Question

The temperature at which the root-mean-square speed of carbon dioxide molecules becomes just sufficient for escaping Earth's atmosphere is approximately
(Mass of carbon dioxide molecule \(=7\times 10^{-26}\ \text{kg}\), Boltzmann's constant \(k_{B}=14\times 10^{-24}\ J/K\), escape velocity from Earth's surface \(=11\ \text{km/s}\))

• A. \(2.016\times 10^{5}\ \text{K}\)
• B. \(4.032\times 10^{5}\ \text{K}\)
• C. \(5.016\times 10^{5}\ \text{K}\)
• D. \(3.012\times 10^{5}\ \text{K}\)

Hint:

Equate the rms speed \(\sqrt{3k_BT/m}\) with the escape speed and solve for T.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The rms speed of a molecule is \(v_{rms}=\sqrt{\dfrac{3k_BT}{m}}\). We need \(v_{rms}\) equal to the escape speed \(v_e=11\ \text{km/s}=1.1\times10^{4}\ \text{m/s}\).

Step 2: Solve for T
\(T=\dfrac{mv_e^{2}}{3k_B}\).

Step 3: Calculation
\(T=\dfrac{7\times10^{-26}\times(1.1\times10^{4})^{2}}{3\times14\times10^{-24}}=\dfrac{7\times10^{-26}\times1.21\times10^{8}}{4.2\times10^{-23}}\)
\(=\dfrac{8.47\times10^{-18}}{4.2\times10^{-23}}\approx2.016\times10^{5}\ \text{K}\).

Final Answer:
The required temperature is about \(2.016\times10^{5}\ \text{K}\). \[ \boxed{\text{(A) }2.016\times 10^{5}\ \text{K}} \]