I Adiabatic expansion work far exceeds isobaric of same volume, why?

[Chestermiller]

Yes

OK. For this adiabatic irreversible process, the first law if thermodynamics gives us $$\Delta U=nC_v(T_f-T_1)=-W$$where $$W=\int_{V_1}^{V_2}{\frac{F_g}{A}dV}=P_{ext}(V_2-V_1)=(14.7)(44.9)(0.113)=74.6\ J$$ and $$n=\frac{P_1V_1}{RT_1}$$Combining these equation, we obtain:$$\frac{P_1V_1}{(\gamma-1)}\left[\frac{T_f}{T_1}-1\right]=-W$$Solving this equation for the ratio of the final temperature to the initial temperature then gives: $$\frac{T_f}{T_1}=1-\frac{W(\gamma-1)}{P_1V_1}=1-\frac{(76.4)(0.4)}{(24.7)(100)(0.113)}=0.8905$$

OK so far?

 

[jack action]

I'm going to chime in again as I feel both of you have difficulties understanding each other. I hope I won't add to the confusion.

For the OP, the piston of cross-sectional area $A$ is connected to something that offers a resistive force $F$ - like a crankshaft for example. The crankcase pressure would be at a steady $P_{atm}$. On the other side of the piston, there is a combustion chamber with pressure $P$ whose volume varies from $V_1$ to $V_2$. In one case, $P$ follows an adiabatic expansion, in the other, it follows an isobaric expansion. So the energy equation would be:
$$\int_{V_1}^{V_2}{PdV}=P_{atm}(V_2-V_1) + F\frac{(V_2-V_1)}{A}$$
(For simplicity, I'm assuming both sides of the piston have the same cross-sectional area.)

The "net work" the OP refers to is $F\frac{(V_2-V_1)}{A}$, i.e. the useful work going to the crankshaft, where $\frac{(V_2-V_1)}{A}$ is the piston displacement.

If the piston has a mass, piston acceleration must also be considered, if any.

Feel free to tell me if I don't understand either of you correctly.

 

[Chestermiller]

I'm going to chime in again as I feel both of you have difficulties understanding each other. I hope I won't add to the confusion.

For the OP, the piston of cross-sectional area $A$ is connected to something that offers a resistive force $F$ - like a crankshaft for example. The crankcase pressure would be at a steady $P_{atm}$. On the other side of the piston, there is a combustion chamber with pressure $P$ whose volume varies from $V_1$ to $V_2$. In one case, $P$ follows an adiabatic expansion, in the other, it follows an isobaric expansion. So the energy equation would be:
$$\int_{V_1}^{V_2}{PdV}=P_{atm}(V_2-V_1) + F\frac{(V_2-V_1)}{A}$$
(For simplicity, I'm assuming both sides of the piston have the same cross-sectional area.)

The "net work" the OP refers to is $F\frac{(V_2-V_1)}{A}$, i.e. the useful work going to the crankshaft, where $\frac{(V_2-V_1)}{A}$ is the piston displacement.

If the piston has a mass, piston acceleration must also be considered, if any.

Feel free to tell me if I don't understand either of you correctly.

You don’t understand what I have been saying correctly. Before you posted this, I thought the OP and I had reached consensus. I hope what you have posted here does not introduce new confusion.

 

[MysticDream]

OK. For this adiabatic irreversible process, the first law if thermodynamics gives us $$\Delta U=nC_v(T_f-T_1)=-W$$where $$W=\int_{V_1}^{V_2}{\frac{F_g}{A}dV}=P_{ext}(V_2-V_1)=(14.7)(44.9)(0.113)=74.6\ J$$ and $$n=\frac{P_1V_1}{RT_1}$$Combining these equation, we obtain:$$\frac{P_1V_1}{(\gamma-1)}\left[\frac{T_f}{T_1}-1\right]=-W$$Solving this equation for the ratio of the final temperature to the initial temperature then gives: $$\frac{T_f}{T_1}=1-\frac{W(\gamma-1)}{P_1V_1}=1-\frac{(76.4)(0.4)}{(24.7)(100)(0.113)}=0.8905$$

OK so far?

Yes.

 

[MysticDream]

I'm going to chime in again as I feel both of you have difficulties understanding each other. I hope I won't add to the confusion.

