Is pressure constant when you heat a balloon?

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Discussion Overview

The discussion revolves around the behavior of gas pressure in a balloon when subjected to temperature changes, particularly focusing on whether pressure remains constant as the balloon is cooled. Participants explore theoretical implications, practical examples, and the differences between various types of balloon materials.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants note that when a balloon is cooled, the volume decreases, which they argue should affect the pressure, unlike a rigid container where pressure can be assumed constant.
  • One participant presents a mathematical relationship involving temperature, volume, and pressure, suggesting that if the ratios of volume to temperature are equal, pressure may also remain constant.
  • Another participant points out that the material of the balloon (Mylar vs. rubber) may influence how pressure is treated in calculations, with Mylar potentially allowing for different assumptions than rubber balloons.
  • Some participants discuss the mechanics of balloon membranes, noting that cooling the gas affects both the tension in the membrane and the pressure differential between the inside and outside of the balloon.
  • There is a mention that rubber balloons hold very little pressure, which may simplify assumptions in calculations, while car tires are described as being close to rigid at operating pressure.
  • A participant questions whether it is reasonable to assume the pressure inside a balloon is always 1 atm for practical calculations, to which another participant agrees.

Areas of Agreement / Disagreement

Participants express differing views on whether pressure can be assumed constant in a cooling balloon scenario. While some suggest that it can be treated as constant under certain conditions, others argue that the dynamics of the balloon's material and the effects of temperature changes complicate this assumption. The discussion remains unresolved regarding the general applicability of these assumptions.

Contextual Notes

Participants highlight the dependence on the type of balloon material and the specific conditions of the scenario, which may affect the validity of assumptions regarding pressure constancy. There is also mention of the limitations of simplified models in accurately representing real-world behavior.

Who May Find This Useful

This discussion may be useful for physics educators, students studying gas laws, and individuals interested in the thermodynamic behavior of gases in flexible containers.

ranger275
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I am teaching Physics and I'm not looking for help for a particular homework problem but on a class. Sorry if this should be in the homework sections.

When I park my car where the sun only hits one side and measure my tire pressure later I find the tires on the sun side have a higher pressure. Yet I keep seeing problems where a sealed balloon is cooled solved assuming the pressure doesn't change. It seems to me as the volume decreases the balloon applies less force to the gas which decreases the pressure. If I had a cylinder with a weight on top then I see how the force and pressure would be constant but I don't see how that works with a balloon. Any help would be appreciated.
 
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Can you show here such a problem?
 
Here is an example.

Krusty, a clown, carries a 2.00 x 10-3 m3 helium filled Mylar balloon from the 295 K heated circus tent to the cold outdoors, where the temperature is 273 K. How much does the volume of the balloon decrease?
 
ranger275 said:
When I park my car where the sun only hits one side and measure my tire pressure later I find the tires on the sun side have a higher pressure. Yet I keep seeing problems where a sealed balloon is cooled solved assuming the pressure doesn't change. It seems to me as the volume decreases the balloon applies less force to the gas which decreases the pressure. If I had a cylinder with a weight on top then I see how the force and pressure would be constant but I don't see how that works with a balloon. Any help would be appreciated.

T\propto VV\propto \frac{1}{P}hence\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}So perhaps if the ratio of \frac{V_1}{T_1} and \frac{V_2}{T_2} were equal, then P_1=P_2?

Like, say V_1=a, T_1=b, V_2=c, T_2=\frac{bc}{a}

In words, maybe it could be explained along the lines of; as temperature decreases, volume decreases proportionally. Hence pressure increases because the gas particles collide more frequently with the inside walls of the balloon. However, each particle now has lower kinetic energy since temperature is a measure of translational KE, so each individual collision exerts less force (hence pressure) and the total pressure remains constant.
 
ranger275 said:
Here is an example.

Krusty, a clown, carries a 2.00 x 10-3 m3 helium filled Mylar balloon from the 295 K heated circus tent to the cold outdoors, where the temperature is 273 K. How much does the volume of the balloon decrease?
And the solution assume constant pressure?
It may be because is Mylar and not rubber. When you start to inflate it, it already have the same surface area, you just need to put enough gas to have a little over atmospheric pressure, to keep its shape. As opposite to a rubber balloon where you need to compensate the elasticity of the stretched surface.
 
The pressure inside the balloon is higher than the air pressure outside the balloon, because the balloon membrane has been stretched and is under tension. When you cool the gas in the balloon, the pressure outside doesn't change, but the gas volume decreases. This causes the radius of curvature of the balloon membrane to decrease, but also causes the amount or stretch and tension in the balloon membrane to decrease. Decreasing the radius of curvature of the membrane tends to allow the balloon to support a greater pressure differential between the inside and outside of the balloon, while decreasing the tension in the balloon membrane tends to allow the balloon to support less of a pressure differential between the inside and outside. So the two effects tend to cancel, and the net effect depends on the stress-strain behavior and thickness of the balloon membrane. In the case of a tire (which is a much stiffer membrane than a balloon), the pressure differential between inside and outside decreases when the gas is cooled, so the lower tension in the tire cords and rubber wins out; the gas pressure inside the tire decreases while the volume doesn't change much.
 
Rubber balloons hold very little pressure, so this is just an easy simplifying assumption that doesn't cause much error (a couple of percent at most).

Car tires, on the other hand are close to rigid at operating pressure.
 
russ_watters said:
Rubber balloons hold very little pressure, so this is just an easy simplifying assumption that doesn't cause much error (a couple of percent at most).

Car tires, on the other hand are close to rigid at operating pressure.

Does that mean I could assume the pressure inside a balloon is 1 atm all the time and get essentially the correct answer?
 
ranger275 said:
Does that mean I could assume the pressure inside a balloon is 1 atm all the time and get essentially the correct answer?
Yes.
 
  • #10
Thanks for all the responses. This site is a great resource for a physics teacher.
 

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