# I Thermodynamics: Gas Expansion with Piston Friction

#### Chestermiller

Mentor
I'm thinking of the following comparison:

System 1: massless piston with friction force F, and a pile of small masses totaling M sitting on top of the piston initially

System 2: frictionless piston having mass m, such that mg for this system is equal to F for system 1, and the same pile of small masses totaling M sitting on top of the piston initially

The initial states of these two systems will be the same, but System 2 will experience an adiabatic reversible expansion and System 1 will experience an adiabatic irreversible expansion when the same pile of small masses is removed quasi-statically from the top of each piston.

#### Robert Davidson

I didn't include all the analysis that led up to the results in post # 20. If you like, I can provide that. The graphs include only the final results of the analysis.

Yes, I would very much appreciate it if you could provide all the analysis.

For the process to be quasi-static, we cannot let the piston rise on its own…..

I was not thinking of a quasi-static process. What I had in mind was to compare the final volume for two irreversible constant pressure adiabatic expansions, one with sliding friction and one without.

#### Chestermiller

Mentor
I didn't include all the analysis that led up to the results in post # 20. If you like, I can provide that. The graphs include only the final results of the analysis.

Yes, I would very much appreciate it if you could provide all the analysis.
I'll post this in a little while.

For the process to be quasi-static, we cannot let the piston rise on its own…..

I was not thinking of a quasi-static process. What I had in mind was to compare the final volume for two irreversible constant pressure adiabatic expansions, one with sliding friction and one without.
So you would consider the same comparison as in my previous post, except with sudden removal of mass M at time zero?

#### Chestermiller

Mentor
The starting equations for the frictional analysis I alluded to in post #20 is:
$$dU=nC_vdT=\frac{F}{A}dV-\frac{nRT}{V}dV\tag{1}$$and $$dU=nC_vdT=-\frac{f}{A}dV\tag{2}$$Dividing Eqn. 1 by $nC_vT$ yields:
$$\frac{dT}{T}=\frac{FdV}{AnC_vT}-\frac{R}{C_v}\frac{dV}{V}=\left[\frac{FV}{AnC_vT}-\frac{R}{C_v}\right]\frac{dV}{V}\tag{3}$$
In Eqn. 3, $$\frac{FV}{AnC_vT}=\frac{FV_0}{nC_vAT_0}\frac{(V/V_0)}{(T/T_0)}=\frac{FR}{AC_v}\frac{V_0}{nRT_0}\frac{(V/V_0)}{(T/T_0)}$$
But, from the ideal gas law, $$\frac{nRT_0}{V_0}=P_0$$
Therefore, $$\frac{FV}{AnC_vT}=\frac{F}{P_0A}\frac{C_v}{R}\frac{(V/V_0)}{(T/T_0)}\tag{4}$$
Substituting Eqn. 4 into Eqn. 3 then yields:
$$\frac{dT}{T}=\frac{C_v}{R}\left[\frac{F}{P_0A}\frac{(V/V_0)}{(T/T_0)}-1\right]\frac{dV}{V}=(\gamma -1)\left[\frac{F}{P_0A}\frac{(V/V_0)}{(T/T_0)}-1\right]\frac{dV}{V}\tag{5}$$
We now make the following dimensionless substitutions:
$$\bar{T}=T/T_0$$
$$\bar{V}=V/V_0$$
This then yields:
$$\frac{d\ln{\bar{T}}}{d\ln{\bar{V}}}=-(\gamma -1)\left[1-\frac{F}{P_0A}e^{(ln{\bar{V}}-\ln{\bar{T}})}\right]\tag{6}$$
This is the differential equation I integrated to get the results in post #20.
To get the work terms, I did the following: From Eqn. 2, if I divide by $nC_vT_0$ and integrate, I obtain:
$$\frac{\left[\int_{V_0}^V{\frac{f}{A}dV}\right]}{nC_vT_0}=1-\bar{T}$$
Similarly, for the integral of the gas pressure, I get:
$$\frac{\left[\int_{V_0}^V{PdV}\right]}{nC_vT_0}=(\gamma-1)\frac{F}{P_0A}(\bar{V}-1)+(1-\bar{T})\tag{7}$$

#### Robert Davidson

Chet,

Thanks for the analysis.

