Question

the average thermal energy of a oxygen atom at room temperature (27c)​

Answer 1

from conservation of energy,

average kinetic energy of an ideal gas = average thermal energy of that gas

we know,

average kinetic energy of an ideal gas is given as, K.E = 3/2 kT

where k is Boltzmann's constant and T is temperature.

here, k = 1.38 × 10^-23 m²kg/s²K

and T = 27°C = 27 + 273 = 300K

now, K.E = 3/2 × 1.38 × 10^-23 × 300

= 450 × 1.38 × 10^-23 J

= 621 × 10^-23 J

= 6.21 × 10^-21 J

so, average thermal energy of an oxygen atom at room temperature is 6.21 × 10^-21 J

Answer 2

$\huge\bold\purple{Answer:-}$

k = 1.38 × 10^-23 m²kg/s²K

and T = 27°C = 27 + 273 = 300K

now, K.E = 3/2 × 1.38 × 10^-23 × 300

= 450 × 1.38 × 10^-23 J

= 621 × 10^-23 J

= 6.21 × 10^-21 J

so, average thermal energy of an oxygen atom at room temperature is 6.21 × 10^-21 J