Question

The mean translational kinetic energy of a gas molecule at a temperature of 47 $^\circ$C is: (Boltzmann constant = 1.38 $\times$ 10$^{-23}$ J K$^{-1}$)

• A. 6.37 $\times$ 10$^{-21}$ J
• B. 138 $\times$ 10$^{-4}$ eV
• C. 414 $\times$ 10$^{-4}$ J
• D. 414 $\times$ 10$^{-4}$ eV

Hint:

Use $KE = \frac{3}{2} k T$ for translational kinetic energy of a gas molecule. Convert temperature to Kelvin and check units (J or eV) carefully.

Answer 1

The Correct Option is A

The mean translational kinetic energy of a gas molecule is given by $KE = \frac{3}{2} k T$, where $k$ is the Boltzmann constant and $T$ is the temperature in Kelvin. 
Temperature: $47^\circ$C = $47 + 273 = 320$ K. 
$KE = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 320$. 
Calculate: $\frac{3}{2} \times 1.38 \times 320 = 1.5 \times 1.38 \times 320 = 2.07 \times 320 = 662.4 \times 10^{-23} = 6.624 \times 10^{-21}$ J. 
The correct answer should be $6.37 \times 10^{-21}$ J