Regarding the new example, I had in mind the following:

System 1: massless piston with friction between the piston and cylinder walls, and a pile of small masses sitting on top of the piston, such that the total downward force due to the masses plus the downward force due to atmospheric pressure equals the upward force due to the gas pressure. Or, the pressure upward exerted by the gas equals the pressure downward exerted by the weights plus atmospheric pressure.

System 2: Same as system 1 except there is no friction.

An equal amount of weight is abruptly removed from each system so that the external downward pressure exerted by the weights plus the atmosphere is ½ the original external downward pressure.

To solve: What will be the final volume of the two systems?

If we can agree on the problem statement, I will give it a try on my own, with the benefit of the analysis you did on our previous example, and then post it.

Bob

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#### Chestermiller

Mentor
Chet,

Thanks for the analysis.

Regarding the new example, I had in mind the following:

System 1: massless piston with friction between the piston and cylinder walls, and a pile of small masses sitting on top of the piston, such that the total downward force due to the masses plus the downward force due to atmospheric pressure equals the upward force due to the gas pressure. Or, the pressure upward exerted by the gas equals the pressure downward exerted by the weights.

System 2: Same as system 1 except there is no friction.

An equal amount of weight is abruptly removed from each system so that the external downward pressure exerted by the weights plus the atmosphere is ½ the original external downward pressure.

To solve: What will be the final volume of the two systems?

If we can agree on the problem statement, I will give it a try on my own, with the benefit of the analysis you did on our previous example, and then post it.

Bob
Bob,

Although I don't think that the two cases are really going to be comparable, I think each of them is interesting on its own. So please proceed and let's see what you come up with. (Please note that, in both these cases, the gas experiences a non-quasistatic deformation, and entropy is generated within the gas in each case).

Chet

#### Chestermiller

Mentor
I've been able to analytically solve the differential equation (Eqn. 6) of post #29 for the dimensionless temperature of the gas as a function of the dimensionless volume ratio. The result, which agrees with the numerical solution I presented in post #20 is:
$$\bar{T}=\frac{(1-\xi)}{\bar{V}^{(\gamma-1)}}+\xi \bar{V}$$where $$\xi=\frac{(\gamma-1)}{\gamma}\frac{F}{P_0A}$$

#### Robert Davidson

Hi Chet,

Haven’t forgot this. So far only tackled the easy case (system 2). Here’s what I have so far.

Going to assume dealing with ideal gas (air) with the following initial conditions and properties:

$P_{i}=10atm$, $T_{i}=300K$, $n=1$, $C_{v}=0.2053\frac{l.atm}{mol.K}$, $R_{u}=0.08205\frac{l.atm}{mol.K}$, $k=1.4$

Applying the ideal gas equation to get the initial volume:

$$V_{i}=\frac{nRT_{i}}{P_{i}}=\frac{(1)(0.08205)(300)}{10}=2.46 l$$

Weight is abruptly removed to reduce the external pressure by one half, or to 5 atm, and the gas allowed to expand non-quasi-statically at constant pressure until equilibrium is reached.

Since the final pressure is known, we can (1) determine the relationship between the final temperature and volume by applying the general gas equation to the initial and final states and (2) substitute the relationship into the first law equation for a constant pressure adiabatic process.

Applying the general gas equation:

$$\frac{P_{i}V_{i}}{T_{i}}= \frac{P_{f}V_{f}}{T_{f}}$$

$$T_{f}=\frac{P_{f}V_{f}T_{i}}{P_{i}V_{i}}$$

$$T_{f}=\frac{(5)(V_{f})(300)}{(10)(2.46)}=61V_{f}$$

Applying the first law for an adiabatic process:

$$\Delta U=-W$$

$$nC_{v}(T_{f}-T_{i})=-P_{f}(V_{f}-V_{i})$$

Substituting $T_{f}=61V_{f}$ and the other known values

$$(1)(0.2053)[61V_{f}-300]=-5(V_{f}-2.46)$$

Which gives us $V_{f}=4.22l$ and $T_{f}=257K$, and the work done is

$$W=5(4.22-2.46)l.atm=892j$$

Comparing to the final volume and work done for a reversible adiabatic expansion with the same initial conditions and same final pressures:

$$P_{i}V_{i}^{k}= P_{f}V_{f}^{k}$$

$$V_{f}=6l$$

$$W_{rev}=\frac{(P_{f}V_{f}-P_{i}V_{i})}{1-k}$$

$$W_{rev}= 13.5 l.atm=1368j$$

Which, as expected, is greater than the irreversible work.

