity and the damping mechanism would be via internal friction.

lt depends on how you define an ideal gas. In a real gas at low densities, the viscosity approaches a function of temperature. We engineers define an ideal gas as the limiting behavior of a real gas at low densities. Bird, Stewart, and Lightfoot. in their classic book Transport Phenomena show a corresponding states plot of viscosity as a function of reduced temperature and reduced pressure. The behavior approaches a limiting line at low reduced pressures.

You can never properly understand irreversible processes that are not quasi-static without allowing for the gas to have viscosity, ideal gas or not.

Back to the problem at hand. So, after an infinite amount of time, what is the amount of work done by the gas on the piston (when it finally comes to rest)?

 

[MichaelWiteman]

(...) Back to the problem at hand. So, after an infinite amount of time, what is the amount of work done by the gas on the piston (when it finally comes to rest)?

So W_g(t) = -mg dx,

If mg = F = PA, then
W=-PA dx = - P dv. Since P is here the external pressure (its derived from mg) we arrived at the equation from my textbook: W_irrev = -P_ext dV

Cool!

 

[Chestermiller]

So W_g(t) = -mg dx,

If mg = F = PA, then
W=-PA dx = - P dv. Since P is here the external pressure (its derived from mg) we arrived at the equation from my textbook: W_irrev = -P_ext dV

Cool!

In this case, $P_{ext}=\frac{F_g}{A}=\frac{mg}{A}+m\frac{v^2}{2A}$

 

[MichaelWiteman]

In this case, $P_{ext}=\frac{F_g}{A}=\frac{mg}{A}+m\frac{v^2}{2A}$

I guess I skipped a step for my own convenience since I didn't know how to formally show it. I removed the 2nd term in the rhs of your equation for W_g(t). The v in that term will be changing direction (oscillation) and magnitude (damping) with time until v=0. The way I see it now, is that work will be also done on the system by the piston/weight which will be equal to mv²/2A. Is this thinking correct?

 

[Chestermiller]

I guess I skipped a step for my own convenience since I didn't know how to formally show it. I removed the 2nd term in the rhs of your equation for W_g(t). The v in that term will be changing direction (oscillation) and magnitude (damping) with time until v=0. The way I see it now, is that work will be also done on the system by the piston/weight which will be equal to mv²/2A. Is this thinking correct?

I'm sorry. I made a mistake. It should read: $$P_{ext}=\frac{F_g}{A}=\frac{mg}{A}+\frac{m}{A}\frac{dv}{dt}$$where $P_{ext}$ includes viscous stresses within the gas at the face of the piston.

 

[zanick]

Hi,

An irreversible gas expansion is often described in textbooks with a compressed gas in a cylinder pushing up a weight (with mass m) via a hypothetical friction-less and weightless piston. It is said the work done by the gas is equal to -mg × h and from this you can derive the work for a irreversible expansion (w = - P_ext×ΔV). This makes sense since work is F × displacement. However, wouldn't the force exerted by the gas on the weight be much larger than mg at the beginning and reach mg only at the end of the expansion? What am I missing here?

Cheers,
Mike

With the weight lifted up by the compressed gas, the entropy might stay constant if the heat energy matches the temp reduction due to the work being done by lifting up the weight on the piston. dS=dQ/T isn't the better example of an irreversible process the free expansion of gas into a vacuum?

 

[Chestermiller]

With the weight lifted up by the compressed gas, the entropy might stay constant if the heat energy matches the temp reduction due to the work being done by lifting up the weight on the piston. dS=dQ/T isn't the better example of an irreversible process the free expansion of gas into a vacuum?

Are you saying that, even in an irreversible process, the change in entropy is calculated by integrating dS=dQ/T for that irreversible process? Are you saying that free expansion of gas into a vacuum is a better example of an irreversible process than the irreversible expansion process described in this thread?

