A sealed container holds 2 moles of nitrogen (N2) gas at a pressure of 2 atmospheres and a temperature of 307 K. The atomic mass of nitrogen is 14 g/mol. The Boltzmann constant is 1.38 × 10-23 J/K and the ideal gas constant is R = 8.315 J/mol•K = 0.0821 L ∙ atm/mol ∙ K. The average translational kinetic energy of a nitrogen molecule, in 10-20 J, is closest to
Question
A sealed container holds 2 moles of nitrogen (N2) gas at a pressure of 2 atmospheres and a temperature of 307 K. The atomic mass of nitrogen is 14 g/mol. The Boltzmann constant is 1.38 × 10-23 J/K and the ideal gas constant is R = 8.315 J/mol•K = 0.0821 L ∙ atm/mol ∙ K. The average translational kinetic energy of a nitrogen molecule, in 10-20 J, is closest to
Solution
The average translational kinetic energy of a molecule in an ideal gas is given by the equation:
KE_avg = 3/2 kT
where k is the Boltzmann constant and T is the temperature in Kelvin.
Given that k = 1.38 × 10^-23 J/K and T = 307 K, we can substitute these values into the equation to find the average kinetic energy:
KE_avg = 3/2 * (1.38 × 10^-23 J/K) * 307 K
Solving this equation gives:
KE_avg = 6.35 × 10^-21 J
However, the question asks for the answer in units of 10^-20 J. To convert from 10^-21 J to 10^-20 J, we divide by 10:
KE_avg = 6.35 × 10^-21 J / 10 = 0.635 × 10^-20 J
So, the average translational kinetic energy of a nitrogen molecule is closest to 0.635 × 10^-20 J.
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