# What does V(p) graph tell me?

1. Jan 15, 2013

### skrat

Hello,

Well I have all sorts of question, so let me start.
In school we had a project where we had to examine, actually predict when the balloon will explode. The following data may not be ideal, but here we go.
In the first attachment (graf1.jpg) you can see blue colored line p(t) and red colored line V(t).

If you ever tried to inflate a balloon with your own lungs than you are very familiar with the first high peak of pressure at the beginning. We all know that it is very hard to start inflating BUT is a lot easier to continue building up pressure till the explosion - the start is even a lot easier if the balloon was at least once inflated before! Are there any parameters that describe this? How? Why? Please help! :)

Than I have no idea to what V(p) tells me? What is dV/dp ? Please note that, it is really hard to determine volumes (we had to do determine the volume from a picture using integrals). For example: The volume of course should never REDUCE in the process (but it does at one or two points) - it just does not make any sense, but the measurements say what they say. Check second attachment for this.

THANKS TO ALL!

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2. Jan 15, 2013

### Whovian

I can answer what dV/dp is.

It's well-known that dV/dr is simply the surface area of a sphere. I imagine representing the volume of a figure of any sort wrt one of its lengths (assuming halving the length just scales down the figure) and differentiating it gets us the surface area.

Anyway, the chain rule tells us dV/dp=dV/dr*dr/dp, so dV/dp is the product of the surface area and dr/dp. What does p represent here?

3. Jan 15, 2013

### skrat

V stands for volume of balloon and p stands for pressure.

So there has to be some relation between volume and pressure. Note that temperature was constant, but volume and pressure are not.

4. Jan 15, 2013

### Whovian

Interesting. I was about to pull out the Ideal Gas Law, but then I realized there ... oh.

The Ideal Gas Law should apply here, since there's an equilibrium between the balloon pushing in and the gas pushing out, so the pressure exerted by the balloon on the surface of the gas should be equal in magnitude to the gas's pressure.

5. Jan 15, 2013

### skrat

Well I was thinking about Ideal Gas Law too, but the volume of air in the system is not constant or in other words - the mass of the gas (air) in the system changes. But I'm not smart enough to describe this....

V is of course volume of the balloon, so volume of gas inside the system, also p is pressure in the system (inside the balloon).

6. Jan 15, 2013

Look up "pressure of a soap bubble". The surface tension and the curvature produce a 1/r term. The more curved the surface is the better can the tension in the rubber shell produce inward pressure. Maybe you can calculate the surface tension depending on the radius and see if it looks like the elasticity curves for rubber.

7. Jan 15, 2013

### Whovian

Shouldn't matter.

Again, shouldn't really matter. It's still an ideal gas.

8. Jan 17, 2013

### Staff: Mentor

I think I can help quite a bit with the interpretation of this problem.

The balloon is a rubber sheet that is undergoing large deformation biaxial stretching, and its "stress-strain behavior" is non-linear. One of the key characteristics of rubber is that it is virtually incompressible, so the volume of the rubber sheet remains constant. If the balloon were a perfect sphere, the surface area of the rubber sheet would be $4\pi r^2$ and its thickness (assumed uniform) would be h, so its volume would be $4\pi r^2h$.

Initially, you would need a small (virtually insignificant) amount of initial pressure to snap the balloon into its initial spherical shape. If the initial radius was r0 and the initial thickness of the rubber was h0, then the initial volume of the balloon rubber would be $4\pi r_0^2h_0$. But, since rubber is incompressible, the initial and final volumes of the rubber would have to be the same, so that $$\frac{h}{h_0}=(\frac{r_0}{r})^2$$So, as the balloon inflates, the thickness of the rubber decreases as the square of the radius.

The amount that the balloon rubber stretches can be characterized by the biaxial stretch ratio. The rubber sheet surface stretches equally in all directions (for a sphere), and the distance along a great circle between any two points on the sphere increases in proportion to the ratio of the present radius to the initial radius. This ratio is called the stretch ratio λ:$$\lambda=\frac{r}{r_0}$$
So, in terms of the stretch ratio, the rubber thickness h is given by:$$h=\frac{h_0}{\lambda^2}$$
In general, rubber is a very non-linear elastic material, and the tensile stress within the sheet σ (force per unit area) will be a non-linear function of the stretch ratio λ:
$$\sigma=\sigma(\lambda)$$

The next step in this development is to do an equilibrium force balance so that the stress can be expressed in terms of the pressure difference between the inside and outside of the balloon, the balloon sheet thickness h, and the balloon radius r. Do you know how to derive that equation? If so, write it down. I will continue the development after you provide the equation.

Incidentally, with all due respect to your hand waving explanation of rubber material properties, it is very puzzling why the rapid pressure decrease occurred (without a corresponding perturbation to the gas volume variation). Several responders made reference to the ideal gas law, which is appropriate to your problem. It seems that the number of moles of gas within the balloon was continually increasing, while the temperature was fairly constant. So, it would seem that, the only way that the pressure could have dropped several percent like that and stayed lower for a while would have been if gas volume had increased by roughly the same percentage. The gas volume variation should have been looked a lot like the inverse of the pressure variation, or possibly even more extreme. But it didn't. Is it possible that you missed part of the gas volume variation?

Chet