# Homework Help: Finding the work done by a gas under quasi-static, isothermal expansion

1. Mar 27, 2010

### jobloggs

1. So, an ideal gas is initially at 293K and 200kPa, and has a volume of .004m3
It undergoes a quasi-static, isotheral expansion until its pressure is reduced to 100kPa
What is the Work done by the gas

2.I know that for isothermal conditions W=nRTln(Pf/Pi)

3. my first attempt was making n=1 and Pf=200kPa and Pi=100kPa
i got W to be 1685w.

My first problem with this is that I don't actually know what n is, and i'm not even sure what Pi and Pf are, can't find it in my lecture notes anywhere.

My next step was to isntead of guessing n=1, make PV=nRt because I have P and V, but i'm not sure which P to use.

I know this dosen't seem like a difficult problem, but i won't be getting the answers until next month, so any guidance will be appreciated

2. Mar 27, 2010

### gabbagabbahey

Both P and V change during the expansion. You don't know the final volume, but you do know the initial volume. So, use the initial pressure as well.

3. Mar 28, 2010

### Andrew Mason

It should be Pi/Pf.

dW = PdV. If you substitute P = nRT/V (where T is constant since it is isothermal) you get:

$$\int_{V_i}^{V_f} PdV = nRT\int_{V_i}^{V_f} dV/V = nRT\ln (\frac{V_f}{V_i})$$

Since, by the ideal gas law, PiVi/Ti = PfVf/Tf, Vf/Vi = Pi/Pf (where Ti and Tf are equal)

You have to find n. Use the ideal gas law:

PV=nRT

You know Pi, Vi and Ti so you can find n. n does not change.

AM