015 (part 1 of 3) 10.0 points The molecular mass of helium is 4 g/mol, the Boltz

Question

015 (part 1 of 3) 10.0 points
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.
How many atoms of helium gas are required
to fill a balloon to diameter 31 cm at 49C and
1.799 atm?

016 (part 2 of 3) 10.0 points
What is the average kinetic energy of each
helium atom?
Answer in units of J

017 (part 3 of 3) 10.0 points
What is the rms speed of each helium atom?
Answer in units of m/s

Explanation / Answer

The molecular mass of He is M = 4g/mol

The Boltzmann’s constant is k = 1.38066*10^-23 J/K

The universal gas constant is R = 8.31451 J/K · mol

Avogadro’s number is N = 6.02214*10^23 /mol

1 atm = 101300 Pa

1)

Applying ideal gas law

PV = nkT

n is number of particles of gas

P = 0.863 atm = 87421.9 Pa

V = 4/3 (0.17)^3

V = 0.02057 m^3

temperature of the gas is T = 61 C = 334 K 

The number of He atoms in the balloon is n = PV/kT

n = 87421.9 Pa * 0.02057 m^3 / 1.38066*10^-23 J/K * 334 K

n = 3.899*10^23

2)

The average kinetic energy of each He atom

KEatom = 1.5*kT

KEatom = 1.5*1.38066*10^-23 J/K * 334 K 

KEatom = 6.9171066*10^-21 J

3)

The RMS speed is given by

vRMS = (3RT/M)

vRMS = [3*8.31451 J/K · mol * 334 K/0.004 kg/mol]

vRMS = 1443.1856 m/s

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