# Is pressure constant when you heat a balloon?

1. Apr 11, 2014

### ranger275

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.

2. Apr 11, 2014

### nasu

Can you show here such a problem?

3. Apr 11, 2014

### ranger275

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?

4. Apr 11, 2014

### rtsswmdktbmhw

$$T\propto V$$$$V\propto \frac{1}{P}$$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.

5. Apr 11, 2014

### nasu

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.

6. Apr 11, 2014

### Staff: Mentor

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.

7. Apr 11, 2014

### Staff: Mentor

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.

8. Apr 11, 2014

### ranger275

Does that mean I could assume the pressure inside a balloon is 1 atm all the time and get essentially the correct answer?

9. Apr 11, 2014

### Staff: Mentor

Yes.

10. Apr 14, 2014

### ranger275

Thanks for all the responses. This site is a great resource for a physics teacher.