Question

The average kinetic energy of a gas molecule at ${27}^{o}C$ is $6.21\times {10}^{-21}J$, then its average kinetic energy at ${227}^{o}C$ is:

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

Answer 1

Answer: (A)

$10.35\times {10}^{-21}J$

Average kinetic energy of gas molecules $\propto$ Temperature (Absolute)
$\frac{K.E\left(at{227}^{o}C\right)}{K.E\left(at{27}^{o}C\right)}=\frac{273+227}{273+27}=\frac{500}{300}=\frac{5}{3}$
$K.E\left({227}^o\right)=\frac{5}{3}\times 6.21\times {10}^{-21}J=10.35\times {10}^{-21}J$