Moving Boundary (Piston/Cylinder) -- Work Done

In summary: It is not necessary to have a deep understanding of molecular structures and polytropic exponents in order to understand this problem. The main equation to use is the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature. This equation can be rearranged to give P = nRT/V.In a constant pressure process, the pressure remains constant while the volume changes. This means that the temperature must also change in order for the ideal gas law to hold. In a constant temperature process, the temperature remains constant while the volume changes. This means that the pressure must also change in order for the ideal gas
  • #1
mech-eng
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Homework Statement


a) [/B]There is a piston-cylinder arrangement; initial pressure is 200 kPa, initial volume is 0.04m^3. Volume of the gas increases to 0.1 m3 while the pressure remains constant, ie a constant pressure process. Calculate the work.

b)
There is a piston-cylinder arrangement; initial pressure is 200 kPa, initial volume is 0.04m^3. Volume increases to 0.1 m3 while its temperature is constant, ie a constant temperature process.

c) There is a piston-cylinder arrangement; initial pressure is 200 kPa, initial volume is 0.04m^3. Final volume is 0.1 m3 but this time PV^1.3=constant during the process ie a polytropic process with n=1.3.


My question is: 1) Does constant pressure process always give the biggest work?
2) What polytropic exponent gives the biggest work?
3) What would be the case for a isentropic process with the same final volume?
4) How can we understand if the energy of the system increases or decreases, taking the
gas as a system?

Homework Equations



1) dw=Pdv; P1V1^n=P2V2^n[/B]

The Attempt at a Solution



1) [/B]
image011.jpg
1) as you can see from the picture that constant pressure gives the biggest work, but I cannot be sure if it is always so?

2) if 1 is correct analyse then the answer to 2 should be polytropic exponent is 0.

3) I do not know this now but let me wait for it. I can answer it later.

4) I am not sure if this can be determined because we do not know what energy is given to the system so that we could use the equation Q-W=ΔE

http://home.iitk.ac.in/~suller/lectures/lec5.htm

Thank you.
 
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  • #2
Your answers look right. With regard to 4, neglecting non-ideal gas effects, the internal energy is a monotonic increasing function only of temperature.
 
  • #3
But I cannot be sure if constant pressure process can be named as polytropic with exponent 0.

For 4, in the process a, b, c, constant pressure expansion, constant temperature expansion, expansion with polytropic exponent 1.3 respectively, only temperature is contant clearly for b by definition, but how can I determine the behavior of temperature for a and c. From what equation?

Thank you.
 
  • #4
mech-eng said:
But I cannot be sure if constant pressure process can be named as polytropic with exponent 0.
Sure. Why not?
For 4, in the process a, b, c, constant pressure expansion, constant temperature expansion, expansion with polytropic exponent 1.3 respectively, only temperature is contant clearly for b by definition, but how can I determine the behavior of temperature for a and c. From what equation?

Thank you.
In c, we know that the temperature decreases for adiabatic reversible expansion. For b, the temperature is constant, but the pressure still decreases. How do you think the b process would have to be changed in order for the pressure not to decrease, as in a?
 
  • #5
Chestermiller said:
Sure. Why not?

In c, we know that the temperature decreases for adiabatic reversible expansion.
Sorry for my poor thermodynamics knowledge but:
For c, the only information is it is polytropic expansion with exponent 1.3 but you say it is adiabatic reversible expansion. It seems only overlaping is "expansion". But does polytropic implies that it is adiabatic? I think no, because as I know it, an adiabatic process is of the kind throught which there is no heat transfer.
Do you understand it is reversible because it should be quasi-equilibrium?

Thank you.
 
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  • #6
mech-eng said:
Sorry for my poor thermodynamics knowledge but:
For c, the only information is it is polytropic expansion with exponent 1.3 but you say it is adiabatic reversible expansion. It seems only overlaping is "expansion". But does polytropic implies that it is adiabatic? I think no, because as I know it, an adiabatic process is of the kind throught which there is no heat transfer.
Do you understand it is reversible because it should be quasi-equilibrium?

Thank you.
Oops. I got it in my head that, even though it said polytropic, I was thinking adiabatic. Actually, n=1.3 is pretty close to being adiabatic. For a monoatomic molecule, if the expansion is adiabatic, n = 1.67; if the molecule is diatomic, n = 1.4. Both these will involve temperature decrease. And, n = 1.0 means isothremal. So, 1.3 will also involve temperature decrease. An n = 0 will have to involve temperature increasing.
 
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  • #7
Chestermiller said:
Oops. I got it in my head that, even though it said polytropic, I was thinking adiabatic. Actually, n=1.3 is pretty close to being adiabatic. For a monoatomic molecule, if the expansion is adiabatic, n = 1.67; if the molecule is diatomic, n = 1.4. Both these will involve temperature decrease. And, n = 1.0 means isothremal. So, 1.3 will also involve temperature decrease. An n = 0 will have to involve temperature increasing.

This information is far above my level. I must do a study for this. I have current thermodynamics books but I do not remember such an information that relates molecular structures of gases, polytropic exponents and types of the processes. I only encountered when ##the## ##exponent=1##, ##it## ##is## ##isothermal## because I know the formula ##PV=mRT##, and when the ##PV## is constant then ##T## is constant. So would you give me an advice that which chapter I should refer to?

