# How to predict the weight of helium?

I have a balloon, which weighs 10 grams. now i fill it with helium (x gram)... since I added helium into the balloon. now the weight should be 10+x grams... but the balloon goes upwards. Shape of the balloon is irregular so I cannot find the volume. How to measure the weight of the helium?

## 2 Answers

First you need to find the volume. Submerge this balloon with helium filled in in a large enough tub completely filled with water and placed in another large outward container. Some water will flow out. Measure its volume. That's the volume of your balloon. Multiply this with the density of helium and you will get its weight.

If you suspect your helium supply is polluted with other stuff then you need to determine the effective density yourself. So do the next part of the experiment in a room (with a ceiling so that the balloon does not escape) with a short string attached to the balloon. Tie a small bag to the other end of this string. Let go of the string, the balloon will get stuck at the ceiling. Then start putting sand slowly in the bag until the balloon starts to come down. Do this slowly because you want to stop as soon as the ballon gradually starts to come down.

Then weigh the sand. Say this weight is $W_s$, the weight of the balloon shell is $W_b$, volume is $V$, weight of helium in the balloon is $W_h$, and density of air is $\rho_a$. Then the following will hold true:

$W_s + W_b + W_h = \rho_a V$

Put a little weight on it so that it is just short of going upwards.Then the wight of helium(x) plus the external weieght (y) equals the weight of the balloon (10 gms).Knowing "y" you can solve for "x". By the way You can find the volume of the balloon by immersing it in water.The volume if water displaced would equal to the volume of the balloon.

• Why does the weight of the helium plus the external weight equal the weight of the balloon? If there are more than 10 grams of helium, wouldn't you need a negative weight for that to work? – Blavius May 2 '17 at 5:30
• I stand corrected.The weight of the balloon + the weight of the helium + external weight would equal the buoyant force given by volume of the balloon multiplied by density of air multiplied by acceleration due to gravity. – Chappy May 2 '17 at 6:08