# Why is a hot air balloon "stiff"?

1) Why is a hot air balloon stiff?
2) Is the pressure inside the balloon higher than the pressure outside (atmospheric pressure)?
3) If the pressure inside is higher than the outside, how is it explained by ideal gas law?

To understand the physics of inflated balloons you have to understand curved membranes under tension.

If there's a membrane under tension $T$ (for simplicity assumed to be isotropic), with fluids on both sides, it will be flat (planar) unless there is a pressure difference between the two fluids. If you notice that it's curved with radius of curvature $R$ (again assumed to be isotropic, like a sphere), then there must be a pressure difference given by $$\Delta p = 2 \frac{T}{R}$$ Of course, the membrane bulges away from the side where the pressure is higher, out into the side where the pressure is lower.

Now that we have this basic concept, the answers to your questions all come from it because they're all related.

The hot air balloon is stiff because it is under tension. If a membrane is not under tension it could have folds or wrinkles because there is no energy penalty for having extra area. But since the material of the balloon is under tension it's going to be as smooth as possible. It would be flat if not for the pressure difference, which forces it to have a smooth curve.

The pressure inside the balloon is indeed higher than the pressure outside, and the difference is given by the formula above.

This pressure difference is not really "explained" by the ideal gas law, but it is consistent with it. If you start with a deflated balloon on the ground and gradually fill it up with hot air, you're increasing $N$ (because there are more air molecules inside the balloon than when you started) and $V$, while $p$ and $T$ remain roughly constant (the $T$ inside being higher than the $T$ outside). When $V$ finally gets to the volume that will make the balloon stiff, the pressure begins to go up slightly above the ambient pressure, with the extra pressure being provided by the tension of the balloon. However, the density of the gas inside the balloon is still less than that of the air around it, because although $p$ is greater (tending to increase the density), $T$ is also greater (tending to decrease the density), and the temperature effect dominates.

• Ah, I finally found out the name for that equation: the Young-Laplace equation. Mar 18, 2011 at 5:35
• Wouldn't the pressure increase inside the balloon causes gas flowing to outside the balloon and hence the pressure remains same as outside the balloon? Mar 22, 2011 at 7:35
• @Rejeev No, the pressure is only equalized at the mouth of the balloon (on the bottom). At other points on the balloon there's a difference in pressure between inside and outside. Georg's explanation of "water in a bucket, but upside down" is exactly right. Mar 25, 2011 at 21:47

I'm going to try to accomplish something the other comments don't quite get.

The challenge here is that we're really talking about a fluids problem, really a hydrostatic problem. You can't just write the ideal gas equation and be done with it, although no one has really written that equation yet, so I'll write it here. But I'm not just going to write it. I'm going to write it in the molar / volumetric form that I personally believe does the most to answer the question.

$$\frac{P}{\rho} = R_{\rm specific}T$$

$\rho$ represents density and $R_{specific}$ is constant. I will stress the point that the total number of molecules in the balloon doesn't matter.

We have established that, yes, the pressure is higher on the inside of the balloon, as evidenced by the observed rigidness. There is nothing in the above equation that gives us an idea as to why. We need to introduce something else. That something else is the fact that the balloon exists in a gravitational field.

(I would add in response to the prior answer that it's not about the "gradient" of $P$ or $T$, but hopefully the rest of my answer will make it clear why)

A non-constant temperature within a gravitational field leads to pressure differences. Specifically, let's talk about going from point $A$ to point $B$. Again, keeping with the hydrostatic assumption, we can fairly easily establish the following for the difference in pressure between the two.

$$P_{A} - P_{B} = \int_A^B \rho(\vec{r}) g d\vec{r}$$

This is my version for the common $\rho g h$ that gives the change in pressure over a certain column of a fluid. I will reduce it back to these terms for the specific case of $A$ being a point on the inside surface of the balloon and $B$ being right next to it on the outside. Another assumption that I make is that the temperature is perfectly constant inside the balloon, $T_b$ and perfectly constant outside the balloon, $T_{air}$.

$$P_{A} - P_{B} = - (\rho g h)_b + (\rho gh)_{air} = g h \left( - \rho_b + \rho_{air} \right)$$

$$\rho = \frac{P}{R_{specific} T}$$

$$P_{A} - P_{B} = \frac{g h P}{R_{specific}} \left( - \frac{1}{T_b} + \frac{1}{T_{air}} \right)$$

