# Work Required for Adiabatic Compression

1. Oct 15, 2012

### chris_avfc

1. The problem statement, all variables and given/known data
One mole of gas compressed to half of its original volume adiabatically, what is the work done?

2. Relevant equations

dE= dW + dQ (the d one the w and q should have the line though it)
pV^ α= constant = k

3. The attempt at a solution

As it is adiabatic dQ = 0
Meaning dE = dW

dW = -pdV = kv^-α dv

Integrating this with the limits 1/2 V and V.

I end up with

-κ[-v^(1-α)/(2-2α)]

Then I am stuck on where to go from there?

2. Oct 15, 2012

### Andrew Mason

Are you given the temperature of the gas initially? Can we assume it is an ideal gas? If it is an ideal gas, can you determine its Cv ?
Substitute nRT/V for P to put the adiabatic condition in terms of T and V.

If you express the adiabatic condition in terms of T and V, it is simply a matter of finding the ratio of final T to initial T. Since W = ΔU you just have to find the change in U.

AM

3. Oct 16, 2012

### chris_avfc

No initial temperatures.
It is an ideal gas.

I was using that p for the substitution for dw=-pdv though, so now I'm confused.

4. Oct 16, 2012

### Andrew Mason

If you substitute P = nRT/V in the adiabatic condition you get:

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

So $TV^{\gamma -1} = \text{constant}$

To find the final T, use:

$$T_f/T_i = \left(\frac{V_i}{V_f}\right)^{\gamma - 1}$$

so $\Delta T = T_f - T_i = T_i(\left(\frac{V_i}{V_f}\right)^{\gamma - 1} - 1)$

Using W = ΔU = nCvΔT you can develop the expression for W. You have to know what the Cv is though. If it is a monatomic ideal gas, Cv = 3R/2

AM

5. Oct 16, 2012

### chris_avfc

Okay, I recognise the TV equation, just never though to combine the other two constants with the one.
The heat capacity equation is also in my notes, so I get all of that.

I end up with:

3/2 RT [2^(gamma-1)- 1]

Where the T is the initial temperature.
Obviously R is a constant that the value is known of.
You aren't given any value for T or gamma, so I am struggling to see how to go any further?

Thank you very much by the way.

6. Oct 16, 2012

### Andrew Mason

You can't. The work needed to compress a mole of gas to half its volume depends on the initial temperature. The hotter it is, the more work that is required.

PS Does the question refer to the ideal gas as monatomic? Your use of 3R/2 for Cv is for monatomic gases. This would mean that $\gamma = 5/3$

AM

Last edited: Oct 17, 2012
7. Oct 17, 2012

### chris_avfc

Yeah a couple of hours after I replied, I was looking over my notes and I saw that value for $\gamma$, substituted it in and it worked out great.

Thanks for all the help mate, it has been really useful.