Find the equilibrium position of this piston

[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.