The next step is to tackle system 1, which adds piston friction. I would like to bring system 2 to the same final pressure and then compare the work on the surroundings to system 2. I recall that in your analysis of the quasi-static expansion with friction it was necessary to apply an external tensile force to overcome static friction in order to reach the same volume as the non-friction system. It appears that will again be the case here, except it would be in order to reach the same final pressure.

I was thinking about a possible alternative approach, such as assuming the surface of the cylinder at some point changes to a smooth service prior to reaching the maximum static friction force so that the expansion could be completed, albeit without friction work done at the end. But that would probably create more complications than it's worth.

#### Chestermiller

Mentor
I approached this problem differently but, of course, arrived at the same results. My focus was on the final temperature, rather than the final volume.
$$nC_v(T_f-T_i)=-P_f(V_f-V_i)=P_f\left(\frac{nRT_f}{P_f}-\frac{nRT_i}{P_i}\right)$$
Solving for $T_f$ then yields:
$$T_f=\left[\frac{1}{\gamma}+\left(1-\frac{1}{\gamma}\right)\frac{P_f}{P_i}\right]T_i$$For $\gamma=1.4$ and $P_f/P_i=0.5$, this yields $T_f=0.8571T_i$. And, for $T_i=300\ K$, $T_f=257.1K$.

The work is given by: $$W=-\Delta U=-nC_v(T_f-T_i)=-(1)(2.5)(8.314)(257.1-300)=892\ J$$
It'll be interesting to compare the approach and results you get for system 2 with those that I obtained.

#### Chestermiller

Mentor
Here's an interesting paradox for you. As $P_f/P_i\rightarrow 0$, $$\frac{T_f}{T_i}\rightarrow\frac{1}{\gamma}$$and then $$W\rightarrow nC_vT_i\left(1-\frac{1}{\gamma}\right)>0$$But how can this be if $P_f/P_i\rightarrow 0$ implies free expansion.

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#### Robert Davidson

Chet
Interesting. Going away on a two week cruise so won't get back to this till I get back. But I'll be thinking about it on the beaches of Barbados!! (Ha Ha).

#### Chestermiller

Mentor
Chet
Interesting. Going away on a two week cruise so won't get back to this till I get back. But I'll be thinking about it on the beaches of Barbados!! (Ha Ha).
Don't laugh. You and I both know that you actually will be thinking about it, even on the beaches of Barbados.

#### Chestermiller

Mentor
To complete this thread, here is the beginning of my solution for system 2 in which there is a friction force F.

Equation for the initial force balance on the piston: $$P_{gi}A-F-P_{atm,i}A=0$$where A is the piston cross sectional area, F is the friction force, $P_{gi}$ is the initial equilibrium gas pressure, and $P_{atm,i}$ is the initial atmospheric pressure.

Equation for the force balance on the piston during the expansion: $$F_g-F-\frac{P_{atm,i}}{2}A=0$$
where $F_g$ is the force that the gas exerts on the inside piston face during this irreversible expansion.

#### Robert Davidson

Hi Chester,

Finally getting back to this.

Thanks for giving the beginning of the solution for system 2. Before I go too far down the road, I need to make sure I’m on the same page with you.

I have no problem with the force balance during expansion (your second equation). I am having difficulties with the force balance for the initial state (the first equation).

I would like to see the initial gas pressure exactly equal to the externally applied pressure (atmosphere plus weights= 10 atm) as we did for system 1. Since there would be no net externally applied force, there would be no fiction force, $F$, to oppose it. In other words, $F=0$ initially.

Once the externally applied force is halved, the difference between the force of the gas and the externally applied force will be the fiction force $F$, as presented in the second equation (with external pressure = 5 atm).

Do you see any issues with this approach?

#### Chestermiller

Mentor
Hi Chester,

Finally getting back to this.

Thanks for giving the beginning of the solution for system 2. Before I go too far down the road, I need to make sure I’m on the same page with you.

I have no problem with the force balance during expansion (your second equation). I am having difficulties with the force balance for the initial state (the first equation).

I would like to see the initial gas pressure exactly equal to the externally applied pressure (atmosphere plus weights= 10 atm) as we did for system 1. Since there would be no net externally applied force, there would be no fiction force, $F$, to oppose it. In other words, $F=0$ initially.

