Verified answer

The following equation relates the average kinetic energy of the gas molecules to the temperature of the gas.

½ * m * v^2 = 1.5 * k * T

m = mass of molecule

v = average velocity of the molecules

k = Boltzmann’s constant = 1.38 * 10^-23 J/˚K

T = temperature in ˚K

˚K = ˚C + 273

Divide both sides by ½ * m

v^2 = 3 * k/m * T

Square root both sides

v = (3 * k/m * T)^0.5

At 20˚C, v = (3 * k/m * 293)^0.5 = √879 * √(k/m)

At 40˚C, v = (3 * k/m * 313)^0.5 = √939 * √(k/m)

Factor of increase = Ratio of the two velocities = √939 / √879 ≈ 1.03

40˚C is not twice as hot as 20˚C

313 / 293 = 1.07

40˚C is 1.07 times as hot as 20˚C

Since temperature is proportional to average kinetic energy of the particles (T ~ Ek = 0.5mV²), if the temperature rises by 20 degrees (Celsius of Kelvin, doesn't matter), than the speed must have risen by √20.

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