Change in entropy as gas expands?

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Homework Statement


http://img846.imageshack.us/img846/5678/problemem.jpg [Broken]


Homework Equations



Pv= nRT
U = Q-W
dS = dQ/T (entropy transferred) + σ (entropy created)

The Attempt at a Solution


Part A.
Assuming that as the gas expands Volume increases and Pressure decreases, thus the Temperature will stay constant.

Part B.
Entropy will increase as the gas expands. Assuming the system is isolated, there will be no entropy transfered in or out.

For entropy created: Because T is constant internal energy U will stay constant, which means that Q = -W = -dPV so dS = -dPV. But the pressure in the problem is not given so I am stuck.

So I was wondering, are my assumptions correct?
  • V and P change but T stays the same
  • no entropy is transfered because the system is isolated
How do I go about solving part b?
 
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Answers and Replies

  • #2
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Any help please?
 
  • #3
Andrew Mason
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Homework Equations



Pv= nRT
U = Q-W
dS = dQ/T (entropy transferred) + σ (entropy created)

The Attempt at a Solution


Part A.
Assuming that as the gas expands Volume increases and Pressure decreases, thus the Temperature will stay constant.
Your explanation is not correct. In an adiabatic reversible expansion pressure decreases and volume increases but temperature also decreases.

Apply the first law. Is the process adiabatic (Q = 0)? Does the gas do any work in this process? Why/why not? What does that tell you about the change in internal energy?

Part B.
Entropy will increase as the gas expands. Assuming the system is isolated, there will be no entropy transfered in or out.

For entropy created: Because T is constant internal energy U will stay constant, which means that Q = -W = -dPV so dS = -dPV. But the pressure in the problem is not given so I am stuck.
Change in Entropy is the integral of dQ/T over a reversible path between the beginning and end states. In order to determine that change you have to first determine a reversible path between the two states. Can you think of such a reversible path? (hint: it is not adiabatic).

AM
 
  • #4
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Thank you for your help. I understand part A now, and I think I can solve Part B also.

Could you check if my answer for part B makes sense?

TdS =INT(dU + PdV)
dS = P/T * INT(dV)
dS = nR/V * INT (dV)
S = nR/V * ln (V2/V1)
S = (0.001 moles * 8.314)/(20cm^3 * (1 m^3/1000000cm^3)) * ln (80/20)
S = 576 J/K

So the overall entropy is increasing, because entropy is being produced by the gas expansion. Entropy is not being transferred in or out of the system.
 
  • #5
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Can anyone help? I have a test tomorrow and the professor said the concepts of this question are going to be in it and I am concerned I am still not understanding entropy.
 
  • #6
Andrew Mason
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Thank you for your help. I understand part A now, and I think I can solve Part B also.

Could you check if my answer for part B makes sense?

TdS =INT(dU + PdV)
dS = P/T * INT(dV)
dS = nR/V * INT (dV)
S = nR/V * ln (V2/V1)
S = (0.001 moles * 8.314)/(20cm^3 * (1 m^3/1000000cm^3)) * ln (80/20)
S = 576 J/K

So the overall entropy is increasing, because entropy is being produced by the gas expansion. Entropy is not being transferred in or out of the system.
Here is how you solve B:

1. determine a reversible path between the beginning and end states. The temperature is constant. The volume increases from 20 to 100 ml. So a reversible path between initial and final states would be an isothermal quasi-static expansion from 20 ml to 100 ml. Such an expansion would do work so there would have to be heat flow into the gas.

2. Apply the first law:

Q = ΔU + ∫PdV (what is ΔU if T is constant?)

3. ΔS = Q/T. Determine ΔS.

AM
 

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