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

Why is it that when you first fill up a balloon, it's hard to get air through, but after inflating it a bit, it becomes much easier to further inflate the balloon?

• Probably has to do with the resistance to stretching of the material of the balloon, which MAYbe decreasing as the stretching itself increases. Of course this is just conjecture on my part. – Dev Kanchen Jun 30 '14 at 16:38
• My guess is that once you've got a stretched section, the boundary of that section is applying a lot of stress to the unstressed section, so your breath doesn't have to apply as much pressure. Are you thinking more of a round balloon or a "sausage" balloon, which typically inflates to some elastic limit at each point along its length, and then "moves on" to expand the next section? – Carl Witthoft Jun 30 '14 at 17:15
• I remember the answer on Mr. Wizard didn't satisfy me as a kid, his explanation was that the smaller the balloon, the thicker the rubber, so the force required to separate that thicker particle layer was greater. – Grady Player Jul 6 '14 at 0:30
• Two-balloon experiment (Wikipedia) covers this phenomenon. – Keith McClary Nov 17 '16 at 16:06

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.

• +1, the larger the balloon gets, the smaller a mouthful of air is relative to that balloon. It's the same reason a year is way longer to a 4-year-old than a 50-year-old... the year is 25% of the kid's life, but only 2% of the adult's. :-) The changing ratio between balloon volume and mouth volume is basically a lever that acts in your advantage more as the balloon gets larger. – The111 Jul 2 '14 at 3:20
• This answer is not correct. Stretch a balloon before inflating it. It returns to almost it's exact starting shape if you don't overdo it. Yet it is now much easier to blow up. Likewise, blow up a balloon and then deflate it. It's true that it will be a little deformed (not enough to matter by this answer's analysis), but it's still easier to blow up the second time. – Brock Adams Jul 4 '14 at 8:27
• @BrockAdams mentions a secondary, but still very important (at least, in the case of old balloons stored in a closet for a long time) point: Stretch a balloon before inflating it. It returns to almost it's exact starting shape if you don't overdo it. Yet it is now much easier to blow up. The stickiness of rubber does affect the difficulty a lot. – Yatharth Agarwal Jul 5 '14 at 11:55
• @BrockAdams It is true that there are other factors that have influence on 'difficulty of inflating' (eg. temperature, atmospheric pressure, pre-stretching). Please note however, that the question was about the difficulty of the first blow, no matter if the balloon is being inflated for the first time or it was inflated before. Stretching the balloon before inflating does help, but it is still harder to make the first blow that the second, no matter how many inflating-deflating cycles you perform. – golem Jul 7 '14 at 8:53
• This answer does not explain much of anything. When we increase the radius of the balloon by an amount $dR$, the work done increasing the surface area is $dW = T 8\pi R dR$ where $T$ is the tension in the balloon. Because $dV = 4\pi R^2 dR$, we have $dW = (2 T/R) dV$. Because $dW/dV$ is the pressure, your answer simply gives us the Young-Laplace formula. However, we do not know the relation between $T$ and $R$ apriori, so claiming that geometry explains the result is specious. – Mark Eichenlaub Jun 20 '16 at 7:03

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.

• Other than using the word (non-word) "huger" I really like this post. ;) +1 – paqogomez Jun 30 '14 at 23:04
• Thanx @paqogomez. It's a community wiki now, so feel free to fix other spelling errors. I'm no native speaker but glad you liked it anyway :) – Giszmo Jun 30 '14 at 23:59
• You lost me at the thread on the wall example. How can you pull on the other end if it is attached to the wall? – Muhd Jul 2 '14 at 1:16
• Only one end of my thread is attached to the wall. The other is attached to my hand. – Giszmo Jul 2 '14 at 2:22
• User "i know nothing"'s answer is actually the shortest valid answer so far I would say. Very intuitive. – Giszmo Jul 2 '14 at 2:32

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.

