The escape velocity from Mercury is much smaller than from Earth and is only 4.25 km/s. At what temperature (in K) would nitrogen molecules (molar mass is equal to 28.01 g/mol) have an average velocity vrms equal to Mercury’s escape velocity?
The Correct Answer and Explanation is:
To find the temperature at which nitrogen molecules (N2\text{N}_2) have an average root-mean-square (rms) speed equal to the escape velocity from Mercury (4.25 km/s), we use the formula for vrmsv_{\text{rms}}: vrms=3kTmv_{\text{rms}} = \sqrt{\frac{3kT}{m}}
Where:
- vrmsv_{\text{rms}} is the root-mean-square speed,
- kk is Boltzmann’s constant =1.38×10−23 J/K= 1.38 \times 10^{-23} \, \text{J/K},
- TT is the temperature in Kelvin,
- mm is the mass of a nitrogen molecule in kilograms.
Step 1: Convert molar mass to mass per molecule
The molar mass of nitrogen is 28.01 g/mol. To convert this to kg per molecule: m=28.01×10−36.022×1023=4.65×10−26 kgm = \frac{28.01 \times 10^{-3}}{6.022 \times 10^{23}} = 4.65 \times 10^{-26} \, \text{kg}
Step 2: Solve for temperature
Given vrms=4.25×103 m/sv_{\text{rms}} = 4.25 \times 10^3 \, \text{m/s}, we solve: T=mvrms23k=(4.65×10−26)(4.25×103)23(1.38×10−23)T = \frac{m v_{\text{rms}}^2}{3k} = \frac{(4.65 \times 10^{-26})(4.25 \times 10^3)^2}{3(1.38 \times 10^{-23})} T=(4.65×10−26)(1.80625×107)4.14×10−23=8.4×10−194.14×10−23≈20,290 KT = \frac{(4.65 \times 10^{-26})(1.80625 \times 10^7)}{4.14 \times 10^{-23}} = \frac{8.4 \times 10^{-19}}{4.14 \times 10^{-23}} \approx 20,290 \, \text{K}
Final Answer:
T≈20, 290 K\boxed{T \approx 20,\!290 \, \text{K}}
The escape velocity from a planet is the minimum speed an object must have to overcome the planet’s gravitational pull without further propulsion. For Mercury, this velocity is 4.25 km/s (4250 m/s), which is significantly lower than Earth’s escape velocity due to Mercury’s smaller mass and radius.
Molecules in a gas move with a distribution of speeds, but the root-mean-square (rms) speed provides a useful average when considering thermal motion. The rms speed of a gas molecule is related to the temperature and molecular mass by the equation: vrms=3kTmv_{\text{rms}} = \sqrt{\frac{3kT}{m}}
To determine the temperature at which nitrogen molecules achieve this escape speed, we set vrmsv_{\text{rms}} equal to 4.25 km/s and solve for TT. First, the molar mass of nitrogen (28.01 g/mol) is converted to the mass of a single molecule in kilograms. Boltzmann’s constant relates the average kinetic energy of particles in a gas to temperature.
Using the derived equation, we calculate the required temperature to be approximately 20,290 K. This is extremely hot and far above Mercury’s actual surface temperatures, which typically range from about 100 K at night to 700 K during the day. Therefore, under natural conditions, nitrogen gas on Mercury would not reach escape velocity just from thermal motion. However, this calculation helps illustrate why lighter gases like hydrogen and helium are more likely to escape a planet’s atmosphere over time—because their rms speeds are higher at any given temperature.
This method is a useful way to explore atmospheric retention and planetary habitability across the solar system.
