# The work done by an adiabatic process?

1. Oct 11, 2011

### JustinLiang

1. The problem statement, all variables and given/known data
On my formula sheet I have this:
W=(C/1-y)(Vf^(1-y)-Vi^(1-y))
y=Cp/Cv

2. Relevant equations

3. The attempt at a solution

I am confused about what the C in C/1-y stands for. My textbook does not even have this formula and I realized that C is not Cp or Cv...

2. Oct 11, 2011

### Andrew Mason

The C is usually expressed as K as in: $PV^\gamma = K$. So it can be written:

$$W = \frac{P_iV_i^\gamma}{1-\gamma}(V_f^{1-\gamma} - V_i^{1-\gamma})$$

AM

3. Oct 11, 2011

### JustinLiang

Ahh ok thanks! But why is it Vi and Pi, why not Vf and Pf? Is there a reason?

4. Oct 11, 2011

### Andrew Mason

You can make it Vf and Pf or any V, P during the adiabatic process if you like:

$$P_iV_i^\gamma = P_fV_f^\gamma = PV^\gamma = K = \text{ constant}$$

AM

5. Oct 11, 2011

### JustinLiang

Okay, so for an adiabatic process, we cannot use the typical ideal gas law P1V1=P2V2 right?

6. Oct 12, 2011

### Andrew Mason

If PV=nRT and PV = constant then T would be constant (assuming n is constant). But in an adiabatic process that does work this cannot be the case: dQ = 0 -> dU = -dW where dW is the work done by the gas. So P1V1 ≠ P2V2.

In a reversible adiabatic process involving an ideal gas, the condition $PV^\gamma = K$ holds. But so does PV=nRT. Substituting P = nRT/V the adiabatic condition becomes:

$$PV^\gamma = nRTV^{\gamma-1} = K$$

and since n and R are constant (assuming no change in the amount of gas) we have:

$$TV^{\gamma-1} = K/nR = K' = \text{constant}$$ and

$$PV/T = nR = \text{constant}$$

AM