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Step-by-Step Solution
Step 1: Identify the required condition
We are asked to find the temperature at which the root mean square (rms) velocity of hydrogen molecules equals their escape velocity from Earth. Hence, we set:
$$
V_{rms} = v_{\text{escape}}.
$$
Step 2: Write down expressions for both velocities
(a) RMS speed of gas molecules: For a gas at temperature $T$,
$$
V_{rms} = \sqrt{\frac{3RT}{M}},
$$
where
• $R$ is the universal gas constant ($8.3\,\text{J\,mol}^{-1}\text{K}^{-1}$),
• $M$ is the molar mass of the gas (for hydrogen, $M = 2 \times 10^{-3}\,\text{kg\,mol}^{-1}$).
(b) Escape velocity from Earth: The escape velocity based on Earth’s radius and gravitational acceleration (assuming $g \approx 10\,\text{m\,s}^{-2}$ and Earth’s radius $R_E = 6.4 \times 10^6\,\text{m}$) is
$$
v_{\text{escape}} = \sqrt{2g\,R_E}.
$$
Substituting the given values,
$$
v_{\text{escape}} = \sqrt{2 \times 10\,\text{m\,s}^{-2} \times 6.4 \times 10^6\,\text{m}}
\approx \sqrt{1.28 \times 10^8}\,\text{m\,s}^{-1}.
$$
Numerically, this is about
$$
11.3 \times 10^3\,\text{m\,s}^{-1} \approx 11.2 \times 10^3\,\text{m\,s}^{-1}.
$$
Step 3: Equate $V_{rms}$ with $v_{\text{escape}}$
We now set:
$$
\sqrt{\frac{3RT}{M}} = 11.2 \times 10^3\,\text{m\,s}^{-1}.
$$
Squaring both sides,
$$
\frac{3RT}{M} = \left(11.2 \times 10^3\right)^2.
$$
Step 4: Solve for $T$
$$
T = \frac{M}{3R} \times \left(11.2 \times 10^3\right)^2.
$$
Substituting $M = 2 \times 10^{-3}\,\text{kg\,mol}^{-1}$ and $R = 8.3\,\text{J\,mol}^{-1}\text{K}^{-1}$,
$$
T =
\frac{2 \times 10^{-3}}{3 \times 8.3}
\times
\left(11.2 \times 10^3\right)^2.
$$
Calculate step by step:
• Inside the square: $(11.2 \times 10^3)^2 = 125.44 \times 10^6.$
• Multiply by the prefactor:
$$
\frac{2 \times 10^{-3}}{3 \times 8.3} \times 125.44 \times 10^6.
$$
Simplifying numerically gives approximately
$$
10^4\,\text{K}.
$$
Step 5: Conclude the answer
Therefore, the temperature at which the rms velocity of hydrogen molecules equals the Earth's escape velocity is about
$$
10^4\,\text{K}.
$$