For the OP, the piston of cross-sectional area $A$ is connected to something that offers a resistive force $F$ - like a crankshaft for example. The crankcase pressure would be at a steady $P_{atm}$. On the other side of the piston, there is a combustion chamber with pressure $P$ whose volume varies from $V_1$ to $V_2$. In one case, $P$ follows an adiabatic expansion, in the other, it follows an isobaric expansion. So the energy equation would be:
$$\int_{V_1}^{V_2}{PdV}=P_{atm}(V_2-V_1) + F\frac{(V_2-V_1)}{A}$$
(For simplicity, I'm assuming both sides of the piston have the same cross-sectional area.)

The "net work" the OP refers to is $F\frac{(V_2-V_1)}{A}$, i.e. the useful work going to the crankshaft, where $\frac{(V_2-V_1)}{A}$ is the piston displacement.

If the piston has a mass, piston acceleration must also be considered, if any.

Feel free to tell me if I don't understand either of you correctly.

Yes, I was referring to the useful work that could be transferred to the crankshaft.

I thought your formula that subtracted the isobaric “contraction” of the external atmosphere from the total work of the gas (using the adiabatic formula) is what would give me the work transferred to a crankshaft. I thought chestermiller would be in agreement with that. If he has further insight to offer, I’m all ears.

 

[Chestermiller]

Yes.

Well, the next step is to calculate the final equilibrium pressure within the cylinder using the ideal gas law: $$\frac{P_fV_2}{T_f}=\frac{P_1V_1}{T_1}$$So, $$P_f=P_1\frac{V_1}{V_2}\frac{T_f}{T_1}=(24.7)\frac{100}{144.9}(0.8905)=15.2\ psia=0.5\ psig$$So by stopping the piston motion before reaching its natural equilibrium position, we obtain a final pressure higher than the pressure against whcihthe gas expanded, 14.7 psia. So, in this case, the final temperature and final pressure are higher than in the adiabatic reversible case.

 

[Chestermiller]

Yes, I was referring to the useful work that could be transferred to the crankshaft.

I thought your formula that subtracted the isobaric “contraction” of the external atmosphere from the total work of the gas (using the adiabatic formula) is what would give me the work transferred to a crankshaft. I thought chestermiller would be in agreement with that. If he has further insight to offer, I’m all ears.

The most fundamental concept of good modeling practice is to start simple, and add complexity later as needed. Why? Well first of all, if you are unable to solve a simple version of the problem, you certainly won't be able to solve a more complex version. Plus, once you solve the simple version, you will have some understanding developed and some quantitative results under your belt in a short period of time and with minimal effect. These results can then provide a basis of comparison as more complexity is added. This is called the KISS principle: Keep It Simple, Stupid.

 

[MysticDream]

Well, the next step is to calculate the final equilibrium pressure within the cylinder using the ideal gas law: $$\frac{P_fV_2}{T_f}=\frac{P_1V_1}{T_1}$$So, $$P_f=P_1\frac{V_1}{V_2}\frac{T_f}{T_1}=(24.7)\frac{100}{144.9}(0.8905)=15.2\ psia=0.5\ psig$$So by stopping the piston motion before reaching its natural equilibrium position, we obtain a final pressure higher than the pressure against whcihthe gas expanded, 14.7 psia. So, in this case, the final temperature and final pressure are higher than in the adiabatic reversible case.

Yes, I understand.

 

[jack action]

You don’t understand what I have been saying correctly. Before you posted this, I thought the OP and I had reached consensus. I hope what you have posted here does not introduce new confusion.

I'm sorry to get involved but this thread is very confusing and my understanding is that you are not explaining the process described in the OP. At first, you seemed to be describing this:

For the adiabatic irreversible isobaric expansion you were considering in the second case, I believe you intended to start out at 10 psi gauge again, but in this case to suddenly drop the external pressure to zero psi gauge and to hold that external pressure constant until the gas had expanded by the same volume as in the first case. So $\Delta V=44.9\ in^3=0.000736\ m^3$ and $P_{ext}=14.7\ psi=101.325\ kPa$, and the work is W = 74.6 J.

The OP answered back to this with:

Sorry for the misunderstanding, but that is not the first calculation I was intending to do. I was expanding a 10 psig volume of gas adiabatically against an external pressure of 0 psig. In both cases the external pressure was 0 psig.

This could be visualized as a cylinder with a piston in it. Both cylinders and pistons are of the same exact size, will travel the same volume while doing the work, and have the same starting pressure and external pressure. The only difference is that one cylinder will expand adiabatically to 0 psig, and the other will maintain pressure.

You say the piston goes down while the pressure is at 0 psig; I believe the OP is saying the piston goes down while the pressure is maintained at 10 psig.

