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Step 1: Identify the Relevant Physical Concepts
The problem involves two key ideas: the escape velocity from Earth's surface and the root mean square (rms) speed of gas molecules from the kinetic theory of gases. The escape velocity is the minimum speed required for an object (or molecule) to overcome Earth's gravitational pull. The rms speed of oxygen molecules at a given temperature can be calculated from the kinetic theory.
Step 2: Write Down the Required Formulas
Escape velocity ($v_{\text{escape}}$) from Earth is given as:
$$ v_{\text{escape}} = 11200 \,\text{m s}^{-1}. $$
Root mean square speed ($v_{\text{rms}}$) of a gas molecule at temperature $T$ is:
$$ v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}, $$
where
$k_B$ is the Boltzmann constant, and
$m$ is the mass of one molecule of the gas (in this case oxygen).
Step 3: Set the RMS Speed Equal to the Escape Velocity
We want the rms speed of oxygen molecules to be just sufficient to escape Earth's atmosphere. Hence, we equate:
$$ v_{\text{rms}} = v_{\text{escape}}. $$
In symbols:
$$ \sqrt{\frac{3 k_B T}{m}} = 11200. $$
Step 4: Substitute the Given Values
From the question, we have:
$m = 2.76 \times 10^{-26}\,\text{kg}$ (mass of one oxygen molecule),
$k_B = 1.38 \times 10^{-23}\,\text{J K}^{-1},$
$v_{\text{escape}} = 11200\,\text{m s}^{-1}.$
Thus,
$$ 11200 = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}}}. $$
Step 5: Solve for the Temperature $T$
Square both sides to remove the square root:
$$ (11200)^2 = \frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}}. $$
Rearrange to isolate $T$:
$$ T = \frac{(11200)^2 \times 2.76 \times 10^{-26}}{3 \times 1.38 \times 10^{-23}}. $$
Perform the calculation (using the given numerical values carefully) to find:
$$ T \approx 8.360 \times 10^4 \,\text{K}. $$
Step 6: State the Final Answer
The temperature at which the rms speed of oxygen molecules equals the escape velocity from Earth's atmosphere is:
8.360 × 104 K.
