Question

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere ?
(Given : Mass of oxygen molecule (m) = 2.76×10^-26 kg
Boltzmann's constant k_B = 1.38×10^-23 J K^-1)

• A. 2.508×10⁴ K
• B. 5.016×10⁴ K
• C. 8.360×10⁴ K
• D. 1.254×10⁴ K

Answer 1

The Correct Option is C

 To determine the temperature at which the root mean square (rms) speed of oxygen molecules is sufficient to escape Earth's atmosphere, we start with the expression for the escape velocity and the rms speed.

The escape velocity (\(v_e\)) for Earth is approximately \(11,200 \, \text{m/s}\).

The formula for the root mean square (rms) speed (\(v_{\text{rms}}\)) of a gas molecule is given by:

\(v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}\)

where:

• \(k_B = 1.38 \times 10^{-23} \, \text{J} \, \text{K}^{-1}\) (Boltzmann's constant)
• \(T\) is the temperature in Kelvin
• \(m = 2.76 \times 10^{-26} \, \text{kg}\) (mass of one oxygen molecule)

For the oxygen molecules to escape Earth's atmosphere, their rms speed should be equal to or greater than the escape velocity:

\(v_{\text{rms}} = v_e\)

Substitute the known values into the equation and solve for \(T\):

\(\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}}} = 11,200\)

Square both sides to eliminate the square root:

\(\frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}} = 11,200^2\)

Simplify and solve for \(T\):

\(T = \frac{11,200^2 \times 2.76 \times 10^{-26}}{3 \times 1.38 \times 10^{-23}}\)

Calculating:

Numerator	\(11,200^2 \times 2.76 \times 10^{-26} = 3.4596 \times 10^{-20}\)
Denominator	\(3 \times 1.38 \times 10^{-23} = 4.14 \times 10^{-23}\)
\(T\)	\(\frac{3.4596 \times 10^{-20}}{4.14 \times 10^{-23}} = 8.360 \times 10^4 \, \text{K}\)

Therefore, the temperature at which the rms speed of oxygen molecules is sufficient for escaping the Earth's atmosphere is 8.360×10⁴ K. This is the correct answer.