1) The molecular mass of helium is 4 g/mol, the Boltzmann’s constant is 1.38066
ID: 1984705 • Letter: 1
Question
1) 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? I got 3.899*10^23 and this is not correct
2) What is the average kinetic energy of each
helium atom?
Answer in units of J
I got 1443.1856 but this is incorrect
3) 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 p(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 = v(3RT/M) vRMS = v[3*8.31451 J/K · mol * 334 K/0.004 kg/mol] vRMS = 1443.1856 m/s
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