Work Done by Gas in Isothermal Expansion of Balloon

AI Thread Summary
The discussion focuses on calculating the work done by an ideal gas during isothermal expansion of a balloon. The initial conditions include a pressure of 645.0 Pa and a volume of 0.10 m^3, with the balloon expanding to six times its original size. The correct formula for work done is W = nRT * ln(V2/V1), where nRT remains constant during the process. A participant mistakenly attempts to simplify the equation by omitting nRT, leading to confusion about the dimensional validity of their answer. Clarification is provided that the initial and final states of pressure and volume must be used to accurately determine the work done, reinforcing the importance of including all relevant factors in the calculation.
Joshb60796
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Homework Statement


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


Homework 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)

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?
 
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Joshb60796 said:
W = nRT*ln(V2/V1)
nRT is constant so my answer should be ln(.6/.1)
How does nRT being constant mean that you can throw it away?
Note that your answer cannot be valid dimensionally. ln(v1/v2) is dimensionless.
 
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).
 
Joshb60796 said:

Homework Statement


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

Homework 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)

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

AM
 
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?
 
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The initial PV is the same as the final PV.
 
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