Q)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.
What is the average kinetic energy of each helium atom?
Answer in units of J.
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.
What is the average kinetic energy of each helium atom?
Answer in units of J.
Sol:
The average kinetic energy is given by:
KEatom = (PV)/N
Number of atoms (N)= Avogadro's number (A) * n
where N = A*n = A*(PV/RT)
where N = A*n = A*(PV/RT)
N = 6.02214 x 10^23 atoms/mole * (96235 Pa * 0.01414
m^3) / (8.31451 J/K-mol * 285.15 K)
N = 3.456 x 10^23 atoms
N = 3.456 x 10^23 atoms
KEatom = (96235*0.01414 /3.456 x 10^23 )J/atom
= 3.94 X 10^-21 J/atom
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