• I think it's interesting how this gives an example where the formal concept of force doesn't match the intuition about what is going on... – Abel Molina Jul 6 '14 at 5:14
• I would be interested in the motivation for the downvote... – Floris Jan 11 '16 at 16:48
• Your first equation is wrong. Force cannot depend on $\Delta r$. – Mark Eichenlaub Jun 20 '16 at 6:57
• @MarkEichenlaub are you saying the force should be independent of the increase in size of the balloon? Because that is the significance of $\Delta r$ - the degree to which the balloon was made bigger. – Floris Jun 20 '16 at 11:42
• It's just Hooke's law - force proportional to displacement. I never said (nor intended) for $\Delta r$ to be infinitesimal - but if it was I would expect the increase in force to also be infinitesimal. I could write $\Delta F$ if you prefer. – Floris Jun 20 '16 at 15:06

Like Dev said above, the material your typical round balloon is made from has a non-linear stress strain curve. When just starting to inflate it is fairly stiff, but then as it starts to blow up the stiffness goes down somewhat until it approaches its maximum size. We measured this in my undergraduate advanced lab class, and while I don't have the data handy there's a web site that shows a stress-strain curve for a balloon.

Edit: Swapped out the original link, unclear if the page had malicious content or not? I had no warnings in recent versions of firefox and chrome, but better safe than sorry.

• The stress/strain curve causes the amount of force when starting to stretch beyond the relaxed state to be non-zero, but I think it only increases beyond that. On the other hand, it doesn't take much air to double the surface area of a barely-stretched balloon; unless the tension also doubles, the pressure would drop. – supercat Jun 30 '14 at 20:59
• I should clarify -- when I measured a balloon the slope of the stress-strain curve went slightly negative in the intermediate region. However, even if it is flat or growing slowly it's still possible that it gets easier to inflate, as per the reason mentioned by @golem below. – ARM Jul 1 '14 at 15:13

The volume of a balloon grows linear, while the surface (which you actually stretch) doesn't. So although you're blowing the same amount of air into a balloon, you don't stretch the surface as much as in the beginning.

Let us first summarize what do we actually experience when inflating the balloon. For the very first bit of volume, we have to exert a lot of energy. Or alternatively, we have to apply a lot of pressure coming from our lungs because for the change of energy $\delta E$, change of volume $\delta V$ and extra pressure $\Delta P$ (that is the difference between the actual and the atmospheric), we have roughly

$$\frac{\delta E}{\delta V} \approx \Delta P$$

ARM, golem and John Bentlin pointed out effects that surely cause a balloon to be more difficult to inflate when it is less inflated rather than a lot. However, it is not entirely clear which of the effects plays the major role in this case.

The effect of an "S-curve" in the tensile response of the balloon is nevertheless significant only around the tensile pressure that is at the pressure where we reach the tip of the S-curve. The tensile pressure of rubber is typically around $10-15 MPa$. So we can ask whether by linear stretching in a typical balloon, we reach the tensile pressure in the first blow or not. Once we do, we throw away the model and say afterwards it is much easier to stretch since now the rubber is "over-stretched".

For the pressure inside a spherical volume of diameter r held by a membrane with surface tension $\sigma$, there is a law called the Laplace law and it reads: $$\Delta P = \frac{2 \sigma}{r}$$ The law can be derived by differential calculus and energy variations due to surface growth and volume growth as slightly touched upon by user golem.

For rubber we have a Young's modulus around $E_Y = 0,01-0,1 GPa$. The membrane surface energy can be again by energy considerations derived as

$$\sigma = E_y d$$,

where $d$ is the thickness of the ball wall. However $d$ gets spread out with growing surface, that is as $r^2$. Without further hesitation we can just write an approximate expresion $$d = d_0 \left(\frac{r_0}{r}\right)^2$$

Where $d_0$ and $r_0$ are the initial thickness and density. When we put all the formulas together, we get $$\Delta P = \frac{2 E_Y d_0}{r} \left(\frac{r_0}{r}\right)^2$$.

