[Cooper]

Hi,

I am confused about this problem...

Consider a system consisting of 2.0 mol CO_2 (g), initially at 25 C and 10 atm and confined to a cylinder of cross-section 10.0 cm^2. It expands adiabatically against an external pressure of 1.0 atm until the piston has moved 20 cm. Assume the gas is perfect and C_(V,m) = 28.8 J K^-1 mol^-1. Calculate Delta S.

So since entropy is a state function, I understand the transition of the system should be broken up into a series of reversible steps in order to calculate the change in entropy. The answer key suggests an isochoric cooling followed by an isothermal expansion. My question is, why should you not have a third, isobaric, step? How do you know how many reversible steps to break a problem in to? I always assumed for every changing state variable you must have a reversible step for it where its change is zero.

Thanks

[mjc123]

The state variables are related, so they can't all be varied independently. The phase rule tells you that there are 2 degrees of freedom, so 2 steps should always be sufficient.

[Cooper]

The state variables are related, so they can't all be varied independently. The phase rule tells you that there are 2 degrees of freedom, so 2 steps should always be sufficient.

That makes sense, I forgot about that, thanks. But how do you know which two state variables to keep constant? You would get different answers depending on what you choose.

[mjc123]

You shouldn't. State variables depend only on initial and final states, not on the route between them. Can you give examples?

[Cooper]

For the question in the original post,

Isothermal:

[tex]\Delta S=\int\frac{dq}{T}=nRln\frac{V_2}{V_1}=0.65\frac{J}{K}[/tex]

Isobaric:

[tex]\Delta S=\int\frac{dq}{T}=\int\frac{dH}{T}=C_pln(\frac{T_2}{T_1})=-0.087\frac{J}{K}[/tex]
Where C_p = C_v + nR

Isochoric:

[tex]\Delta S=\int\frac{dq}{T}=C_vln(\frac{T_2}{T_1})=-0.067\frac{J}{K}[/tex]

If you chose an isobaric and isochoric process you would get a different answer than if you chose an isothermal and isochoric process. I am sure I am misconceptualizing something.

[Cooper]

The answer key uses the isothermal and isochoric conditions to calculate the total entropy change. Does it have to do with the process being irreversible (expansion against constant external pressure) making it so I cannot use the isobaric step? I have seen the isobaric step used in other questions, but those were for reversible processes.

[mjc123]

I still think it shouldn't make a difference, as state functions depend only on the states, not on the route between them.
First, I think you've made some calculation errors. I make the isobaric value -0.75 J/K and the isochoric -0.58 J/K.
Then you have to make sure you're adding the right quantities. Referring to the diagram below
Isothermal + isochoric = AC + CB = 0.67 - 0.58 = 0.09 J/K
Isobaric + isothermal = AD + DB = -0.75 + 0.84 = 0.09 J/K (Note isothermal DB ≠ AC)
Isobaric + isochoric = AE + EB = 2.93 - 2.86 = 0.07 J/k (discrepancy might be rounding error)
Thus e.g. the isochoric in isothermal + isochoric is not the same as in isobaric + isochoric.

[Cooper]

I am not sure about the numerical differences we had, but the picture really helped me. I calculated Delta T = -0.347 K, so I can draw two isotherms, one at T_1 = 298 K and the other at T_2 = 297.653 K. I was trying to do an isobaric step from point A to V_2, which I cannot do because it requires me to know T_2 at that position. I was incorrectly assuming the temperature at that position was 297.653 K. I never did these problems using graphs before, but it really helps. Thanks for your help. 

[mjc123]

How did you get that ΔT? For A, P = 10 atm, T = 298K, V = nRT/P = 4.89 L. ΔV = 200 cm³, so V₂ = 5.09 L. P₂ = P₁(V₁/V₂)^γ = 9.50 atm. T₂ = PV/nR = 295 K.

I was trying to do an isobaric step from point A to V_2, which I cannot do because it requires me to know T_2 at that position

You can calculate it. You know P₁ and V₂, so T = P₁V₂/nR = 310 K.

[Cooper]

How did you get that ΔT?

[tex]\Delta U=q+w=-p_{ext}\Delta V=C_v\Delta T\\\Rightarrow -1.0 atm(\frac{10^5Pa}{atm})(10.0 cm^2)(20 cm)(\frac{m}{10^2 cm})^3=(28.8\frac{J}{K*mol})(2.0 mol)\Delta T\\\Rightarrow \Delta T=-0.347 K[/tex]

P₂=P₁(V₁/V₂)^γ = 9.50 atm.

I was under the impression you can only use this equation for reversible adiabatic processes, is this not true?

[mjc123]

I was under the impression you can only use this equation for reversible adiabatic processes, is this not true?

You are right, I had forgotten that.
Now I get ΔS = 0.60 J/k by all 3 routes