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Musya8 [376]
3 years ago
14

The bond energy for the van der waals bond between two helium atoms is 7.9×10−4ev. assuming that the average kinetic energy of a

helium atom is (3/2)kbt, at what temperature is the average kinetic energy equal to the bond energy between two helium atoms? use kb=8.62×10−5ev/k.
Chemistry
1 answer:
Sonja [21]3 years ago
4 0
Given: van der Waal's bond energy of He = <span>7.9×10−4ev;
and </span><span>kb=8.62×10−5ev/k
</span>
Now, according to Kinetic theory of gases we know that, 
K.E = 3/2 (kb) T
where T = temperature

∴ T = (2 K.E)/(3 x kb)
Now for K.E = 7.9×10−4ev
T = (2  X 7.9×10−4)/ (3 X 8.62×10−5) = 6.1 K

Thus, at temperature of 6.1 K, <span>average kinetic energy equal to the bond energy between two helium atoms</span>
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