But if the answer is easy would you explain how we know the following?

Chestermiller said:
Actually, n=1.3 is pretty close to being adiabatic. For a monoatomic molecule, if the expansion is adiabatic, n = 1.67; if the molecule is diatomic, n = 1.4.

Thank you.
 
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  • #8
OK. Let's start out with the polytropic eqatuion. If we use the ideal gas law to substitute for the pressure in this equation, we get:
$$\frac{mRT}{V}V^n=C$$
From this, it follows that $$TV^{n-1}=C'$$where C' is a different constant. From this equation, what can you tell me about whether T increases or decreases with increasing V if n = 1.3, n = 1.0, and n = 0?
 
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  • #9
Chestermiller said:
OK. Let's start out with the polytropic eqatuion. If we use the ideal gas law to substitute for the pressure in this equation, we get:
$$\frac{mRT}{V}V^n=C$$
From this, it follows that $$TV^{n-1}=C'$$where C' is a different constant. From this equation, what can you tell me about whether T increases or decreases with increasing V if n = 1.3, n = 1.0, and n = 0?

Yes, a) when n=0, equation becomes T/V=C so it is clear that when volume increases temperature is also increases but the curve is not linear.
b) when n=1, equation becomes T=C', I cannot understand this equation because I cannot relate it to volume.
c) when n=1.3 equation becomes TV^0.3=C', because exponent is bigger than 0, then when volume increases then T decreases.
The option b is problem now.

Thank you.
 
  • #10
mech-eng said:
Yes, a) when n=0, equation becomes T/V=C so it is clear that when volume increases temperature is also increases but the curve is not linear.
This is the case where the pressure is constant.
b) when n=1, equation becomes T=C', I cannot understand this equation because I cannot relate it to volume.
As the volume increases, the temperature remains constant. This is the isothermal case. The pressure decreases as the volume increases, such that PV=constant.
 
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  • #11
Chestermiller said:
This is the case where the pressure is constant.

As the volume increases, the temperature remains constant. This is the isothermal case. The pressure decreases as the volume increases, such that PV=constant.
1. Is my comment is right for c?
2. I understand that in the isothermal case the temperature and volume are indipendent of each other but you have also mentioned about ##monoatomic## and ##diatomic## cases and ##adiabatic## process, how can I understand this from following formulas you have given?
Chestermiller said:
OK. Let's start out with the polytropic eqatuion. If we use the ideal gas law to substitute for the pressure in this equation, we get:
$$\frac{mRT}{V}V^n=C$$
From this, it follows that $$TV^{n-1}=C'$$where C' is a different constant. From this equation, what can you tell me about whether T increases or decreases with increasing V if n = 1.3, n = 1.0, and n = 0?

Thank you.
 
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  • #12
mech-eng said:
1. Is my comment is right for c?
I already addressed this in post #6.
2. I understand that in the isothermal case the temperature and volume are indipendent of each other but you have also mentioned about ##monoatomic## and ##diatomic## cases and ##adiabatic## process, how can I understand this from following formulas you have given?
The formulas I gave are adequate to answer the question without considering monoatomic gases, diatomic gases, and adiabatic paths. If you want to learn more about these anyway, continue in your studies of thermodynamics.
 
  • #13
Chestermiller said:
I already addressed this in post #6.

The formulas I gave are adequate to answer the question without considering monoatomic gases, diatomic gases, and adiabatic paths. If you want to learn more about these anyway, continue in your studies of thermodynamics.

1) In post #6 when you mentioned these atomic structures, I was confused. Yes the formulas are adequate not to consider atomic structures. But I still wonder why did you mentioned them?

2) Adiabatic, as I know it, implies that there is no heat transfer through the process but we have analysed temperature and volume behaviors so we have no direct information about heat transfers. Yes we can delay this and after the study we can think again in the future.

Thank you.
 
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What is the concept of moving boundary (piston/cylinder) in thermodynamics?

The concept of moving boundary in thermodynamics refers to the process of changing the volume of a system, such as a gas or liquid, by moving a piston inside a cylinder. This movement of the boundary allows for work to be done on or by the system.

How is work done calculated in a moving boundary system?

In a moving boundary system, work done is calculated by multiplying the force applied to the piston by the distance it moves. This can be represented by the equation W = F x d, where W is the work done, F is the force applied, and d is the distance the piston moves.

What factors affect the amount of work done in a moving boundary system?

The amount of work done in a moving boundary system is affected by the magnitude of the force applied to the piston, the distance the piston moves, and the type of process being carried out (e.g. isothermal, adiabatic, etc.). The type of gas and its properties, such as pressure and temperature, also play a role in determining the work done.

Is the work done in a moving boundary system always positive?

No, the work done in a moving boundary system can be either positive or negative. If the force applied is in the same direction as the movement of the piston, then the work done is positive. However, if the force is in the opposite direction, the work done is negative.

How is moving boundary (piston/cylinder) work related to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. In a moving boundary system, work is a form of energy transfer, so the work done is related to the change in internal energy of the system. This relationship is represented by the equation W = ΔU + Q, where W is the work done, ΔU is the change in internal energy, and Q is the heat added or removed from the system.

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