# What is the average kinetic energy and velocity of nitrogen gas molecules at 273 K and at 546 K?

##### 1 Answer

I got...

#<< K >>_(273)//n = "5.68 kJ/mol"#

#<< K >>_(546)//n = 2<< K >>_(273)//n#

#<< v >>_(273) = "454 m/s"#

#<< v >>_(546) = sqrt2<< v >>_(273)# #"m/s"#

#v_(RMS,273) = "493 m/s"#

#v_(RMS,546) = sqrt2v_(RMS,273)# #"m/s"#

You should note that

#<< K >># , the average kinetic energy, is proportional to temperature#T# in#"K"# , as it should be by definition.#v# , some kind of gas speed, is proportional to#sqrt(T/M)# , with#M# being the molar mass in#"kg/mol"# .

**AVERAGE KINETIC ENERGY**

The [ensemble] **average kinetic energy** *equipartition theorem*:

#<< K >> = N/2nRT = << K >>_(trans) + << K >>_(rot) + << K >>_(vib) + . . . # where:

#N# is the number ofdegrees of freedomthat the particle can move in a certain way (translational, rotational, vibrational, . . . ).For

translation, we have#x,y,z# axes, so#N = 3# .For

rotationin diatomic molecules, we need two angles to describe rotation in three dimensions, so#N = 2# for#"N"_2# .For

vibration, it is more complicated, but for most molecules at ordinary temperatures, we can ignore its contribution to#<< K >># .

#n# is the mols of gas.#R# and#T# are straight from the ideal gas law, except#R = "8.314472 J/mol"cdot"K"# .

As mentioned, there are **rotational** and **vibrational** contributions to

For

#Theta_(rot) = tildeB_e//k_B#

#Theta_(vib) = tildeomega_e//k_B# where:

#tildeB_e# is the ground-state rotational constant in units of#"cm"^(-1)# .#tildeomega_e# is the ground-state fundamental vibrational frequency in units of#"cm"^(-1)# .#<< K >>_B ~~ "0.695 cm"^(-1)"/K"# is the Boltzmann constant in terms of#"cm"^(-1)# .

*These values are the minimum temperatures at which natural rotation or vibration can occur.*

For

#Theta_(rot) = "1.99824 cm"^(-1)/("0.695 cm"^(-1)"/K") = "2.88 K"#

#Theta_(vib) = "2358.57 cm"^(-1)/("0.695 cm"^(-1)"/K") = "3394 K"#

Therefore, it is a good approximation to **ignore** the vibrational degrees of freedom, but **include** the rotational degrees of freedom for the value of

Given that, the **average kinetic energy** for

#<< K >> = << K >>_(trans) + << K >>_(rot) + cancel(<< K >>_(vib))^(~~ 0) + . . . #

#("Ignore other minor contributions")#

#~~ 3/2nRT + 2/2nRT = 5/2nRT#

At those temperatures of

#color(blue)(<< K >>_(273)/n) = 5/2RT#

#= 5/2 cdot "0.008314472 J/""mol"cdotcancel"K" cdot 273 cancel"K"#

#=# #color(blue)("5.68 kJ/mol")#

#color(blue)(<< K >>_(546)/n) = 5/2RT#

#= 5/2 cdot "0.008314472 J/""mol"cdotcancel"K" cdot 546 cancel"K"#

#=# #color(blue)("11.4 kJ/mol")#

**AVERAGE VELOCITY**

*When you say the average velocity, I am taking you literally...*

#<< v >> = sqrt((8RT)/(piM))# where

#M# is the molar mass in#"kg/mol"# and everything else is as defined before.

Thus,

#color(blue)(<< v >>_(273)) = sqrt((8 cdot 8.314472 cancel"kg"cdot"m"^2"/"("s"^2cdotcancel"mol"cdotcancel"K") cdot 273 cancel"K")/(pi cdot 0.028014 cancel"kg""/"cancel"mol"))#

#=# #color(blue)("454 m/s")#

Without any further computation,

#color(blue)(<< v >>_(546)) = color(blue)(454sqrt2 " m/s")# How do I know that the average velocity at twice the temperature is

#sqrt2# times the velocity at the lower temperature?

**ROOT-MEAN-SQUARE SPEED**

Anyways, if you did not want the average velocity, you should ask for the **root-mean-square speed** instead...

#v_(RMS) = sqrt((3RT)/M)#

If we had chosen this instead...

#color(blue)(v_(RMS,273)) = sqrt((3 cdot 8.314472 cancel"kg"cdot"m"^2"/"("s"^2cdotcancel"mol"cdotcancel"K") cdot 273 cancel"K")/(0.028014 cancel"kg""/"cancel"mol"))#

#=# #color(blue)("493 m/s")#

Again, we can say

#color(blue)(v_(RMS,546) = 493sqrt2 " m/s")#

It clearly matters whether you want average, RMS, or most probable speed.

Regardless, in the end we still get that any kind of gas speed among those listed above is proportional to