Once the externally applied force is halved, the difference between the force of the gas and the externally applied force will be the fiction force $F$, as presented in the second equation (with external pressure = 5 atm).

Do you see any issues with this approach?
I find this kind of confusing. We are saying that, initially, there is no friction force, but, as soon as we drop the external pressure, friction kicks in at full force during the entire time that the piston is sliding. And it's still present at the very end. I guess this would be OK, under our assumption that the coefficients of static and kinetic friction are equal. But, during the expansion, the friction force would have to be less than 5 atm times the piston area, or the gas would not be able to expand. OK. Let's see where that takes us.

Chet

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#### Philip Koeck

Hi all.
I'm wondering how certain processes can become irreversible. In the other thread we discussed isothermal processes, but adiabatic processes are just as mysterious to me.
I haven't gone through all the previous posts so I'm entering the discussion somewhat unprepared, but maybe with a new angle.

If I take system 1 depicted in post #6 of this thread and assume a quasistatic, adiabatic expansion of an ideal gas.
Without friction, ΔU of the system will be equal to W done by the system. Q is zero and the process is reversible.

Now I'll assume that the cylinder and piston have zero heat capacity and that the thermal insulation is such that no heat can pass into the environment.
If there's friction between piston and cylinder, the gas has to do extra work to overcome this.
This extra work will be converted directly into heat which goes back into the gas.
I get the impression that nothing has changed compared to the frictionless case.

Does that mean that in this case friction doesn't make the process irreversible?

#### Philip Koeck

... if there is a differential change in volume dV, we have
$$PdV=\frac{mg}{A}dV+\frac{F}{A}dV+\frac{f}{A}dV\tag{1}$$

FIRST LAW OF THERMODYNAMICS APPLIED TO SYSTEM 1, THE GAS:
$$dU=dQ-PdV\tag{2}$$
where dQ is the differential heat transferred from the piston to the gas.

FIRST LAW OF THERMODYNAMICS APPLIED TO SYSTEM 2, THE GAS AND PISTON:
In this case, we need to use the more general form of the first law which includes the change in potential energy of whatever is present with the boundaries of the system (in this case, the piston).
$$dE=dU+d(PE)=dU+\frac{mg}{A}dV=-\frac{f}{A}dV\tag{3}$$where $\frac{F}{A}dV$ is the work done by system 2 on its surroundings. Note that, in this case, there is no heat transfer because the combined system is adiabatic. If we substitute Eqns. 1 into this relationship, we obtain:
$$dU=\frac{F}{A}dV-PdV\tag{4}$$From Eqns. 2 and 4, it follows that the heat transferred from the piston to the gas dQ is given by:
$$dQ=\frac{F}{A}dV\tag{5}$$

Where does the last step of equation 3 come from?

Equation 2 should be dU = dQ - dW in general.

If you write PdV instead of dW, are you not assuming that the expansion is reversible, if P stands for the system's pressure?

#### Chestermiller

Mentor
Hi all.
I'm wondering how certain processes can become irreversible. In the other thread we discussed isothermal processes, but adiabatic processes are just as mysterious to me.
I haven't gone through all the previous posts so I'm entering the discussion somewhat unprepared, but maybe with a new angle.

If I take system 1 depicted in post #6 of this thread and assume a quasistatic, adiabatic expansion of an ideal gas.
Without friction, ΔU of the system will be equal to W done by the system. Q is zero and the process is reversible.

Now I'll assume that the cylinder and piston have zero heat capacity and that the thermal insulation is such that no heat can pass into the environment.
If there's friction between piston and cylinder, the gas has to do extra work to overcome this.
This extra work will be converted directly into heat which goes back into the gas.
I get the impression that nothing has changed compared to the frictionless case.

Does that mean that in this case friction doesn't make the process irreversible?
No. It is reversible for the gas, but not for the combination of gas plus piston. The system and surroundings cannot be restored to the original state without causing a change in something else. Please read the previous posts to come up to speed on this.

#### Chestermiller

Mentor

Where does the last step of equation 3 come from?
$\frac{f}{A}dV$ is the work done by the combination of piston plus gas (the system) on their surroundings (at the outside face of the piston).
Equation 2 should be dU = dQ - dW in general.
No. The general form of the first law of thermodynamics also includes change in the potential energy and kinetic energy of the system: $$\Delta U+\Delta (PE)+\Delta (KE)=Q-W$$The potential energy of the piston changes (case 2).
If you write PdV instead of dW, are you not assuming that the expansion is reversible, if P stands for the system's pressure?
No. The force per unit area at the interface between the system and surroundings (where work is being done by the system, the gas plus piston) is $\frac{f}{A}$.