 

[zanick]

Are you saying that, even in an irreversible process, the change in entropy is calculated by integrating dS=dQ/T for that irreversible process? Are you saying that free expansion of gas into a vacuum is a better example of an irreversible process than the irreversible expansion process described in this thread?

I was saying that i thought dS=dQ/T was more commonly used for a reversible process and it looked like the example here was reversible, based on a weighted piston being lifted some distance as pressure reached equilibrium. Isnt free expansion of a pressurized gas into a vacuum the classic example of an irreversible process? (even though there are many others obviously).

 

[Chestermiller]

I was saying that i thought dS=dQ/T was more commonly used for a reversible process and it looked like the example here was reversible, based on a weighted piston being lifted some distance as pressure reached equilibrium. Isnt free expansion of a pressurized gas into a vacuum the classic example of an irreversible process? (even though there are many others obviously).

The process described here is an irreversible expansion, and would be irreversible even if it were adiabatic with dQ=0. dQ/T gives the entropy change only for reversible processes.

 

[zanick]

The process described here is an irreversible expansion, and would be irreversible even if it were adiabatic with dQ=0. dQ/T gives the entropy change only for reversible processes.

so, with the irreversible process , ...as you said " dQ=0" but T will change, correct? and if the entropy change for a reversible has no change in T, then does Q change?

 

[Chestermiller]

so, with the irreversible process , ...as you said " dQ=0" but T will change, correct? and if the entropy change for a reversible has no change in T, then does Q change?

If I understand you correctly, yes.

 

[zanick]

If I understand you correctly, yes.

so the reason that dS=dQ/T only works with a reversible process as that it incorporates teh surroundings and the system, which would be injecting energy control the "reversibility" of the process. if irreversible, this can't be used due to that equation leaving out other factors showing the energy spreading out in the system. would you say that is a high level , generally correct explanation?

 

[Chestermiller]

so the reason that dS=dQ/T only works with a reversible process as that it incorporates teh surroundings and the system, which would be injecting energy control the "reversibility" of the process. if irreversible, this can't be used due to that equation leaving out other factors showing the energy spreading out in the system. would you say that is a high level , generally correct explanation?

I have no idea what you are saying.

 

[zanick]

Can you briefly describe in your words why the equation wouldn't /doesnt work for irreversible processes. thx

 

[Chestermiller]

Can you briefly describe in your words why the equation wouldn't /doesnt work for irreversible processes. thx

There are two ways by which the entropy of a closed thermodynamics system can change during a process that takes the system from one thermodynamic equilibrium state to another:

1. Heat transfer across the boundary of the system (interface with its surroundings) at the boundary temperature $T_B$, given by the integral of $dq/T_B$. This mechanism is operative during both reversible and irreversible processes.

2. Entropy generation within the system as a result of irreversibilities, including viscous stresses and deformations caused by finite velocity gradients, and internal heat conduction caused by finite temperature gradients. This mechanism is operative only during irreversible processes. (There are other causes of entropy generation, but I'll leave it at there for now).

 

[zanick]

That makes more sense, especially since i was probably thinking of an isolated system rather than closed where heat can be exchanged across the border. in the Irrev process, the entropy is due to irreversibility of the process, while in the reversible process, the entropy rise is due to the exchange of heat across the border. intuitively, i understand the entropy of the irrev process of gas expanding into a vacuum spontaneously, as the pressure drop and no other changes, as that energy that can do work, spreads out . but still would like to know how it relates to S=dQ/T.

it seems much different than what michael talked about above, where the massless piston moves and he was talking about the acceleration being infinite. ( i think he confused or reversed his idea of pressure with force which is not a pressure x area, but pressure being force/area. ) the pressure is a result of the change of the momentum of the molecules of gas hitting themselves or a surface in an elastic collision. their speed dictated by temp, doesn't change, as the massless piston moves outward (against a vacuum), creating a spontaneous movement of the gases to fill the eventual limit of the physical container where only then will the "piston stop" allow for a change of momentum of the molecule at its surface, which causes/is a force. seems like it would act exactly like free expansion in the classic example of a reversible process of pressurized gas in a container with a valve that opens to a vacuum of equal volume. (pressure drops, isothermal, adibiatic, and entropy increases).

