First slide

Kinetic theory of ideal gases

Question

Two gases are at absolute temperatures 300 K and 350 K respectively. The ratio of average kinetic energy of their molecules is

Easy

A

6 : 7
B

36 : 49
C

7 : 6
D

343 : 216

Solution

From the relation,
$\begin{array}{l} \bar{KE} = \frac{3}{2} kT \\ \Rightarrow \frac{({\bar{KE)}}_{1})}{(\bar{KE})_{2}} = \frac{T_{1}}{T_{2}} \end{array}$
Given: $T_{1} = 300 K, T_{2} = 350 K$
$\Rightarrow \frac{({\bar{K E)}}_{1}}{{(\bar{K E})}_{2}} = \frac{300}{350} = \frac{6}{7}$

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