How do you estimate the average kinetic energy of #"2 mols"# of #"O"_2# gas at #"298 K"#?

Answer 1
We need to consider what kinds of kinetic energy contributions there are... and once we do, we should get #"12.4 kJ"#.

And the average kinetic energy at the high-temperature limit is given by the equipartition theorem:

#<< kappa >> = << kappa >>_(tr) + << kappa >>_(rot) + << kappa >>_(vib) + . . . #
#= K/n = N/2 RT#

where:

The high-temperature limit is simply the temperature where the molecule is actually accessing certain energy levels that allow it to move a certain way.

Breaking it down...

We can verify above what temperatures we can use #N/2RT# by using the following:
#Theta_(rot) = (tildeB_e)/(k_B) stackrel(?)(" << ") T#
#Theta_(vib) = (tildeomega_e)/(k_B) stackrel(?)(" << ") T#

where these are the rotational and vibrational temperatures, respectively.

These data are readily available on NIST for diatomic molecules:

#tildeB_e = "1.4376766 cm"^(-1)# #tildeomega_e = "1580.193 cm"^(-1)#

These temperatures are then:

#Theta_(rot) = ("1.4376766 cm"^(-1))/("0.695 cm"^(-1)"/K") = "2.07 K"#
#Theta_(vib) = "1580.193 cm"^(-1)/("0.695 cm"^(-1)"/K") = "2273.7 K"#
Since we are well above #"2.07 K"# but well below #"2273.7 K"#, we are to include #N_(rot) = 2# and ignore #N_(vib)# completely.
Therefore, the average kinetic energy at #"298 K"# is around:
#color(green)(<< kappa >>) = N/2RT#
#= N_(tr)/2RT + N_(rot)/2RT#
#= 3/2RT + 2/2RT = color(green)(5/2RT)# #" J/mol"#

And we then get:

#color(green)(<< kappa >>) = 5/2 cdot "8.314472 J/mol"cdot"K" cdot "298 K"#
#=# #"6194.3 J/mol"#
#=# #color(green)("6.19 kJ/mol")#
Since you want #"2 mols"# of #"O"_2#...
#color(blue)(K) = n<< kappa >>#
#= "2 mols" cdot "6.19 kJ/mol"#
#=# #color(blue)("12.4 kJ")#
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Answer 2

To estimate the average kinetic energy of 2 moles of ( \text{O}_2 ) gas at 298 K, you can use the equation for average kinetic energy:

[ \text{Average Kinetic Energy} = \frac{3}{2} k_b T ]

where:

  • ( k_b ) is the Boltzmann constant (( 1.38 \times 10^{-23} , \text{J/K} )),
  • ( T ) is the temperature in Kelvin.

Plugging in the values:

[ \text{Average Kinetic Energy} = \frac{3}{2} \times 1.38 \times 10^{-23} , \text{J/K} \times 298 , \text{K} ]

[ = 6.21 \times 10^{-21} , \text{J} ]

So, the average kinetic energy of 2 moles of ( \text{O}_2 ) gas at 298 K is ( 6.21 \times 10^{-21} , \text{J} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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