Again, michael was talking a massless piston, but with weights on it... the force would be determined by the pressure / area . if that pressure was 100psi and the area was 1sq", that would be a 100lbf. a mass of 1lbm would have an acceleration of 981m/s^2... but after the volume was doubled, that acceleration would be half that ... and so on. so, it wouldn't be infinite, or instantaneous or constant due to the collisions of the gas molecules based on their velocity (at a constant temp). depending on the size of the container, the max velocity of the piston even with weights would only be the velocity of gas based on its temp. the mass less piston without weight would freely expand, while the massless piston with weights on it, would only accelerate based on a change of momentum of those molecules hitting the surface, to a maximum velocity to that of the velocity of the molecules. Eventually providing a pressure or force on it as it stopped at its limit, proportional to the volume change it created. Q changed and temp stayed the same, so this reversible process still had an increase in entropy due to the heat transfer and 0 entropy gain due to irreversibility.

 

[Chestermiller]

That makes more sense, especially since i was probably thinking of an isolated system rather than closed where heat can be exchanged across the border.

What I'm saying applies both to closed systems and isolated systems.

in the Irrev process, the entropy is due to irreversibility of the process, while in the reversible process, the entropy rise is due to the exchange of heat across the border.

No. In an irreversible process, entropy change can be due to both mechanisms, while for a reversible process, it is only due to entropy transfer by heat across the border.

intuitively, i understand the entropy of the irrev process of gas expanding into a vacuum spontaneously, as the pressure drop and no other changes, as that energy that can do work, spreads out . but still would like to know how it relates to S=dQ/T.

In the free expansion you described, if the container is insulated, there is no heat flow across the border, and all the entropy generation takes place within the gas. In the alternate reversible process, we expand the gas against a restraining force, and heat transfer across the border is now allowed (in order for us to get to the same final state). The work done by the gas W and the heat transferred to the gas Q in the reversible process are equal, and the entropy change is $\Delta S=\frac{Q}{T}$. The initial and final states of the gas are the same in both processes, so the entropy changes are the same also. But, for the irreversible path, $\frac{Q}{T}=0$, The difference is that, for the irreversible path, all the entropy change has been generated within the system (for this particular process) and, for the reversible path, all the entropy has been transferred to the system. For any arbitrary process on a closed system (whether reversible or irreversible), the change in entropy entropy change is given by $$\Delta S=\int{\frac{dQ}{T_B}}+\sigma$$where $T_B$ is the temperature at the border with the surroundings (through which the heat flows) and $\sigma$ is the entropy generated during the process due to irreversibilities.

Again, michael was talking a massless piston, but with weights on it... the force would be determined by the pressure / area . if that pressure was 100psi and the area was 1sq", that would be a 100lbf. a mass of 1lbm would have an acceleration of 981m/s^2... but after the volume was doubled, that acceleration would be half that ... and so on. so, it wouldn't be infinite, or instantaneous or constant due to the collisions of the gas molecules based on their velocity (at a constant temp). depending on the size of the container, the max velocity of the piston even with weights would only be the velocity of gas based on its temp. the mass less piston without weight would freely expand, while the massless piston with weights on it, would only accelerate based on a change of momentum of those molecules hitting the surface, to a maximum velocity to that of the velocity of the molecules. Eventually providing a pressure or force on it as it stopped at its limit, proportional to the volume change it created. Q changed and temp stayed the same, so this reversible process still had an increase in entropy due to the heat transfer and 0 entropy gain due to irreversibility.

What do you think would happen to the piston movement after a long time? Would it continue to move forever with the gas pressure getting lower and lower, even though there are weights on the piston? Wouldn't the force of the weights eventually exceed the force of the gas?