When the OP talks about "external pressure", he means the pressure on the other side of the piston.

I believe the isobaric case in the OP is NOT adiabatic.

Then you came back with:

If the piston is massless and frictionless, then by Newton's 2nd law, the force per unit area exerted by the external atmosphere on the outside face of the piston $P_{ext}$ must be equal in magnitude to the force per unit area exerted by the gas on the inside face of the piston. That means that the work done by the internal gas on the inside face of the piston must be $$dW=P_{ext}dV$$

Here you seem to say that the piston is set free, just being pushed around by the different pressures in the chambers on each side of the piston. I believe the OP expects that one side produces more work than the other takes, the resultant energy being transferred to the piston doing useful work by pushing something else.

My understanding is that the piston is not free. Its motion is restricted by some external mechanism.

The work done by the internal gas on the inside face of the piston is expected to be greater than the work done by the external atmosphere on the outside face of the piston, not equal.

Finally, here you say:

You want to do it for the case in which the piston is stopped once the final volume matches that for the adiabatic reversible case, right?

To which the OP answered "Yes". The case to be compared is supposed to be isobaric by definition. How come you end up in post #56 with a process where the final pressure is different from the initial pressure? To me, both the initial and final pressures are 24.7 psia. That is the constraint of the problem.

If the OP is not confused, I am.

 

[MysticDream]

If the OP is not confused, I am.

I think he is intending to explain everything by starting with basics rather than just answering my initial question, which is fine, although I was confused at first. I believe you answered my initial question already. My misunderstanding was that I thought the adiabatic formula I was using described the useful work that could be done on a piston, but did not realize that work was also being done (unfortunately wasted) on the external atmosphere. By subtracting that from the total work done by the gas, I end up with the useful work that would be converted into kinetic energy (on the piston).

 

[jack action]

I think he is intending to explain everything by starting with basics rather than just answering my initial question, which is fine, although I was confused at first. I believe you answered my initial question already. My misunderstanding was that I thought the adiabatic formula I was using described the useful work that could be done on a piston, but did not realize that work was also being done (unfortunately wasted) on the external atmosphere. By subtracting that from the total work done by the gas, I end up with the useful work that would be converted into kinetic energy (on the piston).

To keep it simple, you should forget about the pressure on the external face of the piston. Whether the energy goes against the pressure on the other side of the piston, fights friction, or produces work somewhere else is irrelevant to your problem.

This means that your calculations for the adiabatic case are correct but not the ones for the isobaric case. I gave you the corrected calculations in post #22. It corresponds to the same answer given in post #21 by @Chestermiller .

In your calculations, the final pressure at the end of your adiabatic process is NOT what you refer to as "external pressure", i.e. the pressure on the other side of the piston.

The fact that the adiabatic case is dropping to the atmospheric pressure in your example - the same as the pressure on the other side of the piston - should be considered coincidental, nothing more. In fact, in an engine where the compression ratio is equal to the expansion ratio (Otto or Diesel cycles for example), the final pressure after the expansion process is always greater than the atmospheric pressure. The Atkinson or Miller cycles do have a larger expansion ratio than their compression ratio, effectively bringing the final pressure to atmospheric pressure.

 

[Chestermiller]

I wouldn't consider pushing back the external atmosphere as useless or "lost work."

If we consider a piston with mass, we need to address the following questions:
1. If the piston eventually comes to rest, what happens to the kinetic energy it had? (a) it is dissipalllllted by viscous damping imposed by the gas in the cylinder (b) part of it is released to the gas as thermal energy resulting from yield deformation in the piston

2'. How do we determine the kinetic energy of the piston as a function of time for a piston that is very massive? Is this needed to determine the final steady state of the gas (and the total work the gas in the cylinder has done)?

3. How do we determine the kinetic energy of the piston as a function of time for a piston of small finite mass or zero mass? Is this needed to determine the final steady state of the gas (and the total work the gas in the cylinder has done)?

 

[MysticDream]

To keep it simple, you should forget about the pressure on the external face of the piston. Whether the energy goes against the pressure on the other side of the piston, fights friction, or produces work somewhere else is irrelevant to your problem.

This means that your calculations for the adiabatic case are correct but not the ones for the isobaric case. I gave you the corrected calculations in post #22. It corresponds to the same answer given in post #21 by @Chestermiller .

In your calculations, the final pressure at the end of your adiabatic process is NOT what you refer to as "external pressure", i.e. the pressure on the other side of the piston.

