I know that helium balloons float because it is less dense than air. I'm not expecting my bike to float, although that would be pretty cool. I just wanna know if replacing normal air with helium in the tires will produce a noticeable effect on its weight. Will the helium 'lift'/reduce the weight force on the bike?
Bike Weight – Will Helium in the Tires of a Bike Make It Lighter?
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Basic idea
Picture in your mind a deep ocean of water. Imagine a column of the water, going from the surface down to a depth $d$. That column of water has some weight $W$. Therefore, there is a downward force of magnitude $W$ on that column of water. However, you know the column of water is not accelerating, so there must be an upward force of magnitude $W$ pushing on that column. The only thing underneath the column is more water. Therefore, the water at depth $d$ must be pushing up with force $W$. This is the essence of buoyancy. Now let's do details.
Details
The weight $W$ of a column of water of cross-sectional area $A$ and height $d$ is
$$W(d) = A d \rho_{\text{water}}$$
where $\rho_{\text{water}}$ is the density of water. This means that the pressure of water at depth $d$ is
$$P(d) = W(d)/A = d \rho_{\text{water}}.$$
Now suppose you put an object with cross sectional area $A$ and height $h$ in the water. There are three forces on that object:
- $W$: The object's own weight.
- $F_{\text{above}}$: The force of the water above the object.
- $F_{\text{below}}$: The force of the water below the object.
Suppose the bottom of the object is at depth $d$. Then the top of the object is at depth $d-h$. Using our results from before, we have
$$F_{\text{below}} = P(d)A=d \rho_{\text{water}} A $$
$$F_{\text{above}}=P(d-h)A=(d-h)A\rho_{\text{water}}$$
If the object is in equilibrium, it is not accelerating, so all of the forces must balance:
$\begin{eqnarray} W + F_{\text{above}} &=& F_{\text{below}} \\ W + (d-h) \rho_{\text{water}} A &=& d \rho_{\text{water}} A \\ W &=& h A \rho_{\text{water}} \\ W &=& V \rho_{\text{water}} \end{eqnarray}$
where in the last line we defined the object's volume as $V\equiv h A$. This says that the condition for equilibrium is that the weight of the object must be equal to its volume times the density of water. In other words, the object must displace an amount of water which has the same weight as the object. This is the usual law of buoyancy.
From this description I believe you can extend to the case of air instead of water, and horizontal instead of vertical pressure gradient.
If you have a "slack" balloon (one with no elasticity, like is used for some extreme altitude work like the one Felix Baumgartner used for the highest free-fall)
then the pressure inside is the same as the pressure outside, and the balloon will not find an equilibrium position due to pressure (the volume of air displaced will change with altitude, and the weight of air displaced will be unchanged). As Peter Green pointed out, if such balloons reach a "fully inflated" state at a certain altitude, then they will stop rising (reach equilibrium) - assuming they don't burst because of the pressure differential that will build up. For weather balloons this allows for sufficiently fine altitude control; but it is nowhere near the 1-5m range you are asking about.
On the other hand, if you had a perfectly rigid container, then it's conceivable that you will see a height at which the "balloon" would stabilize. It's worth noting how small the pressure gradient is. The approximate equation (ignoring certain large-scale effects) is
$$P=P_0 e^{-mgh/kT}$$
The derivative:
$$\frac{dP}{dh}=-\frac{mg}{kT}P$$
From this it follows that the pressure changes by 1% for a height change of approximately 88 m; if you want to stabilize over a range of 1 - 5 m, you need to have a rigid balloon with a mass that is correct to within 0.05%. But of course the ideal gas law still applies, so a 1C change in temperature will change the density by about 0.3%. This means the temperature has to be known (and stable) to about 0.1C before you can even think about it...
But wait - there's more. The pressure inside an ordinary (toy) balloon drops as the balloon increases in size: this means that a "real" balloon with elastic walls that is initially too light will not only rise - it will keep rising, as the buoyancy actually goes up as the balloon gets higher since the elastic skin becomes less capable of sustaining a pressure difference as the balloon gets bigger. Typically the result is that the balloon will eventually burst - hence the need for a "slack" balloon for high altitude work.
Now if you just have a thin string hanging from the bottom of your balloon, and it drags on the ground, then the additional weight of string that the balloon carries will cause it to find a stable height - even without it being tied.
Final note: indoors, there is typically a bit of air current which significantly affects the motion of a "free" balloon. And "cheap" helium balloons have significant leakage, meaning that a balloon that has the correct buoyancy one moment will be too heavy the next. I once did an experiment where I inflated a balloon and hung a box of matches underneath; I then waited until the balloon-with-matchbox landed, would take out one match, and watch it float off again. It would take about 10 minutes for it to "land", and I repeated the cycle. That gave a neat way of estimating the leakage rate for that particular type of balloon, in "match weights per hour".
Best Answer
It will make it lighter, but the effect will be very small. The volume of the tube is probably less than a liter. One mol of an ideal gas is 23 liters at atmospheric pressure. So you have about 0.2 mol of gas in there at 4 bar pressure. Helium weighs 4 g/mol, nitrogen about 28 g/mol. So for 0.2 mol, the weights are 0.8 g and 5.6 g. Cleaning off the dirt from the frame will have a greater effect.
Helium atoms are smaller than nitrogen molecules. Therefore there is a greater rate of diffusion through the bike tires. Your tires will become flat quicker than normal. Therefore it is not really a good idea to use helium.