Were you confused by the signs? I certainly was. You start at $A$ in the balloon and go downward, so the term for the path inside the balloon should be given a negative. Next, why did I take $P$ constant? Doesn't that violate the entire point of the problem? No. Pressure has little impact on the density in this case, but this is a relative statement, and it's relative to the temperature contribution to the density difference. The temperature within the balloon could be $10^{\circ} F$ higher than outside, which would make the temperature difference on the order of $5$% relative to absolute zero. The same is not true for pressure. The pressure difference due to the entire hydrostatic head for the entire balloon height is roughly $0.01 psi$, or $0.1$% of absolute.

I have an explanation for this. I would like to verify if that makes sense. Basically my hypothesis is that whenever there is a temperature gradient, there will be a pressure gradient. In our case there is a temperature gradient between outside the balloon and inside the balloon. This leads to a pressure gradient. This pressure gradient (read pressure difference) causes the balloon to be stiff.
I would like to seek feedback from the community on my hypothesis.

• Nope. Wrong. A pressure gradient is only possible if you have a force or if you permit the gas to flow. Consider a small element of mass $dM$, area of cross section $A$. If it had a pressure gradient from $P$ to $P+dP$, then it would have a force $AdP$, and a finite acceleration $A\frac{dP}{dM}$. In a closed tube, this is not possible as there is not 'outlet' for these accelerated elements to go. So, there must be a force balancing $AdP$. In the atmosphere, this force is gravity ($dMg=AdP$). Feb 22, 2012 at 3:23
• If a pressure gradient is not possible, the only other thing that can have a gradient is density (Due to \$PM=\rho RT). Feb 22, 2012 at 3:25

I have another explanation for this. It is related to Alan's answer. It is related to hydrostatic pressure of atmosphere.

Consider a column of air in a long vertical cylinder (10m height, 1 sq m area). The pressure difference between two layers at h0 and h1, equal to the weight of the air between h0 and h1 per unit area. That is, the pressure gradient is equal to the density of the air.

Now consider two air columns, one at ambient temperature 293 K and other at 373 K. Pressure gradient for first air column is 1.2 Kg/m2 and that of air column 2 is 0.94 kg/m2. (density of air at 293K is 1.2 kg/m3 and that at 373K is 0.94 kg/m3).

Now turn the second cylinder upside down - now that will become a hot air balloon. The pressure at the bottom of the upside down cylinder (mouth of the cylinder) will be equal to the ambient pressure (otherwise, air will move in either direction to make the pressure same). Atmospheric pressure at ambience is, 10000kg/sq m.

Now consider the pressure at 10m height in cylinder 2, as the pressure gradient is 0.94 kg/m2, at 10m height, the pressure difference is 9.4 kg/m2. Whereas in cylinder 1 the pressure difference is 12kg/m2. Hence, at height 10m, the pressure in cylinder 1 is 9988 kg/m2 (10000-12) and pressure in cylinder 2 is 9990.6kg/m2. So at every height above h0, pressure in cylinder 2 is greater than that in the cylinder 1. Cylinder 1 is equivalent to ambience, as the temperature and pressure are the same as in ambience. Hence, pressure in cylinder 2 is higher than that of ambience at every point above the mouth of the upside cylinder.

As we have noted an upside down cylinder is equivalent to hot air balloon, the same is applicable to hot air balloon. Few important points can be noted from the above explanation - the pressure difference is not the same at each point on the balloon, it is greater at the higher end of the balloon and no pressure difference at the mouth of the balloon. The lift of the balloon is not depending on the volume of the balloon, rather it depends on the height of the balloon and area of the balloon, probably that is why balloons are shaped like an onion rather than sphere.

# Lift of the balloon

Now consider the upside down cylinder again, consider the base of the cylinder (now at the top as the cylinder is upside down), air inside the cylinder pushes it up with 9990.6 kg/m2 whereas the air above the cylinder pushes it down 9988 kg/m2. Hence, there is a total lift force of 2.6kg/m2 (for a balloon with 10m height).

If a balloon shaped like T (rather a bolt in three-dimensional shape) with area of the head 200 sq m and height of 20m, can provide a lift of 1040 kg (even if the volume of the balloon is fraction of a conventional balloon).