Question

The average kinetic energy of a gas molecules at $27^{\circ} C$ is $6.21 \times 10^{-21} J$. Its average kinetic energy at $227^{\circ}C$ will be

• A. $10.35 \times 10^{-21} J$
• B. $52.2 \times 10^{-21} J$
• C. $5.22 \times 10^{-21} J$
• D. $11.35 \times 10^{-21} J$

Answer 1

Answer: (A)

We know that the $K.E$ of one mole of a gas molecules at a temperature $T$ is given by $K =\frac{3}{2} K T$
Now,$T_{1}=27^{\circ} C=300\, K $
$K_{1}=6.21 \times 10^{-21}\, J $
$T_{2}=227^{\circ} C=500\, K $
$K_{2}=? $
We have, $\frac{K_{1}}{K_{2}}=\frac{T_{1}}{T_{2}}$
$ \Rightarrow K_{2}=\frac{T_{1}}{T_{2}} \times K_{1}=\frac{500}{300} \times 6.21 \times 10^{-21}$
$=10.35 \times 10^{-21}\, J$