Work Done by Gas in Isothermal Expansion of Balloon

In summary: So, it doesn't matter which you use. However, if you use the initial PV, you will get a negative value for work, indicating that the work is done by the gas. If you use the final PV, you will get a positive value for work, indicating that work is done on the gas. Either way, the magnitude of the work will be the same.In summary, in this conversation, the question was asked about the amount of work done by an ideal gas in a balloon that is expanding to 6.0 times its original size while being kept in thermal equilibrium with its surroundings. The equation PV = nRT was used to calculate the work, with the final PV being equal to the initial PV.
  • #1
Joshb60796
62
0

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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  • #2
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.
 
  • #3
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
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
 
  • #5
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:
  • #6
The initial PV is the same as the final PV.
 

Related to Work Done by Gas in Isothermal Expansion of Balloon

1. What is the definition of work done by gas in isothermal expansion of balloon?

The work done by a gas in an isothermal expansion of a balloon refers to the energy transferred to the surroundings as the gas expands and pushes against the external pressure. This work is typically measured in joules and is represented by the symbol W.

2. How is the work done by gas in isothermal expansion of balloon calculated?

The work done by gas in an isothermal expansion of a balloon can be calculated by multiplying the external pressure by the change in volume of the gas. This can be represented by the equation W = PΔV, where P is the external pressure and ΔV is the change in volume.

3. Why is the work done by gas in isothermal expansion of balloon important?

The work done by gas in an isothermal expansion of a balloon is important because it helps us understand the energy transfer that occurs during this process. It also has practical applications, such as in the design and functioning of engines and other machines that use gas expansion to do work.

4. How does the work done by gas in isothermal expansion of balloon differ from adiabatic expansion?

The main difference between isothermal and adiabatic expansion is that in isothermal expansion, the temperature of the gas remains constant, while in adiabatic expansion, there is no heat transfer between the gas and its surroundings. This means that the work done by gas in an isothermal expansion will be greater than in an adiabatic expansion, as the temperature remains constant and allows for more energy to be transferred as work.

5. What factors affect the work done by gas in isothermal expansion of balloon?

The work done by gas in an isothermal expansion of a balloon is affected by the external pressure, the initial and final volumes of the gas, and the temperature. It is also influenced by the type of gas being used and any external forces acting on the balloon during expansion.

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