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A balloon is filled with helium at a pressure of 2.4 x 105 Pa. The balloon is at a

temperature of 18 oC and has a radius of 0.25 m. (a) How many helium atoms are

contained in the balloon? (b) Suppose we double the number of helium atoms in the

balloons, keeping the pressure and the temperature fixed. By what factor does the radius

of the balloon increase?

Verified Answer :

Edufirst

1)
p = 2.4 10^5 Pa
T = 18° C + 273.15 = 291.15 k
r = 0.25 m => V = [4/3]π(r^3) = [4/3]π(0.25m)^3 = 0.06545 m^3 = 65.45 L

Use ideal gas equation: pV = nRT => n = pV / RT = [2.410^5 Pa 0.06545 m^3] / [8.31 J/kmol 291.15k] = 6.492 mol

Avogadro number = 1 mol = 6.022 10^23 atoms

Number of atoms = 6.492 mol 6.022 10^23 atom/mol = 39.097 10^23 atoms = 3.91 10^24 atoms

2) Double atoms => double volume

V2 / V1 = r2 ^3 / r1/3

2 = r2 ^3 / r1 ^3 => r2 ^3 = 2* r1 ^3

r2 = [∛2]r1

The factor is ∛2

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