# The work done by a perfect gas

1. Feb 22, 2014

### Emspak

1. The problem statement, all variables and given/known data
An ideal gas at initial temperature T1 and pressure P1 is compressed by a piston to half its original volume. The temperature is varied so that the relation P=AV always holds and A is a constant. What is the work done n the gas in terms of n (moles of gas) R (gas constant) and T1?

2. Relevant equations

Using ideal gas law, PV= nRT

and assuming the work done is $W = \int^{V_b}_{V_a} P dV$

3. The attempt at a solution

OK, simple enough, right?

Vb = (1/2) Va

So V = nRT/P and $V_a = \frac{nRT}{P_1}$ and $V_b = \frac{nRT}{2P_1}$

Plug this into the integral above. Since P=AV (in the givens) it should be

$W = \int^{V_b}_{V_a} (AV) dV = A \int^{V_b}_{V_a} V dV = A\frac{V^2}{2}|^{V_a}_{V_b}$

Which leaves me with

$=\frac{A}{2} \left[\left(\frac{nRT}{2P_1}\right)^2 - \left(\frac{nRT}{P_1}\right)^2\right] = \frac{A}{2}\left(\frac{nRT}{P_1}\right)^2(-3/4) = \frac{-3A}{8}\left(\frac{nRT}{P_1}\right)^2$

Yet the answer is listed as $\frac{-3A}{8}\left(\frac{nRT}{P_1}\right)$

So I am trying to figure out how I got the extra factor in there. I think I did everything right, but is there some stupid mathematical error I made?

Anyhow, it's possible there's a typo in the book's answer too. But...

any help is appreciated, even though I anticipate it will be trivial...

2. Feb 22, 2014

### Andrew Mason

The given answer is not dimensionally correct. The answer has to have dimensions of energy (work) ie. PV or nRT. Since A has dimensions of P/V, the correct answer has to have dimensions of AV2. The given answer has dimensions of AV = P.

AM

3. Feb 22, 2014

### Emspak

Thanks lots, I knew there was something stupid I was missing. I looked at it again and the given answer was yours, but I forgot about just plugging P/V in for A. D'OH!!!