Work Done by Gas in Isothermal Expansion of Balloon

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Homework Help Overview

The discussion revolves around the work done by an ideal gas during the isothermal expansion of a balloon. The problem involves calculating the work when the balloon expands to six times its original volume, given specific initial conditions of pressure and volume.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the work formula W = nRT*ln(V2/V1) and question the implications of nRT being constant during the process. There is also a focus on the dimensional validity of the logarithmic term and the need for specific values of n and T.

Discussion Status

Some participants have provided guidance on the importance of including the nRT factor in the calculations. There is an ongoing exploration of which values of pressure and volume should be used in the context of the work formula, indicating a productive direction in clarifying the problem.

Contextual Notes

Participants note the absence of specific values for the number of moles (n) and temperature (T), which are crucial for the calculations. This lack of information is acknowledged as a potential source of confusion in the problem-solving process.

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?
 
Last edited:
The initial PV is the same as the final PV.
 

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