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The molecular mass of helium is 4 g/mol, the Boltzmann’s constant is 1.38066 × 10−23 J/K, the universal gas constant is 8.31451 J/K · mol, and Avogadro’s number is 6.02214 × 1023 1/mol. Given: 1 atm = 101300 Pa.

1.How many atoms of helium gas are required to fill a balloon to diameter 36 cm at 49◦C and 0.902 atm?
2. What is the average kinetic energy of each helium atom? Answer in units of J.3. What is the rms speed of each helium atom? Answer in units of m/s.

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greenkelvin5

Answer:

1.) N = 1.69×10²³ atoms

2.) K = 6.88×10⁻²¹ J

3.) v = 1.44×10³ m/s

Explanation:

1)  Using the molecular formula pV = nRT , where P=  pressure, V= volume, n= number of atom, R= Avogaro's constant and K = temperature in Kelvin

(0.934 atm)(101300 Pa/atm)(4π/3)(0.25 m / 2)³ = (N / 6.02214×10²³ mol^-1)(8.31451 J/mol-K)(59 + 273.15K)

N = 1.69 ×10²³ atoms

2)

K = (3/2)(kB)(T)

K = (3/2)(1.38066e-23 J/K)(59 + 273.15K)

K = 6.88×10⁻²¹ J

3)

K = (1/2)mv²

6.88×10⁻²¹ J = (1/2)(4 g/mol)(mol / 6.02214ײ³)(kg / 1000 g)v²

v = 1.44×10³ m/s

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