Question

A sealed container holds 7 moles of nitrogen (N2) gas at a pressure of 10 atmospheres and a temperature of 338 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​

Answer 1

The average translational kinetic energy of a nitrogen molecule at a temperature of 338 K is approximately 7.01 × 10^-21 J, which can be expressed as 70.1 × 10^-20 J.

Step-by-step explanation:

The question relates to the application of kinetic theory and thermodynamics principles in physics to calculate the average translational kinetic energy of a nitrogen molecule.

According to the kinetic molecular theory, the average translational kinetic energy (KE) of a single gas molecule is given by the formula KE = (3/2)kT, where k is the Boltzmann constant and T is the absolute temperature in Kelvin.

Using the provided values, the average translational kinetic energy of a nitrogen molecule can be calculated as:

KE = (3/2) x (1.38 × 10-23 J/K) x (338 K)

After plugging in the numbers and calculating, you'll find that the average kinetic energy is approximately 7.01 × 10-21 J, which is equal to 70.1 × 10-20 J.

So, when expressed in units of 10-20 J, the average translational kinetic energy of a nitrogen molecule is closest to 70.1.