Answer:
v₂ = 395.83 m/s
Explanation:
given,
Average speed of nitrogen molecule, v₁ = 475 m/s
Temperature in summer, T₁ = 300 K
Average speed of nitrogen molecule in winter, v₂ = ?
Temperature in winter, T₂ = 250 K
The relation of average speed with temperature
[tex]v\ \alpha\ \sqrt{T}[/tex]
now,
[tex]\dfrac{v_2}{v_1} = \dfrac{\sqrt{T_2}}{\sqrt{T_1}}[/tex]
[tex]\dfrac{v_2}{475} = \dfrac{\sqrt{250}}{\sqrt{300}}[/tex]
v₂ = 0.833 x 475
v₂ = 395.83 m/s
The average speed on a frigid winter day is equal to v₂ = 395.83 m/s