# Isothermal Process Help.

1. Oct 6, 2009

### mrmonkah

Hey all,

If i have an isothermal process of an ideal monotomic gas, IE. deltaT = 0 and i have an initial volume (V1) and an initial pressure (P1). Then im given the work done BY the gas (W = -### as it is the environment providing the energy)

How do i calculate the final volume and the final pressure? Ive been toying with different equations of state and like the look of w = nRT ln(V2/V1).... it is probably just my algebra that needs to improve in order for me to re-arrange for V2...... am i on the right track?

Dan

2. Oct 6, 2009

### rock.freak667

P1V1=nRT

so you have W=P1V1ln(V2/V1)

you can get V2 now I believe.

3. Oct 6, 2009

### Andrew Mason

If you are given the work, you can find the volume change using:

$$W= \int PdV = nRT\int dV/V = nRTln(V2/V1)$$

as you have suggested. I am not sure why you are hesitating or unsure.

AM

4. Oct 6, 2009

### mrmonkah

okay cool, so the equation i have is right, im just un-sure about re-arranging for V2 now, what with the ln(v2/v1) business....... am i right in thinking that i can take ln(v2) - ln(v1) as being equal to ln(v2/v1)?

Like i say its my calculus/algebra that is poor at the moment.

5. Oct 7, 2009

### Andrew Mason

You don't need to do that. Just take the anti-log of both sides:

$$e^{W/nRT} = \frac{V_2}{V_1}$$

You are given W, and nRT so this is just a matter doing the algebra to calculate V2

AM

6. Oct 8, 2009

### mrmonkah

Hi Andrew,

Thank you very much for the pointer, i was slowly coming to that conclusion i think once i knuckled down and did some revision. Now i can hopefully conquer the rest of the question.

Ill post any issues i have, but hopefully ill get there under my own steam. Best way to learn right!

Cheers,

Dan