The fact that the adiabatic case is dropping to the atmospheric pressure in your example - the same as the pressure on the other side of the piston - should be considered coincidental, nothing more. In fact, in an engine where the compression ratio is equal to the expansion ratio (Otto or Diesel cycles for example), the final pressure after the expansion process is always greater than the atmospheric pressure. The Atkinson or Miller cycles do have a larger expansion ratio than their compression ratio, effectively bringing the final pressure to atmospheric pressure.

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I believe it was actually post #25 where you answered my question. I wanted to know why my calculations for the isobaric case was lower than the adiabatic case. I now understand that I was not subtracting the work done on the external atmosphere. The P*V formula I was using for the isobaric case was essentially calculating only the work done on the piston, so it was accurate. The formula I was using for the adiabatic case was calculating the total work done by the gas, not only on the piston, which is what I needed to do for an accurate comparison. Now after you pointed that out, I now see that the "useful work" done on the piston is lower for the adiabatic case compared to the isobaric case which is what one would expect to see.

Yes, I understand that the adiabatic case dropping to atmospheric pressure is coincidental. I was not using the final pressure inside the cylinder as the external pressure.

 

[MysticDream]

I wouldn't consider pushing back the external atmosphere as useless or "lost work."

If we consider a piston with mass, we need to address the following questions:
1. If the piston eventually comes to rest, what happens to the kinetic energy it had? (a) it is dissipalllllted by viscous damping imposed by the gas in the cylinder (b) part of it is released to the gas as thermal energy resulting from yield deformation in the piston

2'. How do we determine the kinetic energy of the piston as a function of time for a piston that is very massive? Is this needed to determine the final steady state of the gas (and the total work the gas in the cylinder has done)?

3. How do we determine the kinetic energy of the piston as a function of time for a piston of small finite mass or zero mass? Is this needed to determine the final steady state of the gas (and the total work the gas in the cylinder has done)?

1. That is an interesting question.

2. Would not the work done on the piston be the same regardless? A massive piston would accelerate slower than a lighter one but the gas would do the same amount of work, correct? This is because the amount of force on the piston would be the same.
Force = mass * acceleration
work = force * distance

3. For a piston of zero mass, would it not be the speed of sound?

 

[Chestermiller]

1. That is an interesting question.

2. Would not the work done on the piston be the same regardless? A massive piston would accelerate slower than a lighter one but the gas would do the same amount of work, correct? This is because the amount of force on the piston would be the same.
Force = mass * acceleration
work = force * distance

No. If the piston is very massive, then its acceleration and velocity will be very low throughout the process, as will the rate of change of gas volume in the cylinder. So for the gas, its expansion at short times will approach adiabatic reversible expansion, as will the amount of work that the gas does on the inside face of the piston. Of course, the expansion and subsequent oscillation will take a very long time, and when the piston oscillation motion has finally damped to a stop, all its kinetic energy will be dissipated. and the same final state of the gas will result as that for a massless piston.

3. For a piston of zero mass, would it not be the speed of sound?

No way. For a piston of zero mass, at time t during the irreversible expansion, $F_g=P_{ext}A$, and the work will be $W=P_{ext}(V_2-V_1)$.

 

[jack action]

The formula I was using for the adiabatic case was calculating the total work done by the gas, not only on the piston, which is what I needed to do for an accurate comparison. Now after you pointed that out, I now see that the "useful work" done on the piston is lower for the adiabatic case compared to the isobaric case which is what one would expect to see.

What I wanted you to see is that work from an isobaric process is always greater than work from an adiabatic process, period. That is what post #22 shows. You don't need to introduce extra forces acting on the piston to prove that. It doesn't matter what this energy is used for, whether it is for compressing the gas on the other side of the piston or pushing the piston itself or both, or anything else one could imagine.

 

[MysticDream]

What I wanted you to see is that work from an isobaric process is always greater than work from an adiabatic process, period. That is what post #22 shows. You don't need to introduce extra forces acting on the piston to prove that. It doesn't matter what this energy is used for, whether it is for compressing the gas on the other side of the piston or pushing the piston itself or both, or anything else one could imagine.

Understood. I also just wanted to understand how to calculate the work done for a "real life" example. Hypothetically, it would be easier to test something like that at atmospheric pressure rather than doing it in a vacuum chamber. I was disappointed to learn that much of this expansion energy is wasted with atmospheric pressure on the opposing side of the piston. I suppose this is one aspect of calculating efficiency.