So you can see that the pressure is dropping for higher $r$ as $r^{-3}$. The only chance the S-curve would play a role is if we were close to the tip even for initial values of thickness and $r$. By putting a small $r_0=1cm$, $d_0=1mm$ and $E_Y = 0.1 GPa$, we get an initial pressure of $$\Delta P = 20 kPa$$,

which is by all means less than $15 MPa$, so the S-curve will for sure play no role in the first blow.

To conclude, the pressure is highest for the first blow because both the rubber gets spread out and a larger surface requires less of an investment of energy to contain more volume.

The effects of pre-spreading your balloon just give you a larger $r_0$ and the balloon eventually breaks only because the wall gets too thin and small imperfections cause it to break down even for the very small pressure.

• Don't confuse the pressure inside the sphere with the stress on the rubber - to get from one to the other you need a factor $(2\pi r t / \pi r^2 = 2t/r$ - depending on the dimensions of the balloon, this can get you in the neighborhood of 15 MPa I think. – Floris Jul 2 '14 at 16:04
• Your answer says that the more you stretch the balloon, the less tension there is in it. This is wrong by common experience. Your claim that tension is directly proportional to thickness is wrong. The thickness is not zero when the tension is zero, for example. – Mark Eichenlaub Jun 20 '16 at 6:59

The material of an unstretched baloon does resist stretching; this can largly be overcome by manually pre-stretching it (something I learned from my mom when I was about 5 years old). The other thing I'd consider (and that I don't see mentioned so far) is the cumulitive force exerted on the inner surface of the baloon as it's inflated, calculated by Pressure x Area. Example: a 2" diameter round baloon has in surface area of about 50 sq in. at 2 pounds sq in, that's 100 pounds of force total. At 4" diameter, it has about 200 sq in; at the same 2 pounds per sq in, that's now 400 pounds of force total. The difference in the force is the square of the change in diameter.

• -1 for not using SI (derived) units on a scientific website. – Guntram Blohm Jul 1 '14 at 23:35

The pressure curve for a rubber balloon , summarized in the Wikipedia article Two-balloon experiment, derives the pressure using a theoretical stress equation based on the thermodynamic theory of elasticity for an ideal rubber. They find the approximate formula $$P= \frac{C}{r_0^2 r}\left[1 - \left(\frac{r_0}{r}\right)^6\right]$$ where $r_0$ is the balloon's uninflated radius.

This lecture uses the "two-dimensional, plane stress generalized Hooke’s law with respect to the Eulerian strain measure" (which they call a a very rough approximation) to derive a curve ($\nu = 1/2$ for incompressible).

Both derivations take into account the thinning of the (assumed incompressible) rubber as the membrane stretches.

Because the rubber is thicker initially. Thick rubber is harder to stretch than thin rubber, in proportion to its thickness. And the thickness of the rubber in a balloon is inversely proportional to its surface area.

• Not true in general, as seen on the stress strain curve. As it gets beyond a certain thickness it becomes harder to inflate even though the balloon material is still becoming thinner. – ARM Jun 30 '14 at 20:11

The elastic nature of rubber is varying on pressure/temperature inversely, which makes it harder when cool and softer when applied pressure.

So to inflate a balloon in a normal condition we need to put more pressure initially. As when it expands, the pressure inside increases which reduces the elasticity of rubber making it easier to blow later on.

The Elasticity of a material is defined as the tendency of solid material to return to their original shape after being deformed.

We have noticed that when balloon bursts , its pieces becomes cool.This is the reverse effect as when the pressure inside reduces on a sudden it absorbs the room temperature and retains its elasticity.

Actually it depends on the material of balloon. If the balloon is made of less elastic material then it will become even more difficult to blow it even after little inflation cause it will reach its elasticity limit very early and this might not be true in case of more elastic balloons.

Intuitively, you can think of it this way:

Pressure is measured as Force / Area, e.g. Pounds per Square Inch (PSI), or Newtons per Square Meter (Pascals). Initially, the surface area inside the deflated balloon is small, meaning more pressure will be required to inflate it. Once you begin inflating the balloon, the surface area inside grows larger and larger. This means that less pressure will be required to inflate it, even though the balloon is resisting with more force as it is stretched.

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