First slide

Law of equipartition of energy

Question

At which of the following temperatures would the molecules of a gas have twice the average kinetic energy they have at $20^{0} C$ ?

Easy

A

$40^{0} C$
B

$80^{0} C$
C

$313^{0} C$
D

$586^{0} C$

Solution

$E \propto T ∴ \frac{E_{2}}{E_{1}} = \frac{T_{2}}{T_{1}} \Rightarrow \frac{2 E_{1}}{E_{1}} = \frac{T_{2}}{(20 + 273)}$
$\Rightarrow T_{2} = 293 \times 2 = 586 K = 313^{0} C$

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