Question

The average speed of molecules in a sample of ozone gas is $1.84 \times 10^4 \text{ cm s}^{-1}$. The average translating kinetic energy per molecules is xJ. The value of $\frac{x}{2} \times 10^{22}$ is

• A. 8
• B. 16
• C. 4
• D. 2

Answer 1

Answer: (A)

8

Answer 2

The average speed of gas molecules is given by $v_{avg} = \sqrt{\frac{8kT}{\pi m}}$. The average translational kinetic energy per molecule is $KE_{avg} = \frac{3}{2} kT$. We can rewrite the average speed formula as $v_{avg} = \sqrt{\frac{8}{3\pi}} \sqrt{\frac{3kT}{m}}$. Given $v_{avg} = 1.84 \times 10^4 \text{ cm s}^{-1} = 1.84 \times 10^2 \text{ m s}^{-1}$ and $\sqrt{\frac{8}{3\pi}} = 0.92$. Substituting these values: $1.84 \times 10^2 \text{ m s}^{-1} = 0.92 \sqrt{\frac{3kT}{m}}$ $\frac{1.84 \times 10^2}{0.92} \text{ m s}^{-1} = \sqrt{\frac{3kT}{m}}$ $2 \times 10^2 \text{ m s}^{-1} = \sqrt{\frac{3kT}{m}}$ Squaring both sides: $(2 \times 10^2 \text{ m s}^{-1})^2 = \frac{3kT}{m}$ $4 \times 10^4 \text{ m}^2\text{ s}^{-2} = \frac{3kT}{m}$ $3kT = m \times (4 \times 10^4 \text{ m}^2\text{ s}^{-2})$ Since $KE_{avg} = \frac{3}{2} kT$, we have $kT = \frac{2}{3} KE_{avg}$. Substituting this into the equation for $3kT$: $3 \left( \frac{2}{3} KE_{avg} \right) = m \times 4 \times 10^4 \text{ m}^2\text{ s}^{-2}$ $2 KE_{avg} = m \times 4 \times 10^4 \text{ m}^2\text{ s}^{-2}$ $KE_{avg} = m \times 2 \times 10^4 \text{ m}^2\text{ s}^{-2}$ The problem states that the average translational kinetic energy per molecule is $x$ J. Therefore, $x = KE_{avg}$ in Joules. The molar mass of $O_3 = 3 \times 16 = 48 \text{ g/mol} = 0.048 \text{ kg/mol}$. Avogadro's number $N_A = 6 \times 10^{23} \text{ mol}^{-1}$. Mass of one molecule $m = \frac{\text{Molar mass}}{N_A} = \frac{0.048 \text{ kg/mol}}{6 \times 10^{23} \text{ mol}^{-1}} = 8 \times 10^{-26} \text{ kg}$. Now, calculate $x$: $x = (8 \times 10^{-26} \text{ kg}) \times (2 \times 10^4 \text{ m}^2\text{ s}^{-2}) = 16 \times 10^{-22} \text{ J}$. The question asks for the value of $\frac{x}{2} \times 10^{22}$. $\frac{x}{2} \times 10^{22} = \frac{16 \times 10^{-22} \text{ J}}{2} \times 10^{22} = 8 \times 10^{-22} \times 10^{22} = 8 \times 10^0 = 8$.