At what temperature would the root-mean-square speed of hydrogen, $\mathrm { H } _ { 2 }$ molecules equal 11.2 km/s (the earth's escape speed)? The mass of a hydrogen atom is $1.67 \times 10 ^ { - 27 } \mathrm {~kg}$ and the Boltzmann constant is $1.38 \times 10 ^ { - 23 } \mathrm {~J} / \mathrm { K }$
A) $1.01 \times 10 ^ { 8 } \mathrm {~K}$
B) $1.01 \times 10 ^ { 6 } \mathrm {~K}$
C) $1.01 \times 10 ^ { 4 } \mathrm {~K}$
D) $1.01 \times 10 ^ { 2 } \mathrm {~K}$

Question

At what temperature would the root-mean-square speed of hydrogen, $\mathrm { H } _ { 2 }$ molecules equal  11.2 km/s  (the earth's escape speed)? The mass of a hydrogen atom is $1.67 \times 10 ^ { - 27 } \mathrm {~kg}$ and the Boltzmann constant is $1.38 \times 10 ^ { - 23 } \mathrm {~J} / \mathrm { K }$
A) $1.01 \times 10 ^ { 8 } \mathrm {~K}$
B) $1.01 \times 10 ^ { 6 } \mathrm {~K}$
C) $1.01 \times 10 ^ { 4 } \mathrm {~K}$
D) $1.01 \times 10 ^ { 2 } \mathrm {~K}$

Quizplus · Accepted Answer

The root-mean-square speed, $v_{rms}$, of a gas molecule can be calculated using the formula $v_{rms} = \sqrt{\frac{3kT}{m}}$, where $k$ is the Boltzmann constant, $T$ is the temperature in Kelvin, and $m$ is the mass of a molecule. For $\mathrm{H}_2$, the molecular mass is twice the mass of a hydrogen atom, so $m = 2 \times 1.67 \times 10^{-27} \mathrm{~kg}$. Setting $v_{rms} = 11.2 \times 10^3 \mathrm{~m/s}$ and solving for $T$ gives a temperature of approximately $1.01 \times 10^4 \mathrm{~K}$.

At What Temperature Would the Root-Mean-Square Speed of Hydrogen H2\mathrm { H } _ { 2 }H2​

At What Temperature Would the Root-Mean-Square Speed of Hydrogen $\mathrm { H } _ { 2 }$