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Adiabatic, Ideal gas, changing heat capacity, work calculation

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1. Propane vapour (1kmol) at a pressure of 40 bar and 230 C expands adiabatically to 0.25 bar and 95 C. Determine a)W, b)Delta S, c)The ammount of work obtained if the expansion were done reversibly from the same initial conditions to the final pressure of 0.25bar



2. I am not sure how to calculate work for a irreversible process. Is it zero, a gas that expands adiabatically in an irreversible process loses no kinetic energy right? so then work has to be zero cause heat is zero as well right? Im not sure about delta S either, I think in reversible delta S is zero but in irreversible its always increasing? I think theres is a formula for delta S for ideal This stuff can get confusing, can anyone help me?



The Attempt at a Solution

 

Answers and Replies

  • #2
Andrew Mason
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1. Propane vapour (1kmol) at a pressure of 40 bar and 230 C expands adiabatically to 0.25 bar and 95 C. Determine a)W, b)Delta S, c)The ammount of work obtained if the expansion were done reversibly from the same initial conditions to the final pressure of 0.25bar



2. I am not sure how to calculate work for a irreversible process. Is it zero, a gas that expands adiabatically in an irreversible process loses no kinetic energy right? so then work has to be zero cause heat is zero as well right? Im not sure about delta S either, I think in reversible delta S is zero but in irreversible its always increasing? I think theres is a formula for delta S for ideal This stuff can get confusing, can anyone help me?
[tex]W = \int PdV[/tex]

What is the change in volume? How would you determine that? What is the relationship between P V and T?

AM
 
  • #3
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Change in volume would be determined by the Ideal Gas law. Would you replace P with nRT/V? In this case what is constant because the gas is going to new temperature and volume. Is Temp constant when you replace P?
 
  • #4
If you replace the P from the integration formula a few posts earlier with (nRT/V) and integrate the formula, you will get W=nRT*ln[V(init)/v(fin)]
 
  • #5
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the temperature is changing. I dont see how you can do that. I think I figured it out. From the first law. dU = dQ + dW. If its adiabatic the dQ = 0 and dU = dW.

We can say dU = C*dT for an ideal gas.
 
  • #6
Andrew Mason
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If you replace the P from the integration formula a few posts earlier with (nRT/V) and integrate the formula, you will get W=nRT*ln[V(init)/v(fin)]
That is only the case if T remains constant. But since work is being done and there is no heat flow, internal energy and therefore T must decrease.

AM
 
  • #7
sorry, that was isothermal. adiabatic expansion work can be found by W = (3/2)nR(deltaT)
 
  • #8
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did I figure it out Andrew that internal energy is equal to work and we can say dU = C*dT for Ideal gas.
 
  • #9
Andrew Mason
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Change in volume would be determined by the Ideal Gas law.
In looking at this again, you can only determine the work from the change in volume if this is a reversible process. You have to first determine the change in internal energy. What is the relationship between change in internal energy and work in an adiabatic process?

AM
 
  • #10
Andrew Mason
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did I figure it out Andrew that internal energy is equal to work and we can say dU = C*dT for Ideal gas.
Yes. Change in internal energy = - Work done by gas. [itex]dU = nC_vdT[/itex]

AM
 
  • #11
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this is confusing. It is an irreversible process. I do not have a constant heat capacity. We defined in class a reversible work formula for adiabatic conditions with a constant heat capacity. How do I use that if my heat capacity is not constant.
 
  • #12
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so basically the work from a reversible adiabatic process in an ideal gas with a non constant heat capacity is the same as the work in an irreversible process in an ideal gas with a non constant heat capacity
 
  • #13
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but I thought that if a process is reversible. no work is done because it goes back to its original state. but then i remember that free expansion of gas is an irreversible process so the process cant be reversible. I feel like I am going in circles.
 
  • #14
Andrew Mason
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this is confusing. It is an irreversible process. I do not have a constant heat capacity. We defined in class a reversible work formula for adiabatic conditions with a constant heat capacity. How do I use that if my heat capacity is not constant.
How does the heat capacity vary with temperature? Plug that into:

[tex]\Delta U = \int nC_vdT[/tex]


AM
 
  • #15
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the heat capacity is a function of temperature. Pretty easy integration. its like. Cp/R = A + BT + CT^2. Where A, B, C are constants.
 
  • #16
Andrew Mason
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the heat capacity is a function of temperature. Pretty easy integration. its like. Cp/R = A + BT + CT^2. Where A, B, C are constants.
You are determining internal energy: you must use Cv not Cp. Cv = Cp-R

AM
 

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