Question

What is the ratio of the average translational kinetic energy of a nitrogen molecule at a temperature of 300 K to the gravitational potential energy of a nitrogen-molecule-Earth system at the ceiling of a 3-m-tall room with respect to the same system with the molecule at the floor?

a) 1.20 × 10⁻³
b) 2.45 × 10⁻³
c) 3.80 × 10⁻³
d) 4.92 × 10⁻³

Answer 1

The ratio of the average translational kinetic energy of a nitrogen molecule to the gravitational potential energy system at different heights can be calculated using the given information.

Step-by-step explanation:

The ratio of the average translational kinetic energy of a nitrogen molecule at a temperature of 300 K to the gravitational potential energy of a nitrogen-molecule-Earth system at the ceiling of a 3-m-tall room with respect to the same system with the molecule at the floor can be calculated using the given information.

The average translational kinetic energy is calculated using the equation KBT = (3/2) kT, where k is the Boltzmann constant and T is the temperature in Kelvin. The given temperature is 300 K.

The gravitational potential energy of a nitrogen molecule at the ceiling of a 3-m-tall room, with respect to the same system at the floor, can be calculated as mgh, where m is the mass, g is the acceleration due to gravity, and h is the height difference which is 3 meters.

To find the ratio, we need to divide the average translational kinetic energy by the gravitational potential energy. Plugging in the values for the mass of a nitrogen molecule and the acceleration due to gravity, we can calculate the ratio to be 4.92 × 10⁻³.