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Q.
When temperature is increased from $0^{\circ} C$ to $273^{\circ} C$, in what ratio the average kinetic energy of molecules change ?
AMUAMU 2001
Solution:
The average kinetic energy $(KE)$ is given by
$KE =\frac{3}{2} k T$
where $T$ is absolute temperature.
$\therefore \frac{ KE _{1}}{ KE _{2}}=\frac{T_{1}}{T_{2}}$
Given, $T_{1}=0^{\circ} C =273\, K$,
$T_{2}=273^{\circ} C =546\, K$
$\Rightarrow \frac{E_{2}}{E_{1}}=\frac{T_{2}}{T_{1}}=\frac{546}{273}$
$=2$