# Determine balloon diameter needed to lift a 7 pound weight with various gases

My question is regarding setting up this equation. The givens are as follows:

• Gases are helium, methane and hydrogen. (Test all three)
• Assume balloon is spherical.
• Assume temperature is $25~\mathrm{^\circ C}$

So I need to find the diameter of the balloon needed to lift a 7lb weight. I setup my equation as follows. $$m=\frac{PVM}{RT}$$

Next, I subtract my first gas ($\ce{He}$) from ambient air and convert my $m$ to equal grams in the weight. so I set it up like so.

$$3730.15~\mathrm{g}=\frac{PVM}{RT}-\frac{PVM}{RT}$$

Next I factor out the pressure and volume and fill in my numbers. $$3730.15~\mathrm{g}=PV\left(\frac{(28.9~\mathrm{g~mol^{-1}})}{(0.0821~\mathrm{L~atm^{-1}mol^{-1}K^{-1}})(298~\mathrm{K})}-\frac{8~\mathrm{g~mol^{-1}}}{(0.0821~\mathrm{L~atm^{-1}mol^{-1}K^{-1}})(298~\mathrm{K})}\right)$$

This is where I am stuck at as I don't know what makes sense to use for pressure so I can solve for volume? Or maybe this is all setup wrong and makes no sense?

• You don't really have to go to this extent to solve the problem, but I think it will work either way. You can assume that the balloon is a thin flexible membrane with negligible mass, separating the gasses from the rest of the atmosphere. As such, the gas inside the balloon will be equal to atmospheric pressure. How about answering your own question? – Nicolau Saker Neto Jun 5 '15 at 21:05
• That's what I was stuck at. I finished the problem but was wondering if that was appropriate to use 1atm as the pressure for this equation. You are saying it is appropriate? If there is an easier way to do it I am more than happy to hear it! I spent a long time working on this two different ways and I just am having a hard time wrapping my head around buoyancy. – DanBaba Jun 6 '15 at 3:17