Question

What is the kinetic energy of $1\text{g}$ of $O_2$ at ${47}^{\circ }\text{C}$?

• A. $2.17\times {10}^2\text{J}$
• B. $2.24\times {10}^2\text{J}$
• C. $1.24\times {10}^2\text{J}$
• D. None of these

Answer 1

The correct option is C $1.24\times {10}^2\text{J}$

$T=47+273=320\text{K}$
$K.E=\frac{3}{2}RT$ (For 1 mole)
$=\frac{3}{2}\times \left(8.314{\text{JK}}^{-1}{\text{mol}}^{-1}\right)\times 320K$
$=3990.7{\text{J mol}}^{-1}$
Molar mass of $O_2=32{\text{g mol}}^{-1}$
$\therefore KE$ (per gram of $O_2$)$=\frac{3990.7}{32}{\text{J g}}^{-1}$
$=124.7{\text{J g}}^{-1}$
$=1.247\times {10}^2{\text{J g}}^{-1}$
So, the kinetic energy of 1 gram of $O_2$ will be $1.24\times {10}^2\text{J}$.