Question

calculate the total kinetic energy of 1kg of oxygen at 27℃ (R=8.31×10^3 J/mole/K)

Answer 1

$v = \sqrt{ \frac{3rt}{m} } \\ \frac{m {v}^{2} }{2} =( 3/2)rt \\ ke = (3/2)\times 8.31 \times {10}^{3} \times 300 \\ = 3739.5 \times \times {10}^{3}$
hope it helps

Answer 2

Total kinetic energy = n (3/2 RT0

where n = Number of moles of the gas

R = Gas constant

T = Absolute temperature

Molecular weight of methane,

CH4 = 12 + 4 * 1 = 16

∴ Number of moles of methane in 8.0 gm of methane

= 8.0/16.0 = 0.5

R = 8.314 joules/K/mole

T = 27 + 273 = 300 K

∴ Total kinetic energy of the molecules in 8.0 gm of methane at 27oC = InI * 3/2 RT

= 0.5 * 3/2 * 8.314 * 300 = 1870.65 joules

∴ Average kinetic energy = 1870.65/6.023 *1023 *0.5

= 6.21 * 10-21 joules/molecule