#### Philip Koeck

No. It is reversible for the gas, but not for the combination of gas plus piston. The system and surroundings cannot be restored to the original state without causing a change in something else. Please read the previous posts to come up to speed on this.
I specified that the cylinder and piston have zero heat capacity. To keep things really simple I would also give the piston zero mass or orient the cylinder horizontally so that the piston doesn't change its potential energy.
I distinguish between two situations: In the first all friction heat goes back into gas, in the second all friction heat goes into the surroundings.

If the cylinder and piston have zero heat capacity they can't receive heat and they can't have any inner energy so they shouldn't be part of the thermodynamic state of either the system (gas) or the surroundings.

In the first case the friction heat goes back into the gas. To me that looks like the first case is indistinguishable from the frictionless case and should be reversible. That could be an example of reversibility despite friction.

In the second case all the friction heat goes into the surroundings and the entropy of the surroundings increases. The system behaves just like in the friction-less case. In sum that should mean that the second case is irreversible.

Could it be that whether or not a particular process is irreversible depends only on where the friction heat goes?

#### BvU

Homework Helper
If the cylinder and piston have zero heat capacity
Meaning they would evaporate due to extremely high temperature ?
If there is friction there is material. If there is material there is non zero heat capacity.

#### Chestermiller

Mentor
I specified that the cylinder and piston have zero heat capacity. To keep things really simple I would also give the piston zero mass or orient the cylinder horizontally so that the piston doesn't change its potential energy.
I distinguish between two situations: In the first all friction heat goes back into gas, in the second all friction heat goes into the surroundings.

If the cylinder and piston have zero heat capacity they can't receive heat and they can't have any inner energy so they shouldn't be part of the thermodynamic state of either the system (gas) or the surroundings.

In the first case the friction heat goes back into the gas. To me that looks like the first case is indistinguishable from the frictionless case and should be reversible. That could be an example of reversibility despite friction.

In the second case all the friction heat goes into the surroundings and the entropy of the surroundings increases. The system behaves just like in the friction-less case. In sum that should mean that the second case is irreversible.

Could it be that whether or not a particular process is irreversible depends only on where the friction heat goes?
In the first case where the frictional heat goes into the gas, the entropy of the gas does increase, and thus, the entropy of the system and surroundings increases. Such a process is, of course, irreversible. If there were no friction, then the entropy of the gas would not increase (but this is a different process, and it has a different final state).

#### Philip Koeck

Meaning they would evaporate due to extremely high temperature ?
If there is friction there is material. If there is material there is non zero heat capacity.
It's a thought experiment! Perfect thermal contact with gas, so the cylinder and piston have the same temperature as the gas. Quasistatic expansion, so the friction heat produced per second is zero. Very exotic material! Just don't try to do it in practice!
I completely understand your concerns. I'm just simplifying the description, hopefully not too much.

#### Chestermiller

Mentor
Quasistatic expansion, so the friction heat produced per second is zero.
Quasistatic does not mean that cumulative friction heat produced is zero.

#### Philip Koeck

I'll make another attempt, this time with equations.
I'll treat only the ideal gas as system and assume a quasistatic, adiabatic expansion.
I'll assume that the cylinder and piston have negligible mass and heat capacity and that the thermal insulation is such that no friction heat can pass into the environment.
I'll discuss two cases, one without friction and one with friction between cylinder and piston.
I'll use the same letters for the forces as in post #8.

The first law for the gas (the system) alone is
dU = dQ - dW

Frictionless case:
dQ = 0
dW = f/A dV
Therefore: dU = - f/A dV

Case with friction:
Here all the friction work is converted to heat that goes into the gas!
dQ = F/A dV
dW = F/A dV + f/A dV
Therefore: dU = - f/A dV

If I assume that the initial state of the system was the same in both cases, then the final state must also be the same, since U changed by the same amount. Therefore dS for the system is the same in both cases.

In the frictionless case dS = 0 for both system and environment.

In the case with friction dS = 0 for the system because of the reasons given above and for the surroundings it's also 0 because no heat flows into them.

I conclude that the case with friction is also reversible.

"Thermodynamics: Gas Expansion with Piston Friction"

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