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

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When heating a helium-filled balloon that is fully expandable, the thermodynamic process is isobaric, as the pressure inside the balloon remains constant at approximately 1 atm, matching the external pressure. Although heating increases the gas's randomness and collision frequency, the balloon's expansion counteracts any pressure increase, maintaining equilibrium. The key factor is that the balloon membrane does not develop significant tension during expansion. This scenario aligns with the ideal gas law, where volume increases proportionally with temperature without changing pressure. Thus, the process can be classified as isobaric under these specific conditions.
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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.

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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.
 
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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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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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