Adiabatic Free Expansion and Reversible Isothermal Path

[says]

Homework Statement

Consider an ideal monatomic gas that undergoes an adiabatic free expansion starting from equilibrium state A with volume 500 cm³, pressure 40 kPa and temperature 300K to state B, which has a final equilibrium volume of 1000 cm³.

1. Construct an reversible isothermal path that joins the initial and final states and show it on a schematic p-V diagram.
2. Calculate the total entropy change between the two states using the reversible path.
3. Would you expect this to be the same as the entropy change for the irreversible free expansion?

Homework Equations

1) P₁*V₁ = P₂*V₂
ΔT for free adiabatic expansion = 0

2) ΔS = nRln(V_f/V_i)

3) Ideal Gas Law pV=nRt

The Attempt at a Solution

1) 40*500 = P₂ * 1000
P₂ = 20 kPa

I've attached a P-V diagram. What I'm failing to understand is how i draw a reversible isothermal path that joins the states. If ΔT in an adiabatic free expansion = 0, and then I'm asked to draw a reversible isothermal path that joins the initial and final states together then wouldn't this be the exact same path because ΔT for isothermal path = 0?

2) P₁ = 40,000Pa
V₁ = 0.0005m³
T = 300 K
R (gas constant) = 8.3144
∴ = PV / RT = n = 0.008018217

ΔS = nRln(V_f/V_i)

ΔS = (0.008018217)*(8.3144)* ln(0.0005/0.001)
= - 0.046209810 J/K

3) The entropy change in the adiabatic free expansion is the same as in question two, only it is positive because entropy is increasing.

 

[ehild]

Homework Statement

Consider an ideal monatomic gas that undergoes an adiabatic free expansion starting from equilibrium state A with volume 500 cm³, pressure 40 kPa and temperature 300K to state B, which has a final equilibrium volume of 1000 cm³.

1. Construct an reversible isothermal path that joins the initial and final states and show it on a schematic p-V diagram.
2. Calculate the total entropy change between the two states using the reversible path.
3. Would you expect this to be the same as the entropy change for the irreversible free expansion?

Homework Equations

1) P₁*V₁ = P₂*V₂
ΔT for free adiabatic expansion = 0

2) ΔS = nRln(V_f/V_i)

3) Ideal Gas Law pV=nRt

The Attempt at a Solution

ΔS = nRln(V_f/V_i)

ΔS = (0.008018217)*(8.3144)* ln(0.0005/0.001)

Vi and Vf are interchanged.[/QUOTE]

 

[andrevdh]

A reversible process is exactly what it says, that is you can at any stage stop and reverse the process
and the system will be exactly like before. This means that the process need to be well-behaved or as it
is called - in equilibrium - at all stages, no funny stuff going on like friction or tubulence is allowed. Free
expansion is an example of an irreversible process, since the gas expands in vacuum. This means that the
pressure is probably varying through the gas while it is expanding and it definitely is not in equilibrium during the expansion.

 

[Chestermiller]

Also, for the free expansion, if P represents the force per unit area applied by the surroundings to the gas, then the P-V diagram for the free expansion is incorrect.

Chet

 

[says]

Also, for the free expansion, if P represents the force per unit area applied by the surroundings to the gas, then the P-V diagram for the free expansion is incorrect.

Chet

Thanks! I understand that the adiabatic p-v diagram is incorrect now. I only have to provide the isothermal p-v diagram.

 

[Chestermiller]

Thanks! I understand that the adiabatic p-v diagram is incorrect now. I only have to provide the isothermal p-v diagram.

Aside from the correction that ehild indicated, very nice analysis.

Chet

 

[says]

I think I've misread question 1. I don't think it's asking for isothermal compression, but isothermal expansion. In this case:

ΔS = nRln(Vf/Vi)

ΔS = (0.008018217)*(8.3144)* ln(0.001/0.0005)
= 0.046209810 J/K (positive entropy and not negative)