 

[MysticDream]

No. If the piston is very massive, then its acceleration and velocity will be very low throughout the process, as will the rate of change of gas volume in the cylinder. So for the gas, its expansion at short times will approach adiabatic reversible expansion, as will the amount of work that the gas does on the inside face of the piston. Of course, the expansion and subsequent oscillation will take a very long time, and when the piston oscillation motion has finally damped to a stop, all its kinetic energy will be dissipated. and the same final state of the gas will result as that for a massless piston.

No way. For a piston of zero mass, at time t during the irreversible expansion, $F_g=P_{ext}A$, and the work will be $W=P_{ext}(V_2-V_1)$.

That is what I was saying. A massive piston would accelerate more slowly and a lighter piston would accelerate faster, but either way Force = Mass * Acceleration, so the force would be the same whether it's slow or fast, as long as there is no heat transfer. I don't understand why we're talking about oscillations as we're only focused on the initial adiabatic expansion of the gas and the work done by the gas up until that point.

I don't quite understand the piston of zero mass example. I thought gas released to a lower pressure unrestrained will expand at the speed of sound(at that pressure). Knowing that, you could calculate the time it would take a volume of gas to expand fully.

 

[Chestermiller]

That is what I was saying. A massive piston would accelerate more slowly and a lighter piston would accelerate faster, but either way Force = Mass * Acceleration, so the force would be the same whether it's slow or fast, as long as there is no heat transfer. I don't understand why we're talking about oscillations as we're only focused on the initial adiabatic expansion of the gas and the work done by the gas up until that point.

But the force the gas exerts at a given volume change is different for a large piston mass than for a small mass. So the acceleration is not simply inversely proportional to the mass.

We are not focused on the initial adiabatic expansion of the gas. We are focused on the final equilibrium state when the piston is no longer moving.

I don't quite understand the piston of zero mass example. I thought gas released to a lower pressure unrestrained will expand at the speed of sound(at that pressure). Knowing that, you could calculate the time it would take a volume of gas to expand fully.

This is not a correct description of the wave mechanics in a gas enclosed in a cylinder and it also neglects the viscous dissipation in the gas enclosed in the cylinder. In any event, it will not tell you the work done by the gas on the piston.

 

[MysticDream]

But the force the gas exerts at a given volume change is different for a large piston mass than for a small mass. So the acceleration is not simply inversely proportional to the mass.

We are not focused on the initial adiabatic expansion of the gas. We are focused on the final equilibrium state when the piston is no longer moving.

This is not a correct description of the wave mechanics in a gas enclosed in a cylinder and it also neglects the viscous dissipation in the gas enclosed in the cylinder. In any event, it will not tell you the work done by the gas on the piston.

Why would the force be different? You would start with the same exact pressure on the piston face. Do you mean the Pressure-Volume curve would be different?

 

[MysticDream]

But the force the gas exerts at a given volume change is different for a large piston mass than for a small mass. So the acceleration is not simply inversely proportional to the mass.

We are not focused on the initial adiabatic expansion of the gas. We are focused on the final equilibrium state when the piston is no longer moving.

This is not a correct description of the wave mechanics in a gas enclosed in a cylinder and it also neglects the viscous dissipation in the gas enclosed in the cylinder. In any event, it will not tell you the work done by the gas on the piston.

I understand the irreversible work done on the atmosphere will not be affected by the pistons mass, but the total reversible work should be the same regardless of the pistons mass, should it not?

 

[Chestermiller]

Why would the force be different? You would start with the same exact pressure on the piston face. Do you mean the Pressure-Volume curve would be different?

Do you think that an ideal gas satisfies the ideal gas law in a rapid irreversible expansion (not at thermodynamic equilibrium)?

 

[MysticDream]

Do you think that an ideal gas satisfies the ideal gas law in a rapid irreversible expansion (not at thermodynamic equilibrium)?

Yes.

 

[MysticDream]

Why is more work done on a heavier piston?

How do you calculate the work done on a piston with mass and the work done on the atmosphere that is wasted?

 

[Chestermiller]

Yes.

This is incorrect. The ideal gas law is valid only for thermodynamic equilibrium conditions or for a reversible process (which consists of a continuous sequence of thermodynamic equilibrium states).

