Question

Consider 3.01*10^24 molecules of oxygen (atomic mass is 16.0 g/mole) in a container at a temperature of 22.0 degrees celsius. If this gas can be considered as an ideal gas, calculate: 1.The Average translational kinetic energy of the gas. 2.The average translational kinetic energy per molecule 3. The root mean square speed of a gas molecule. 4. The most probable speed of the gas molecules.

Answer 1

Answer:

See explanation

Explanation:

The tranlational energy of a molecule is obtained by; ET =3/2 kT

T is the temperature while k is the Boltzmann constant (k)

The Boltzmann constant (kB) relates temperature to energy.

Substituting values;

ET = 3/2 (1.3807 × 10−23 joule/K) * (22 + 273)K

ET = 6.11 * 10^-21 J

The average translational kinetic energy per molecule=  6.11 * 10^-21 J/ 3.01*10^24 molecules = 2.03 * 10^-45 J per molecule

vrms = √3RT/M

vrms = √3 * 8.314 * 295/32

vrms = 15.2 m/s

Most probable speed =  √2RT/M

Most probable speed = √2 * 8.314 * 295/32

Most probable speed = 12.38 m/s