Calculating number of molecules in a balloon?
Consider a balloon containing nitrogen molecules at room temperature (20 C) and atmospheric pressure. If the balloon has a spherical shape with radius of 13 cm, what is the number of molecules in the balloon?
- pisgahchemistLv 710 years agoFavorite Answer
Balloons turn out to be a very poor model for numerical calculations involving the gas laws. The reason is that the elasticity of the rubber balloon itself contributes to the forces involved. Therefore, the pressure of the gas inside the balloon is not the same as atmospheric pressure.
We assume that blowing up a balloon means that the balloon expands as molecules are added, and that we we stop the pressure on the inside is equal to the pressure outside. That is not the case. In reality, the pressure inside is greater and equal to the outside pressure PLUS the elastic force needed to stretch the balloon. And that latter value we do not know, and so we don't know the pressure inside unless there is some kind of gauge attached to the balloon.
Your question is one of the few that is actually worded reasonably well because it states the pressure inside the balloon rather than outside. To solve it, simply use the ideal gas equation to get the moles and convert to the number of particles.
Get the volume in cubic decimeters, which are equivalent to liters. 1 cm is equal to 0.1 dm. 13 cm is 1.3 dm. Volume is 4/3(pi)r^3. So V = 9.20L
PV = nRT
n = PV / RT
n = 1.00 atm x 9.20L / 0.0821 Latm/molK / 293K
n = 0.383 mol N2
Convert moles to molecules by multiplying by Avogadro's number:
0.383 mol N2 x (6.022x10^23 molecules N2 / mol N2) = 2.30x10^23 molecules N2
- 10 years ago
You need a value for the atmospheric pressure in your problem.
Use the formula PV=nRT to solve for the number of moles of N2. Then multiply by Avogadro's number.
First you need the volume of the balloon. V= 4/3*π*r^3 because you have a sphere. Plug in your r=13 and you get:
V of the balloon = 4/3*π*2197 V=9202.77208 cm^3. You need this volume to be in L because the pressure in your problem is in atmospheres. So you use:
1 cm^3=1 ml 1 L=1000ml 9202.77208 cm^3=9202.77208ml/1000=9.20277208 L
Plug into PV=nRT: (P in atm)(9.20277208L) = n(0.08206)(293.15K)
So n = (P)(9.20277208L)/(0.08206)(293.15K)
Once you get your number of moles from this, just multiply the value by Avogadro's number (6.022X10^23) and that will give you the number of molecules in the balloon.
# molecules = n*6.022X10^23
- Anonymous10 years ago
p = 1 atm
va = (4/3)(pi)(0.13)^3 = 0.0092 m3
va = 0.0092 m3 air
v = 0.0092 * 78% = 0.007 178 m3 Nitrogen
R = 8.205 746E-5 m3-atm/K/mol picked from wiki based on units used
T = 20 + 273.15 = 293.15K
pv = nRT
n = pv / RT
plug in the data
n = 0.2225 mol
molecules = 0.2225 * 6.022E+23
molecules = 1.34E+23
- Anonymous4 years ago
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- Anonymous10 years ago
The Obama administration has earmarked millions of tax payer dollars to study this perplexing scientific issue. To pay for this campaign promise, a moderate tax increase is on the books. Those who gleefully voted Democratic in the last presidential election have no basis to complain and will gladly support whatever the president wants.