Adiabatic, Ideal gas, changing heat capacity, work calculation

In summary, the conversation discusses the determination of work (W) and change in entropy (Delta S) for the adiabatic expansion of propane vapor. The formula for work is determined to be W = nRT*ln[V(init)/v(fin)], and the formula for change in internal energy is dU = nCv*dT. The relationship between change in internal energy and work in an adiabatic process is dU = -W. Additionally, the conversation touches on the confusion around the use of heat capacity (Cp or Cv) in determining work for an irreversible process, and the need for a constant heat capacity for a reversible process. The conversation concludes with a mention of the heat capacity being a function of temperature
  • #1
dhoyda
10
0
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 amount 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? I am not sure about delta S either, I think in reversible delta S is zero but in irreversible its always increasing? I think there's is a formula for delta S for ideal This stuff can get confusing, can anyone help me?



The Attempt at a Solution

 
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  • #2
dhoyda said:
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 amount 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? I am not sure about delta S either, I think in reversible delta S is zero but in irreversible its always increasing? I think there's 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
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
the temperature is changing. I don't 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
matthew1991 said:
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
did I figure it out Andrew that internal energy is equal to work and we can say dU = C*dT for Ideal gas.
 
  • #9
dhoyda said:
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
dhoyda said:
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
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
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
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 can't be reversible. I feel like I am going in circles.
 
  • #14
dhoyda said:
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
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
dhoyda said:
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
 

1. What is adiabatic process in thermodynamics?

Adiabatic process refers to a thermodynamic process in which there is no heat exchange between the system and its surroundings. This means that the change in internal energy of the system is equal to the work done on the system.

2. What is an ideal gas in thermodynamics?

An ideal gas is a theoretical gas that follows the ideal gas law, which states that the pressure, volume, and temperature of the gas are all directly proportional to each other. In an ideal gas, there are no intermolecular forces and the molecules have zero volume.

3. How does changing heat capacity affect a system?

Changing heat capacity of a system can affect the temperature change of the system in response to a given amount of heat. A higher heat capacity means that the system can absorb more heat before experiencing a significant change in temperature, while a lower heat capacity means that the system will experience a larger temperature change with the same amount of heat added.

4. How is work calculated in thermodynamics?

In thermodynamics, work is calculated as the product of the force applied on a system and the distance over which the force is applied. Mathematically, work is represented as W = F * d, where W is work, F is force, and d is distance.

5. What is the relationship between work and energy in thermodynamics?

In thermodynamics, work is a form of energy transfer. When work is done on a system, it increases the energy of the system, and when work is done by a system, it decreases the energy of the system. This relationship is described by the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted from one form to another.

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