# Isothermal Expansion

1. Apr 24, 2013

### Joshb60796

1. The problem statement, all variables and given/known data
An Ideal gas in a balloon is kept in thermal equilibrium with it's constant-temperature surroundings. How much work is done by the gas if the outside pressure is slowly reduced, allowing the balloon to expand to 6.0 times it's original size? The balloon initially has pressure 645.0 Pa and volume 0.10m^3. The ideal gas constant is R=8.314 J/mol*K

2. Relevant equations
P1V1=P2V2
W= p∫ dv from v1 to v2 --> pV=nRT --> p=nRT/V --> nRT∫1/V dV from V1 to V2 --> nRT*ln(V2/V1)
W = nRT*ln(V2/V1)

3. The attempt at a solution

Since this is isothermal and the amount of gas isn't changing, nRT is constant so my answer should be ln(.6/.1) but my answer key states 120J, maybe I'm doing a conversion wrong?

2. Apr 25, 2013

### haruspex

How does nRT being constant mean that you can throw it away?

3. Apr 25, 2013

### Joshb60796

Hmm...now that you pointed that out, it's glaringly obvious. I made the assumption based on the fact that those bits of information weren't given (n, and T).

4. Apr 25, 2013

### Andrew Mason

You are forgetting the nRT factor. Since PV = nRT, W = PV ln(6) (rounded to 2 sig. figs).

AM

5. Apr 25, 2013

### Joshb60796

Thank you so much for your reply. I completely understand the train of thought but am puzzled as to which p and which V replaces the nRT, the initial or the final...or do I look at the difference of p and V?

Last edited: Apr 25, 2013
6. Apr 25, 2013

### Staff: Mentor

The initial PV is the same as the final PV.