# Why is filling a balloon from your mouth much harder initially?

I think that most of the answers here are incorrect since it has nothing to do with decreasing resistance of rubber. In fact, the force required to stretch the balloon increases, not decreases while inflating. It's similar to stretching a string, ie. the reaction force is proportional to the increase in length of the string - this is why there is a point when you can no longer stretch a chest expander.

The real reason that initially it's hard to inflate the balloon is that in the beginning, ie. with the first blow, you increase the total surface of the balloon significantly, thus the force (pressure on the surface) increases also significantly. With each subsequent blow, the increase of the total surface is smaller and so is the increase of force. This is the result of two facts:

- constant increase of volume with each blow
**volume**of the balloon is proportional to the**cube of radius**while**surface**of the balloon is proportional to the**square of the radius**

For a sphere you have:

$$ A={4}\pi R^2 \\ V={4\over3}\pi R^3 $$ The equations says that the amount of work required to increase the volume of the balloon by one unit is smaller if the balloon is already inflated.

Take a strip of balloon rubber and pull it. It will get harder the more you pull. So why is it that inflating the balloon gets easier (at least long before the breaking point)?

The balloon starts with very high curvature, so the air pressure is distorting each spot on its surface a lot relative to its 1cm neighbors for example. All the rubber's tension pulls inwards at a relatively sharp angle. With a larger balloon, this angle becomes flatter.

Imagine you have a thread attached to the wall. You hang a weight from the centre of the thread and pull the other end away. Now pulling gets harder and harder as the angle between the ends of your thread gets wider. The impact of the weight is getting bigger, even though the weight is not changing. Conversely, if you pull with a consistent force on your thread, you need a much bigger weight to produce a sharp angle than to produce a wide angle.

This effect is largely overcompensating the actually increasing tension in the rubber.

Try it out with http://www.calculatoredge.com/calc/sphere.htm It's not perfect, mainly as it doesn't provide reasonable numbers to start with but find some and then change pressure and volume to see the effects on the stress. Twice the radius would mean twice the stress, so conversely you need half the pressure if the stress staid the same, to inflate a twice as big balloon.

When in doubt, use mathematics.

Imagine the balloon as a sphere (close enough for this answer) of initial radius $r_0$ and thickness $t$. Let's inflate it just a little bit from the uninflated state (to radius $r_0 + \Delta r$). Now we can take a look at what happens by taking a cut through the equator of the sphere. The total circumference at the equator is $2\pi r$; with the thickness $t$ the area of rubber we're working against is $2\pi r t$. Stretching the balloon's radius by $\Delta r$ increases the circumference by a fraction of $\frac{\Delta r}{r}$ - that is the strain. Now if we accept that rubber is a perfectly elastic material (constant Young's modulus E), then the force we need to exert is $$\begin{align}F &= E\cdot2 \pi \cdot r \cdot t \cdot \frac{\Delta r}{r}\\ &=2\pi \cdot E \cdot t \cdot \Delta r\\ \end{align}$$

so *force is independent of radius* - although it does depend on the degree of stretching ($\Delta r$).

Now the force on the rubber is generated by the pressure in the balloon divided by the area at the equator:

$$\begin{align} F &= P A\\ &= \pi r^2P\\ \end{align}$$

Combining these two, you get

$$P = \frac{2 \cdot E \cdot t \cdot \Delta r}{r^2}$$

Because there is an $r^2$ term in the denominator, this shows that pressure will be less when the balloon gets bigger - in other words, blowing a balloon is initially harder, as is the general experience.

But wait - there's more. The thickness of the balloon becomes less as the balloon stretches - for a sphere this is a slightly complex quantity involving the Poisson ratio of the material. But the point is that $t$ will become smaller as $r$ gets larger: this will make pressure drop even faster with radius.

Finally, the modulus of elasticity isn't quite constant - in particular, when rubber is stretched beyond a certain point it becomes much stiffer. This is the reason that the balloon, having initially become easier to inflate, finally become quite hard - and continuing to blow it further may make it pop.