Find the equilibrium position of this piston

[Chestermiller]

The sign of Patm must agree with the sign of mg.

Yes, I've gone back and corrected that algebraic error.

I see you have taken x as measured from the bottom of the cylinder making the terms mg + Patm A always negative which makes sense for work done by the gas but am wrestling a bit with the positive piston kinetic energy but find no other issue with the equation. Interesting approach.

So you see that the work done on the gas, the compression of the gas $(l+H)-x(t)$, and the kinetic energy of the piston are all interacting functions of time during the compression. What do you think happens to the kinetic energy of the piston at very long times when the system equilibrates thermodynamically?

 

[Chetty]

We know the kinetic energy dissipates. But it will all come down to how you distribute the work done by the gas to the internal properties of the gas. To assume that when everything comes to equilibrium, the internal energy of the gas is the only thing that prospers by assigning the work done by the gas or on the gas all goes to Mcv(Tf-T) and none goes to the integral of pdv leaves me wondering.

 

[Chestermiller]

We know the kinetic energy dissipates. But it will all come down to how you distribute the work done by the gas to the internal properties of the gas. To assume that when everything comes to equilibrium, the internal energy of the gas is the only thing that prospers by assigning the work done by the gas or on the gas all goes to Mcv(Tf-T) and none goes to the integral of pdv leaves me wondering.

The integral of pdv is $(mg+P_{atm}A)[x_{final}-(l+H)]$ which is minus the compressional work done by the piston on the gas after infinite time, where $x_{final}$ is the final position of the piston. So what's the problem? From the first law of thermodynamics, since the cylinder is insulated and Q = 0, this will be equal to the change in internal energy of the gas.

 

[Chetty]

If I do what is considered normal adiabatic compression from l+h to l+h-x, I essentially get very close to the same or somewhat higher resulting temperatures than you get from your analysis. So what role are the oscillations playing? I would have expected the temperatures to be somewhat higher because of the dissipation of the oscillations.

 

[Chestermiller]

If I do what is considered normal adiabatic compression from l+h to l+h-x, I essentially get very close to the same or somewhat higher resulting temperatures than you get from your analysis. So what role are the oscillations playing? I would have expected the temperatures to be somewhat higher because of the dissipation of the oscillations.

I have no idea what you did. Here are my results for comparison:

The final height of the piston is $$x_{final}=l+\frac{(\gamma-1)}{\gamma}H$$and the final temperature is $$T_{final}=\frac{x_{final}}{l}T_1$$

What did you get?

 

[Chetty]

I get exactly what you get and the example I am using shows a temperature rise of from 50 F to 86 F compared to an normal adiabatic compression of 50 F to 63 F. And maybe you can tell me how to put math expressions on this page? Much impressed with your condensing of the formula to the above.

 

[Chestermiller]

I get exactly what you get and the example I am using shows a temperature rise of from 50 F to 86 F compared to an normal adiabatic compression of 50 F to 63 F. And maybe you can tell me how to put math expressions on this page? Much impressed with your condensing of the formula to the above.

We recommend using the LaTex Editor. Here is a tutorial: https://www.physicsforums.com/help/latexhelp/

 

[Chetty]

Do you have any suggestions as to how one would go about computing the change in entropy for this process?

 

[Chetty]

Do I need to download some software to use this LaTex or do I need Java?

 

[Chestermiller]

Do you have any suggestions as to how one would go about computing the change in entropy for this process?

Sure. Just use the standard formula for the entropy change of an ideal gas:
$$\Delta S=C_V\ln{(T_f/T_i)}+R\ln{(V_f/V_i)}$$or equivalently $$\Delta S=C_P\ln{(T_f/T_i)}-R\ln{(P_f/P_i)}$$These are per mole.

 

[Chestermiller]

Do I need to download some software to use this LaTex or do I need Java?

Neither. Just do the tutorial. It's very easy.

 

[Chetty]

Back to the change in entropy, both equations that you site are developed by using the perfect gas expression and integrating over the region from point A to B. In the process we are working on, the perfect gas expression does not hold during the oscillations so how are we justified in traversing this region with an expression that is not appropriate?
And in using laTex, I write \int x^2 e^x\,dx and preview and right click and nothing happens?

 

[Delta2]

And in using laTex, I write \int x^2 e^x\,dx and preview and right click and nothing happens?

You have to put the latex code inside $$ (for the latex equation to appear in its own separate line with bigger letters) or inside ## (for the latex equation to appear inside the rest of the text with smaller letters) for example

##\int \frac{1}{2}\pi e^x dx##

will give $\int \frac{1}{2}\pi e^x dx$
Note the double # (or double $) at start AND end of latex code. Also you don't have to type the 1 at the start of the code.

 

[Chestermiller]

Back to the change in entropy, both equations that you site are developed by using the perfect gas expression and integrating over the region from point A to B. In the process we are working on, the perfect gas expression does not hold during the oscillations so how are we justified in traversing this region with an expression that is not appropriate?
And in using laTex, I write \int x^2 e^x\,dx and preview and right click and nothing happens?

You are aware that, to get the change in entropy suffered by a system that was subjected to an irreversible process, you need to devise an alternate reversible path between the same pair of end states and calculate the integral of dq/T for that reversible path, right?

 

[Chetty]

Thanks. No, I was not aware of that. Where does that come from? And who is to say that such a process is valid.

 

[Chestermiller]

Thanks. No, I was not aware of that. Where does that come from? And who is to say that such a process is valid.

