Q158 / 180 · COMEDK 2022 · Physics
The average kinetic energy of a molecule in air at room temperature of $20^{\circ} \mathrm{C}$
- A$6 \times 10^{-22} \mathrm{~J}$
- B$7.06 \times 10^{-21} \mathrm{~J}$
- Correct answerC$6.07 \times 10^{-21} \mathrm{~J}$
- D$6.70 \times 10^{-21} \mathrm{~J}$
Solution
The average kinetic energy of a molecule is given by the formula $K_{avg} = \dfrac{3}{2} k_{B} T$, where $k_{B}$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.
Given temperature $T = 20^{\circ} \mathrm{C} = 20 + 273.15 = 293.15 \mathrm{~K}$.
The Boltzmann constant $k_{B} \approx 1.38 \times 10^{-23} \mathrm{~J/K}$.
Substituting the values into the formula:
$K_{avg} = \dfrac{3}{2} \times (1.38 \times 10^{-23} \mathrm{~J/K}) \times 293.15 \mathrm{~K}$
$K_{avg} = 1.5 \times 1.38 \times 293.15 \times 10^{-23} \mathrm{~J}$
$K_{avg} = 2.07 \times 293.15 \times 10^{-23} \mathrm{~J}$
$K_{avg} \approx 606.82 \times 10^{-23} \mathrm{~J}$
$K_{avg} \approx 6.0682 \times 10^{-21} \mathrm{~J}$
Rounding to three significant figures, we get $6.07 \times 10^{-21} \mathrm{~J}$.
Answer: $6.07 \times 10^{-21} \mathrm{~J}$
Given temperature $T = 20^{\circ} \mathrm{C} = 20 + 273.15 = 293.15 \mathrm{~K}$.
The Boltzmann constant $k_{B} \approx 1.38 \times 10^{-23} \mathrm{~J/K}$.
Substituting the values into the formula:
$K_{avg} = \dfrac{3}{2} \times (1.38 \times 10^{-23} \mathrm{~J/K}) \times 293.15 \mathrm{~K}$
$K_{avg} = 1.5 \times 1.38 \times 293.15 \times 10^{-23} \mathrm{~J}$
$K_{avg} = 2.07 \times 293.15 \times 10^{-23} \mathrm{~J}$
$K_{avg} \approx 606.82 \times 10^{-23} \mathrm{~J}$
$K_{avg} \approx 6.0682 \times 10^{-21} \mathrm{~J}$
Rounding to three significant figures, we get $6.07 \times 10^{-21} \mathrm{~J}$.
Answer: $6.07 \times 10^{-21} \mathrm{~J}$
Hence, the correct answer is option: 'C'.