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

Question

Natalie Anton · Accepted 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

AVERAGE KINETIC ENERGY

The [ensemble] average kinetic energy #<< K >># at these fairly high temperatures can sometimes be approximated via the equipartition theorem:

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

where:

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

For rotation in 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 >>#.

As mentioned, there are rotational and vibrational contributions to #<< K >>#, and those depend on what the appropriate high temperature limits, or "classical limits", for #"N"_2# are with regards to rotation and vibration.

For #"N"_2#, using data from NIST, we can determine their relative importance by calculating the rotational temperature(s) #Theta_(rot)# and vibrational temperature(s) #Theta_(vib)#:

#Theta_(rot) = tildeB_e//k_B#

#Theta_(vib) = tildeomega_e//k_B#

where:

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

For #"N"_2#, #tildeB_e = "1.99824 cm"^(-1)# and #tildeomega_e = "2358.57 cm"^(-1)#, so:

#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 #N#.

Given that, the average kinetic energy for #"N"_2# far above #Theta_(rot)# but far below #Theta_(vib)# would be:

#<< K >> = << K >>_(trans) + << K >>_(rot) + cancel(<< K >>_(vib))^(~~ 0) + . . . # #("Ignore other minor contributions")#

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

At those temperatures of #"273 K"# and #"546 K"#,

#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

For particles following the Maxwell-Boltzmann distribution, there are "most probable" velocities, "root-mean-square" velocities, and "average" velocities.

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 #sqrt(T/M)#.

Henry Antrim · Answer

undefined The average kinetic energy ($KE_{avg}$) of gas molecules can be calculated using the formula:

$$ KE_{avg} = \frac{3}{2} kT $$

Where:
- $ k $ is the Boltzmann constant ($1.38 \times 10^{-23} \, \text{J/K}$)
- $ T $ is the temperature in Kelvin

The root mean square velocity ($v_{rms}$) of gas molecules can be calculated using the formula:

$$ v_{rms} = \sqrt{\frac{3kT}{m}} $$

Where:
- $ m $ is the molar mass of the gas

Given that nitrogen gas ($N_2$) has a molar mass of approximately $28.02 \, \text{g/mol}$, we can calculate the average kinetic energy and velocity at the given temperatures:

At $273 \, \text{K}$:
- $ KE_{avg} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 273 $
- $ v_{rms} = \sqrt{\frac{3 \times (1.38 \times 10^{-23}) \times 273}{0.02802}} $

At $546 \, \text{K}$:
- $ KE_{avg} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 546 $
- $ v_{rms} = \sqrt{\frac{3 \times (1.38 \times 10^{-23}) \times 546}{0.02802}} $

After calculation, you will get the values for both $KE_{avg}$ and $v_{rms}$ at $273 \, \text{K}$ and $546 \, \text{K}$.