 

[zanick]

What do you think would happen to the piston movement after a long time? Would it continue to move forever with the gas pressure getting lower and lower, even though there are weights on the piston? Wouldn't the force of the weights eventually exceed the force of the gas?

the piston with the weights will move as long there is a net force acting on them. as the pressure reduces the force reduces and the piston will stop moving when equilibrium has been reached (when the pressure on the piston / area (F) = the weight on the piston's weight. the force will be reduced proportional with distance where at the end there would an oscillation due to restoring force (assuming a frictionless environment)

 

[Chestermiller]

the piston with the weights will move as long there is a net force acting on them. as the pressure reduces the force reduces and the piston will stop moving when equilibrium has been reached (when the pressure on the piston / area (F) = the weight on the piston's weight. the force will be reduced proportional with distance where at the end there would an oscillation due to restoring force (assuming a frictionless environment)

Are you saying that the piston will oscillate forever with no friction? Because, if that's the case, the gas will never reach thermodynamic equilibrium, and its entropy change can not be established.

 

[zanick]

Are you saying that the piston will oscillate forever with no friction? Because, if that's the case, the gas will never reach thermodynamic equilibrium, and its entropy change can not be established.

that was one of my questions to you. if no friction, I am leaning toward it never coming to rest, but i really don't know. It should act like a spring mass system. i suppose that the entropy calculation would have to assume it comes to rest at some point.

 

[zanick]

No. In an irreversible process, entropy change can be due to both mechanisms, while for a reversible process, it is only due to entropy transfer by heat across the border.
In the free expansion you described, if the container is insulated, there is no heat flow across the border, and all the entropy generation takes place within the gas. In the alternate reversible process, we expand the gas against a restraining force, and heat transfer across the border is now allowed (in order for us to get to the same final state). The work done by the gas W and the heat transferred to the gas Q in the reversible process are equal, and the entropy change is $\Delta S=\frac{Q}{T}$. The initial and final states of the gas are the same in both processes, so the entropy changes are the same also. But, for the irreversible path, $\frac{Q}{T}=0$, The difference is that, for the irreversible path, all the entropy change has been generated within the system (for this particular process) and, for the reversible path, all the entropy has been transferred to the system. For any arbitrary process on a closed system (whether reversible or irreversible), the change in entropy entropy change is given by $$\Delta S=\int{\frac{dQ}{T_B}}+\sigma$$where $T_B$ is the temperature at the border with the surroundings (through which the heat flows) and $\sigma$ is the entropy generated during the process due to irreversibilities.

If the entropy is taking place in the gas, in the irrev example, how it is easily calculated? if you can't use S=dQ/T, can you use a boltzman equation instead? S=k.logW ?

 

[Chestermiller]

that was one of my questions to you. if no friction, I am leaning toward it never coming to rest, but i really don't know. It should act like a spring mass system. i suppose that the entropy calculation would have to assume it comes to rest at some point.

It comes to rest as a result of viscous damping (dissipation) by the gas. This is an irreversible effect, and is a major contributor to entropy generation. In a reversible process, it is negligible.

 

[zanick]

that makes sense..."viscous damping" and so does it contribution to entropy of both a reversible and Irrev process.

 

[Chestermiller]

If the entropy is taking place in the gas, in the irrev example, how it is easily calculated? if you can't use S=dQ/T, can you use a boltzman equation instead? S=k.logW ?

To calculate the amount of entropy generation, you need to first calculate the overall entropy change from the reversible path. Then you subtract $$\left[\int{\frac{dq}{T}}\right]_{at\ boundary}$$ for the irreversible path. This indirectly determines for you the amount of entropy that was generated.

To get the entropy generation directly, you would need to solve the partial differential equations for fluid flow and heat transfer within the system to determine detailed temperatures and velocities of the gas as a function of time and spatial position throughout the irreversible change. This is a very complicated calculation.