When we say that $$\Delta S=\int{\frac{dq_{rev}}{T}}$$that's what we mean. Entropy is a function of state (i.e., it is a physical property of the system, not the process path), and its change depends only on the two end states. For a cookbook recipe on how to determine the entropy change for an irreversible process, see my Physics Forums Insights article https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

 

[Chetty]

Thanks for your reply. I will read with great interest. Also want to thank the individual that responded with the correction needed to use LaTex.

 

[Chetty]

Are you able to tie this all together with the public use of the term entropy? I mean some descriptions are the state of disorder and with that description it would certainly seemed to be path dependent. Is the term misused in the public sector or how can all this be tied together or can it be?

 

[Chetty]

Forget the above request. I understand your comment about state variable and if so it is path independent.

 

[Chetty]

$s2-s1=R log \left( \frac {1+(γ-1)λ)} γ\right)^\left(\frac γ {γ-1} \right)$ $\frac 1 λ \$
$s2-s1=R log\left[ \left( \frac {1+(γ-1)λ)} γ\right)^\left(\frac γ {γ-1} \right) \frac 1 λ \right]$

$T2=\frac {(ϒ-1)λ+1} ϒ\ T1$ ; λ is the pressure ratio
I wanted to let you know that you helped me resolve an issue I had with a textbook printed in 1957. The first expression above for change in entropy was what was printed in the book as the answer to a problem very similar to the one we have been discussing. This expression just did not make sense. The second expression is the correction based on redoing the problem with your help. Now I suppose one could guess what they meant, but until I understood where the expression came from that was not sufficient. I have never before approached a problem by solving a simple representation and backed out the general expression. I always solved for the general expression and then applied it to the simple case. Following your lead changed the approach and solved the problem.
Incidentally, the problem in this case was an additional weight was added to a stable piston and a new equilibrium found. Thank you.

 

[Chestermiller]

$s2-s1=R log \left( \frac {1+(γ-1)λ)} γ\right)^\left(\frac γ {γ-1} \right)$ $\frac 1 λ \$
$s2-s1=R log\left[ \left( \frac {1+(γ-1)λ)} γ\right)^\left(\frac γ {γ-1} \right) \frac 1 λ \right]$

$T2=\frac {(ϒ-1)λ+1} ϒ\ T1$ ; λ is the pressure ratio
I wanted to let you know that you helped me resolve an issue I had with a textbook printed in 1957. The first expression above for change in entropy was what was printed in the book as the answer to a problem very similar to the one we have been discussing. This expression just did not make sense. The second expression is the correction based on redoing the problem with your help. Now I suppose one could guess what they meant, but until I understood where the expression came from that was not sufficient. I have never before approached a problem by solving a simple representation and backed out the general expression. I always solved for the general expression and then applied it to the simple case. Following your lead changed the approach and solved the problem.
Incidentally, the problem in this case was an additional weight was added to a stable piston and a new equilibrium found. Thank you.

Looks like you made some good progress learning LaTex. Here's some additional info:

To do natural log, \ln{} with the argument between the squiggly brackets

To do a subscript, s_2

To do delta, \Delta S (there must be a space between the delta and the letter symbol.

 

[Chetty]

Your writing the integral for the pressure force from the gas is extremely helpful in gaining understanding. But it caused me some pause when I wrote the expression from a different reference point and it seemed to change the piston kinetic energy from a plus contribution to a minus. I think the following resolves the issue. You have probably done all this before but I thought it a good opportunity for me to develop my LaTex skills.

With reference point +x1 up from the bottom of the cylinder:

$m \frac{dv} {dt}=-(mg+PatmA)+PA$

Here I use $\frac {dv} {dt}=\frac {dv} {dx}\frac {dx} {dt}$

$\int PA\,dx1 =(mg+Patm A)[x1-(l+h)]+\frac 1 2\ mv^2$

Using $V1=Ax1\ and\ dV1=Adx1$

$\int PA\,dV1 =-(mg+Patm A)[(l+h)-x1]+\frac 1 2\ mv^2$

Now with reference point +x2 down from the top of the cylinder:

$m \frac{dv} {dt}=(mg+PatmA)-PA$

$\int PA\,dx2 =(mg+Patm A)[x2]-\frac 1 2\ mv^2$

Using $V2=A(l+h-x2)\ and\ dV2=-Adx2$

$\int PA\,dV2 =-(mg+Patm A)[x2]+\frac 1 2\ mv^2$

$x1+x2=l+h$

Everything is a function of time until the kinetic energy dissipates.
So as you have said, this expression is indeed the pdv expression of the gas.

 

[Chetty]

I just read your write up on entropy and have another question. When I asked earlier about the change in entropy for the piston problem we were working on, you mentioned that entropy was a state variable which seems accurate and I interpret it to mean that for an equilibrium condition there is one value of entropy or if this is too specific, then at least there is only one value for the difference in entropy between two states of equilibrium. In your write up, if I understand, any irreversible path will give a lower value for the change than will a reversible path. What point am I missing?

 

[Chestermiller]

I just read your write up on entropy and have another question. When I asked earlier about the change in entropy for the piston problem we were working on, you mentioned that entropy was a state variable which seems accurate and I interpret it to mean that for an equilibrium condition there is one value of entropy or if this is too specific, then at least there is only one value for the difference in entropy between two states of equilibrium. In your write up, if I understand, any irreversible path will give a lower value for the change than will a reversible path. What point am I missing?

Any irreversible path between the same two end states will give a lower value of the integral of $dq/T_B$ than a reversible path, where $T_B$ is the temperature at the boundary between the system and surroundings (i.e., through which dq passes) along the path.