# Homework Help: Moving Boundary (Piston/Cylinder) -- Work Done

1. Apr 15, 2017

### mech-eng

1. The problem statement, all variables and given/known data
a)
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?

2. Relevant equations

1) dw=Pdv; P1V1^n=P2V2^n

3. The attempt at a solution

1)

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.

Last edited by a moderator: Apr 15, 2017
2. Apr 16, 2017

### Staff: Mentor

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. Apr 16, 2017

### mech-eng

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. Apr 16, 2017

### Staff: Mentor

Sure. Why not?
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. Apr 16, 2017

### mech-eng

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.

Last edited: Apr 16, 2017
6. Apr 16, 2017

### Staff: Mentor

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.

7. Apr 16, 2017

### mech-eng

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?

Thank you.

Last edited: Apr 16, 2017
8. Apr 16, 2017

### Staff: Mentor

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?

9. Apr 16, 2017

### mech-eng

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. Apr 16, 2017

### Staff: Mentor

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.

11. Apr 17, 2017

### mech-eng

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?
Thank you.

Last edited: Apr 17, 2017
12. Apr 17, 2017

### Staff: Mentor

13. Apr 17, 2017

### mech-eng

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.

Last edited: Apr 17, 2017