The compressive stresses for a gas experiencing a rapid irreversible deformation are given by Newton's law of viscosity (in correct 3D tensorial form). For such a deformation, we learn in fluid mechanics that the principal components of the compressive stress tensor are given locally by: $$\sigma_1=\frac{RT}{V_m}-\left(\frac{4}{3}\mu\right)\frac{\partial v_1}{\partial x_1}-\left(\frac{2}{3}\mu\right)\frac{\partial v_2}{\partial x_2}-\left(\frac{2}{3}\mu\right)\frac{\partial v_3}{\partial x_3}$$$$\sigma_2=\frac{RT}{V_m}-\left(\frac{4}{3}\mu\right)\frac{\partial v_2}{\partial x_2}-\left(\frac{2}{3}\mu\right)\frac{\partial v_1}{\partial x_1}-\left(\frac{2}{3}\mu\right)\frac{\partial v_3}{\partial x_3}$$$$\sigma_3=\frac{RT}{V_m}-\left(\frac{4}{3}\mu\right)\frac{\partial v_3}{\partial x_3}-\left(\frac{2}{3}\mu\right)\frac{\partial v_2}{\partial x_2}-\left(\frac{2}{3}\mu\right)\frac{\partial v_1}{\partial x_1}$$where $V_M$ is the local molar volume of the gas, $\mu$ is the gas viscosity, and the subscripts 1, 2, and 3 represent the 3 principal directions of stress and rate of strain. Note that the first term in each of these equations is the local pressure that one would calculate locally from the ideal gas law at the local molar volume and local temperature. The additional terms are viscous contributions that approach zero in a reversible deformation.

 

[Chestermiller]

Why is more work done on a heavier piston?

How do you calculate the work done on a piston with mass and the work done on the atmosphere that is wasted?

Are you talking about at each point in time during a rapid irreversible expansion or only at final steady state when the piston has come to rest and the gas is again at equilibrium?

 

[MysticDream]

Are you talking about at each point in time during a rapid irreversible expansion or only at final steady state when the piston has come to rest and the gas is again at equilibrium?

Each point in time, but more specifically at max volume of the gas.

 

[Chestermiller]

Each point in time, but more specifically at max volume of the gas.

Thermodynamics alone can't determine the work the gas does on the piston at each point in time during an irreversible expansion because of the existence of the viscous contributions to the force exerted by the gas. It can however determine the work done by the gas on the piston through final equilibrium when the piston kinetic energy has been damped to zero (provided we know the pressure imposed by the external atmosphere). To get the work the gas does on the piston at each point int time during an irreversible expansion, one must solve the set of partial differential equations represented by the Navier Stokes equations for viscous flow in conjunction with the partial differential equation for overall energy conservation. The independent variables in this fluid mechanics endeavor are spatial position within the cylinder and time, and the dependent variables are temperature, velocity components, and pressure (all of which are functions of the independent variables).

 

[MysticDream]

Thermodynamics alone can't determine the work the gas does on the piston at each point in time during an irreversible expansion because of the existence of the viscous contributions to the force exerted by the gas. It can however determine the work done by the gas on the piston through final equilibrium when the piston kinetic energy has been damped to zero (provided we know the pressure imposed by the external atmosphere). To get the work the gas does on the piston at each point int time during an irreversible expansion, one must solve the set of partial differential equations represented by the Navier Stokes equations for viscous flow in conjunction with the partial differential equation for overall energy conservation. The independent variables in this fluid mechanics endeavor are spatial position within the cylinder and time, and the dependent variables are temperature, velocity components, and pressure (all of which are functions of the independent variables).

Can it be approximated with a simple formula? Example: Total work done by gas = work on atmosphere + work on piston.

I still don’t understand why more work would be done on a heavier piston.

 

[Chestermiller]

Can it be approximated with a simple formula? Example: Total work done by gas = work on atmosphere + work on piston.

No, because you are not going. to know the work on the piston (.e., its kinetic energy) without directly determining the work done by the gas, which involves solving that complicated set of equations interior to the cylinder, and matching the piston velocity to the gas velocity at the inside face of the piston.

I still don’t understand why more work would be done on a heavier piston.

That's only initially during the expansion. At final thermodynamic equilibrium, the work will be the same.

 

[MysticDream]

No, because you are not going. to know the work on the piston (.e., its kinetic energy) without directly determining the work done by the gas, which involves solving that complicated set of equations interior to the cylinder, and matching the piston velocity to the gas velocity at the inside face of the piston.

That's only initially during the expansion. At final thermodynamic equilibrium, the work will be the same.

I appreciate your insight. You’ve given me a lot to think about.

 

[Chestermiller]

I appreciate your insight. You’ve given me a lot to think about.

Check out Transport Phenomena by Bird, et al, Chapter 1, Section 1.2 Generalization of Newton's Law of Viscosity.