How do you estimate the average kinetic energy of #"2 mols"# of #"O"_2# gas at #"298 K"#?
And the average kinetic energy at the high-temperature limit is given by the equipartition theorem:
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...
where these are the rotational and vibrational temperatures, respectively.
These data are readily available on NIST for diatomic molecules:
These temperatures are then:
And we then get:
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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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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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