Entropy change of a non-perfect gas

1. May 17, 2004

alexbib

how could one evaluate a numerical value for the entropy change of a non-perfect gas?

Also,which system has more entropy, a container with x molecules in it, or the same container with 2x of the same molecules?

2. May 17, 2004

Bystander

Entropy change for what process?

3. May 17, 2004

alexbib

well, for example, the change in entropy for a free expansion where the temperature decreases (because gas is not ideal and the Joule-Thomson effect) (assuming you can measure Vi, Vf, Ti, Tf)

4. May 17, 2004

MiGUi

Working with thermal machines, entropy don't depends on material which do the cycle, almost directly... the entropy is:
$$dS \leq \frac{\delta Q}{T}$$
being equal if the process is reversible, and less if the process is not reversible.

Well, if you work with a simple (without interchange of mass) expansive system the entropy will deppend on variations of internal energy:
$$TdS = dU + pdV$$

In your case, the Joule-Kelvin effect:
"Change in temperature of a thermally insulated gas* when it is forced through a small hole or a porous material"
This gas must be a real gas, because ideal gas can't be detained with a porous material since its particles have not volume.

Lets have a open adiabatic recipient with a pair of adiabatic and mobile pistons, one in each side of the open recipient and in the middle we have a porous material.

When we start, the second piston is over the porous material, so all the gas is "at left". When we begin to move the left piston, the gas will be compressed and the right piston will advance to the right, letting particles move across the porous material.

At the end, the left piston will be near the porous material and all the gas will be in the right of the recipient.

Calling p1,V1,T1 the conditions of the gas when it is at left, and p2,V2,T2 at right, lets have:
From 1st. principle, we have:
$$\Delta U = Q - W = -W_{adiabatic}$$
Because the gas crosses the tube in adiabatic conditions, so:

$$W = p_{1}(0-V_{1}) + p_{2}(V_{2}-0) = p_{2}V_{2} - p_{1}V_{1} = - \Delta U = - (U_{2} - U_{1})$$
then...
$$U_{2} + p_{2}V_{2} = U_{1} + p_{1}V_{1} ...$$
$$H_{2} = H_{1}$$
So the entalpy is the same at the begin than at the end, but that process is not reversible so its not isoentalpic.

To calculate the variation of entropy in that proccess, we can use that entropy is an state function to build a isoentalpic proccess between state 1 and 2. Using the definition of entalpy (without interchange of material with the exterior, to simplify): $$dH = -TdS - Vdp$$ and because dH = 0, lets have:
$$TdS = Vdp$$

Solving that differential equation between states 1 and 2 you will have the variation of entropy. Obviously you need the relation f(p,V,T) (that is, the thermal equation) to do.

MiGUi

(If a term is not correctly spelt or so, I'm sorry but I'm not english, I do what I can :) )

Last edited: May 17, 2004
5. May 17, 2004

TALewis

Isn't entropy a state function, so that the change of entropy of a gas between two states is independent of the process?

That being said, coming from an engineering background, if I wanted to find the change in entropy for a gas between two states, I would find some appropriate gas tables with s (specific entropy) values, and subtract s1 from s2.

6. May 17, 2004

MiGUi

Entropy is a state function, so it not deppends on process. Ok, being exact we must imagine a reversible process between state 1 and state 2. For example:

To calculate the variation of entropy of 1 mol of (ice) water at state 1: T = 270 K, p = 1 atm and state 2: (liquid) T = 275 K p = 1 atm we can't do it directly, because the fusion is not a reversible process, so we must imagine a reversible process like this:

We use the intermediate state "0" to avoid the change of phase consequences.

We heat water from T = 270 K to T = 273 K, the variation of entropy is:
$$\Delta S_{1-0} = nRln\frac{T_1}{T_2}$$
Then, we give the latent heat of fusion, to melt ice... so the variation of entropy is:
$$\Delta S_{0} = -nL_f$$
( being with - because is heat that we must give to the system)
And finally we heat from T=273 K to T = 275 K, and the variation of entropy is:
$$\Delta S_{0-2} = nRln\frac{T_1}{T_2}$$

You can test that if you use directly the formula of the variation of entropy in a isobaric process is different from the value that we test.

7. May 17, 2004

MiGUi

Entropy is a state function, so it not deppends on process. Ok, being exact we must imagine a reversible process between state 1 and state 2. For example:

To calculate the variation of entropy of 1 mol of (ice) water at state 1: T = 270 K, p = 1 atm and state 2: (liquid) T = 275 K p = 1 atm we can't do it directly, because the fusion is not a reversible process, so we must imagine a process like this:

We heat water from T = 270 K to T = 273 K, the variation of entropy is:
$$\Delta S_{1-0} = nRln\frac{T_1}{T_2}$$
Then, we give the latent heat of fusion, to melt ice... so the variation of entropy is:
$$\Delta S_{0} = -nL_f$$
( being with - because is heat that we must give to the system)
And finally we heat from T=273 K to T = 275 K, and the variation of entropy is:
$$\Delta S_{0-2} = nRln\frac{T_1}{T_2}$$

You can test that if you use directly the formula of the variation of entropy in a isobaric process is different from the value that we test.

8. May 17, 2004

alexbib

in the last equation, R is what? the ideal gas constant? I've seen this equation before, but instead or R it was Cv

9. May 18, 2004

MiGUi

R is the ideal gas constant. You are right. It must be Cv instead, but that expression is only for ideal gases. I was wrong... forget the formulae then, I wanted to explain the concept without anymore.