# Adiabatic, Ideal gas, changing heat capacity, work calculation

1. Feb 4, 2011

### dhoyda

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?

3. The attempt at a solution

2. Feb 5, 2011

### Andrew Mason

$$W = \int PdV$$

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

AM

3. Feb 5, 2011

### dhoyda

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. Feb 5, 2011

### matthew1991

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. Feb 5, 2011

### dhoyda

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. Feb 5, 2011

### Andrew Mason

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. Feb 5, 2011

### matthew1991

sorry, that was isothermal. adiabatic expansion work can be found by W = (3/2)nR(deltaT)

8. Feb 5, 2011

### dhoyda

did I figure it out Andrew that internal energy is equal to work and we can say dU = C*dT for Ideal gas.

9. Feb 5, 2011

### Andrew Mason

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. Feb 5, 2011

### Andrew Mason

Yes. Change in internal energy = - Work done by gas. $dU = nC_vdT$

AM

11. Feb 5, 2011

### dhoyda

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. Feb 5, 2011

### dhoyda

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. Feb 5, 2011

### dhoyda

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. Feb 5, 2011

### Andrew Mason

How does the heat capacity vary with temperature? Plug that into:

$$\Delta U = \int nC_vdT$$

AM

15. Feb 5, 2011

### dhoyda

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. Feb 5, 2011

### Andrew Mason

You are determining internal energy: you must use Cv not Cp. Cv = Cp-R

AM