What Thermodynamic Process Occurs When Heating a Helium-Filled Balloon?

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SUMMARY

When heating a helium-filled balloon that is fully expandable, the thermodynamic process is classified as isobaric. This is because the pressure inside the balloon remains nearly equal to the external atmospheric pressure of 1 atm, provided the balloon does not develop significant tension during expansion. The first law of thermodynamics indicates that there is no change in internal energy, as the work done by the balloon and the change in heat are both zero. The discussion emphasizes the importance of the balloon's material properties in determining the behavior of the gas within.

PREREQUISITES
  • Understanding of the first law of thermodynamics
  • Familiarity with isobaric processes
  • Knowledge of gas laws, particularly the ideal gas law (PV=nRT)
  • Basic concepts of pressure and volume relationships in thermodynamics
NEXT STEPS
  • Study the implications of the ideal gas law in different thermodynamic processes
  • Explore the properties of rubber materials and their impact on gas behavior in expandable containers
  • Research the concept of free expansion in thermodynamics
  • Examine real-world applications of isobaric processes in engineering and physics
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Students preparing for competitive exams like IIT JEE, educators teaching thermodynamics, and anyone interested in the practical applications of gas laws and thermodynamic processes.

AdityaDev
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If I heat a rubber balloon filled with helium slowly and if the balloon is fully expandable and (the balloon) can be assumed to require no energy in its expansion,what type of thermodynamic process is taking place? Is it isobaric?
Since the balloon expands the pressure exerted by the gas on balloon should increase. Then how can it be isobaric?
It can't be isochoric nor isothermal.
It is either adiabatic or isobaric.
 
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its called free expansion . work done by the balloon is zero and so is the change in heat . this results into no change in internal energy by the first law of thermodynamics .
 
proton007007 said:
its called free expansion . work done by the balloon is zero and so is the change in heat . this results into no change in internal energy by the first law of thermodynamics .
It's given isobaric... This concept came in IIT
Exam... The toughest exam in India for admission to B.tech. IIT IS THE MOST REPUTED COLLEGE IN MY COUNTRY.
 
AdityaDev said:
It's given isobaric... This concept came in IIT
Exam... The toughest exam in India for admission to B.tech. IIT IS THE MOST REPUTED COLLEGE IN MY COUNTRY.
Yes, it's isobaric. If the balloon is fully expandable (i.e., doesn't develop significant tension as it expands), the pressure inside the balloon is always nearly equal to the pressure outside the balloon. And, the pressure outside the balloon is 1 atm. (constant). So the pressure inside the balloon is constant at about 1 atm, and the process is isobaric. The key to all this is that the balloon membrane doesn't develop significant tension.

Chet
 
If rubber tension is negligible it is isobaric otherwise we can consider P=f(V), and f depend on rubber specification and balloon dimensions.

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mmeftahpour said:
If rubber tension is negligible it is isobaric otherwise we can consider P=f(V), and f depend on rubber specification and balloon dimensions.

https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/59/59222-ef0af70f5c70ec1351bf0a5e6d506887.jpg I
Actually, it would be the pressure difference across the rubber membrane ΔP that you can calculate from the rubber properties and the balloon shape.

Chet
 
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Chestermiller said:
Yes, it's isobaric. If the balloon is fully expandable (i.e., doesn't develop significant tension as it expands), the pressure inside the balloon is always nearly equal to the pressure outside the balloon. And, the pressure outside the balloon is 1 atm. (constant). So the pressure inside the balloon is constant at about 1 atm, and the process is isobaric. The key to all this is that the balloon membrane doesn't develop significant tension.

Chet
when you are heating a gas its randomness increases so it will collidewith the walls of the balloon more freaquently. hence pressure inside th balloon should increase right?
 
AdityaDev said:
when you are heating a gas its randomness increases so it will collidewith the walls of the balloon more freaquently. hence pressure inside th balloon should increase right?
Not if volume increases so that the pressure matches the outside pressure.

Chet
 
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Chestermiller said:
Not if volume increases so that the pressure matches the outside pressure.

Chet
so you mean that the area of contact increases and since P=F/A pressure increase gets countered by expansion of balloon.
 
  • #10
AdityaDev said:
so you mean that the area of contact increases and since P=F/A pressure increase gets countered by expansion of balloon.
No. Pressure stays the same. You're familiar with the ideal gas law, correct?
 
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  • #11
Chestermiller said:
No. Pressure stays the same. You're familiar with the ideal gas law, correct?
Pv=nRT. as volume temperature of balloon increases pressure should increase.
if pressure exerted by gas on balloon doesn't change, how does it expand.
 
  • #12
AdityaDev said:
Pv=nRT. as volume temperature of balloon increases pressure should increase.
From your understanding of mathematics, if the volume v increases in direct proportion to the temperature T, how does the pressure vary?
if pressure exerted by gas on balloon doesn't change, how does it expand.
We already said the we are looking at a limiting case in which the stiffness of the balloon membrane is negligible. Under these circumstances, the pressure inside the balloon must match the pressure outside the balloon, and the pressure outside the balloon is constant at 1 atm.

Chet
 
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