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 $K=\frac{3}{2}KT$
Now,$T_1=27^{\circ }C=300K$
$K_1=6.21\times 10^{-21}J$
$T_2=227^{\circ }C=500K$
$K_2=?$
We have, $\frac{K_1}{K_2}=\frac{T_1}{T_2}$
$\Rightarrow K_2=\frac{T_1}{T_2}\times K_1=\frac{500}{300}\times 6.21\times 10^{-21}$
$=10.35\times 10^{-21}J$