# Thermodynamics Problem: Cylinder-Piston System with Friction

• MathError
In summary: But I'm not sure why. Can you please help me?In summary, In the first case the work done by the envirorment on the system is less because the friction force is not considered. In the second case the work done by the envirorment on the system is the same because the friction force is considered.
MathError
Homework Statement
Find the expression of work during the compression of a gas in presence of friction froce between piston and cylinder
Relevant Equations
P_e=external pression;
P_i=pression of the gas inside the cylinder (uniform pression);
F_f=friction force;
L=work on the system
L_a=friction work
Q=exchanged heat
E_t=total energy;
mass of piston=0;
A=area of the piston
Hi guys,
First of all I'm sorry for my bad english I'll try to be as clear as possible.
I have tried to solve this problem to understand the First Law of Thermodinamics: Q+L=ΔE_t
In fact I know L (in the current convention) is the work which the envirorment does on the system but I don't understand which contributes I have to include in L.
let we start from simple case: in this case F_f=0, so there is no friction force between piston and cylinder. In the hypothesis of quasistatic process the resulting force on the piston is zero so:
P_e=P_i.
The work made by envirornment on the system can be calculated as:
L=P_e*A*Δx=P_e*ΔV=P_i*ΔV
If there is friction force the problem becomes more complicated.
First all since the resulting force on the piston is still zero, in the hypothesis of compression we can find that:
P_e=P_i+F_f/A;
Also, there are two friction forces: the first one which is applied on the piston and it's considered in the previous equation and a second one which is applied on the cylinder but which doesn't produce work because it's applied to fixed points.
Now we have to choose our system: we can consider only the gas (A) or system which includes gas and piston (B).

Case A
In this case, in my opinion, the force on the system is:
F=P_i*A
In fact external pressure and friction force aren't applied to the surface of control of the system but to the piston.
As you can see in the photo, the the resulting work is:
L=P_e*ΔV+L_a

Case B
Including the piston in the system, the external forces applied to the surface of control are:
F=P_e*A-F_f
In this case the resulting work is different from the previous one and it values:
L=P_e*ΔV-L_a

I think both the results are wrong, anyway it's impossibile the two works are different because I only changed the point of view of the system. I hope I have been enough clear and
you can give me an help.
Thanks you

Last edited by a moderator:
Thanks Chestrmiller for your answer. I have read the discussion and I think I have understood better the problem.
First of all I'm happy to see I have made a few good observations, friction forces on the cylinder didn't do work because cylinder is fixed. Then also work expression seems correct in both cases.
However I made a few mistakes in the chooice of the system. In particular I think the two systems are the same system indeed. When I consider gas only or both gas and piston, it's the same case since piston has to be considered part of surface of control.
Anyway the biggest error is not consider friction heat absorbed by system during compression.
So I have repeated the exercise and I have got these results.
However I'have still have a doubt. If I look to the work which envirorment do on system I have the following cases:
L=P_i*ΔV (Gas Only) (1)
L=P_e*ΔV (Gas+Piston+Cylinder) (2)
At a frist sight I could think that work done in first case is lower than in second one indeed in first the envirorment surrenders heat to system so that lost energy is equal in the two cases and the First Law becomes:
P_e*ΔV=ΔE_tot (3)
It's clear the first member of this eqaution is a kind of work and it's tha same in both cases ("it's the work we are looking for") so I would like to know what the difference is between the work which figures in the First Law (3) and the work written in the equations (1) and (2) which are different.

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MathError said:
Thanks Chestrmiller for your answer. I have read the discussion and I think I have understood better the problem.
First of all I'm happy to see I have made a few good observations, friction forces on the cylinder didn't do work because cylinder is fixed. Then also work expression seems correct in both cases.
However I made a few mistakes in the chooice of the system. In particular I think the two systems are the same system indeed. When I consider gas only or both gas and piston, it's the same case since piston has to be considered part of surface of control.
Anyway the biggest error is not consider friction heat absorbed by system during compression.
So I have repeated the exercise and I have got these results.
However I'have still have a doubt. If I look to the work which envirorment do on system I have the following cases:
L=P_i*ΔV (Gas Only) (1)
L=P_e*ΔV (Gas+Piston+Cylinder) (2)
At a frist sight I could think that work done in first case is lower than in second one indeed in first the envirorment surrenders heat to system so that lost energy is equal in the two cases and the First Law becomes:
P_e*ΔV=ΔE_tot (3)
It's clear the first member of this eqaution is a kind of work and it's tha same in both cases ("it's the work we are looking for") so I would like to know what the difference is between the work which figures in the First Law (3) and the work written in the equations (1) and (2) which are different.
I don't understand what you are asking. All three of your equations are correct. Eqn. 1 applies to the system consisting of the gas only, while Eqns. 2 and 3 apply to the system consisting of the combination of gas and piston.

I have understood almost each point of the problem. My dubst is that equation (1) and (2) could pull to think the two works are different and they are indeed. However in the first case envirorment work is lower but thare is also heat trasmission.
Thank you for the help

## 1. What is a cylinder-piston system with friction?

A cylinder-piston system with friction is a thermodynamic problem that involves a cylinder filled with gas and a piston that can move inside the cylinder. The friction between the piston and the cylinder walls affects the movement and energy transfer of the gas.

## 2. How does friction affect the thermodynamics of this system?

Friction in this system causes energy to be lost in the form of heat, which decreases the efficiency of the system. It also affects the pressure and volume of the gas, leading to changes in temperature and work done by the gas.

## 3. What is the first law of thermodynamics and how does it apply to this problem?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. In this problem, the energy lost due to friction is converted into heat, which affects the overall energy balance of the system.

## 4. How does the second law of thermodynamics apply to this problem?

The second law of thermodynamics states that the total entropy of a closed system always increases over time. In this system, the friction between the piston and cylinder walls increases the entropy of the gas, leading to a decrease in the overall efficiency of the system.

## 5. How can this problem be solved using thermodynamic principles?

This problem can be solved by applying the principles of thermodynamics, such as the first and second laws, to analyze the energy transfer, work done, and changes in temperature and entropy of the system. Mathematical equations and calculations can be used to determine the effects of friction and find solutions to optimize the system's efficiency.

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