The molecular mass of helium is 4 g/mol, the Boltzmanns constant is 1.38066Ý1023

Question

The molecular mass of helium is 4 g/mol, the Boltzmanns constant is 1.38066Ý1023J/K, the universal gas constant is 8.31451 J/K·mol, and Avogadros number is 6.02214Ý10^231/mol. Given: 1 atm =101300 Pa. How many atoms of helium gas are required to fill a balloon to diameter 21 cm at 5?C and 0.584 atm? The molecular mass of helium is 4 g/mol, the Boltzmanns constant is 1.38066Ý1023J/K, the universal gas constant is 8.31451 J/K·mol, and Avogadros number is 6.02214Ý10^231/mol. Given: 1 atm =101300 Pa. How many atoms of helium gas are required to fill a balloon to diameter 21 cm at 5?C and 0.584 atm?

Explanation / Answer

1) PV = nRT
n = PV/RT
Number of atoms (N)= Avogadro's number (A) * n
N = A*n = A*(PV/RT)
P = ,584 * 101300 Pa
P = 591592
V = (4/3) pi R^3
R = 0.105 m
V = (4/3) pi (0.14 m)^3
V = 0.004849m^3
T = 273.15 + 5
T = 278.15 K

N = 6.02214 x 10^23* (591592 * 0.004849) / (8.3145 * 278.15 )
N = 7.4698197 x 10^23 atoms

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