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Entropy for an irreversible adiabatic process?

by faceoclock
Tags: adiabatic, entropy, irreversible, process
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faceoclock
#1
Mar8-10, 03:24 PM
P: 5
1. The problem statement, all variables and given/known data
Calculate the entropy for an irreversible adiabatic expansion of an ideal gas which goes from (Vi, Pi) = (9.01L, 3.00 bar) to (Vf, Pf) = (15.44L, 1.50 bar)


2. Relevant equations

[tex]\Delta[/tex]S = [tex]\int dq_{reversible}/T[/tex]

3. The attempt at a solution
I understand that entropy is a state function, and thus independent of path. So the above equation can be used to calculate the change in entropy for an irreversible adiabatic expansion. However, what I don't understand is how the entropy can possibly be non-zero. I mean, in a reversible adiabatic expansion which has the same (Vi,Pi) and (Vf,Pf) as the irreversible case, q = 0 by definition. So if [tex]dq_{reversible}[/tex] = 0 then how can [tex]\Delta[/tex]S be non-zero?


Any help as to what I'm missing here will be highly appreciated!

Edit: having written this question out, I think I might be on to something. I don't think it's possible for the reversible expansion to be adiabatic because U is a path function, and must be the same for reversible and irreversible case. In irreversible case, delta U = w. In the reversible case, w is different, so the remainder of the energy must be necessarily in form of heat, which is q(reversible). Am I on the right track?
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passingthru
#2
Mar21-10, 02:23 PM
P: 19
Your problem doesn't state how the volume was increased, whether it was done by the gas, or by opening a divider of some sort. . The first law for non-diffusive systems is $d U = d Q -d W$. In this problem, $d U = 0$ so that $d Q = d W = T d S = P dV$. From the ideal gas law, $PV = nRT$. Using this to substitute for $P$, we have $dS = \frac{nRdV}{V}$. Integrating yields $\Delta S = nR\;ln \left( \frac{V_f}{V_i} \right)$. Thus, $Q$ and $S$ are not zero.

The reason I asked about how your volume was increased is because there is an irreversible process called free expansion where a divider is removed in order to increase the volume. The gas does no work in increasing the volume. The kinetic energy of the molecules stays the same. This is a question in the GRE Physcis practice bulletin, question 47. I have three different thermo texts that do not agree. It seems to me that the molecules hit the inner surface of the container with the same amount of force as before, but now the force is applied to a larger inner surface area, so the pressure, which is force per unit area, would go down. Since the kinetic energy of the molecules stays the same, it would seem that the temperature must stay the same because of the kinetic theory of gases. What stumps me is that the texts say that $dQ$ and $dW$ are zero, so the above calculation would not be correct, but they give this result as the correct answer to the question. Can anyone explain this?
Andrew Mason
#3
Mar23-10, 08:52 AM
Sci Advisor
HW Helper
P: 6,654
Quote Quote by faceoclock View Post
1. The problem statement, all variables and given/known data
Calculate the entropy for an irreversible adiabatic expansion of an ideal gas which goes from (Vi, Pi) = (9.01L, 3.00 bar) to (Vf, Pf) = (15.44L, 1.50 bar)


2. Relevant equations

[tex]\Delta[/tex]S = [tex]\int dq_{reversible}/T[/tex]

3. The attempt at a solution
I understand that entropy is a state function, and thus independent of path. So the above equation can be used to calculate the change in entropy for an irreversible adiabatic expansion. However, what I don't understand is how the entropy can possibly be non-zero. I mean, in a reversible adiabatic expansion which has the same (Vi,Pi) and (Vf,Pf) as the irreversible case, q = 0 by definition. So if [tex]dq_{reversible}[/tex] = 0 then how can [tex]\Delta[/tex]S be non-zero?
The reversible path between these two states involves doing work (ie expansion occurs against an external pressure infinitessimally lower than the internal gas pressure). Since the temperature, hence internal energy does not change ([itex]\Delta U = 0[/itex]), [itex]\Delta Q = W[/itex]. So heat must be added to reach the final state. If heat is added, entropy increases: dS = dQ/T. Simply put, the reversible path between an adiabatic irreversible expansion is not adiabatic.

AM

Count Iblis
#4
Mar23-10, 09:28 AM
P: 2,158
Entropy for an irreversible adiabatic process?

but they give this result as the correct answer to the question. Can anyone explain this?
http://en.wikipedia.org/wiki/Talk:Work_(thermodynamics)


consider doing the free expansion experiment in infintessimal steps. In the enclosure that is filled up to volume V_0 and th rest is vacuum. We add an extra boundary at V_0 + delta V_0. The space between V_0 and V_0 + delta V_0 is vacuum. We then release the gas so that it fills the volume V_0 + delta V_0. We choose Delta V_0 extremely small, much smaller than the collision length between the molecules. CLearly, in this case P is almost well defined during the expansion and can be taken to be the intitial pressure with negligible error.
Now, what is the meaning of p dV here? The answer is as follows. We are doing an infinitessimal expansion. Before and after the expansion the gas can always be described by thermodynamics. But because in this case the change is infintessimal, the equation
dE = T dS - P dV
holds. This is not because P dV is work and T dS is heat, but simply because E is a state variable that can be specified by S and V. We have a function E(S, V) and the partial derivatives S and V are T and -P (because in the special case where the changes happen in a reversible we know what they are). In this case the change is irreversible and P dV is definitely not work and T dS not the heat supplied. Instead what we do know is that dE = 0, therefore we must have that
T dS = P dV
So, we see that the entropy of the system increases, despite the fact that no heat is supplied to the gas. So, the equation dS = dQ/T doesn't hold here. Note that dS = dQ/T holds when supplying heat in quasistatic way in which internal equilibrium is always maintained. In this case the gas in not in thermal equilibrium during the infinitessimal expansion.
We can, however, still think of the P dV term as work that we could have extracted from the system, had the gas bumped into a movable piston. However, since the "piston" in this case is locked this energy gets dissppated in the gas itself and this is then effectively the same as adding the same amount of energy to the system as heat. So, that's why we get dS = P dV/T.
From the mechanical POV, you could say that since we don't extract work from the system, we should have p = 0 in some sense, that p not being the same as the thermodynamic pressure. But then p is the force exerted on the gas and V is taken to be te volume of the extended volume V_0 + delta V_0. Then the pressure is zero, until the gas bumps into the boundary of V_0 + delta V_0, but the volume doesn't change so no work is done. Count Iblis (talk) 15:17, 17 March 2010 (UTC)



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