Problem 75 - College Physics

Question

The escape speed from the Earth is about \(11000 \mathrm{~m} / \mathrm{s}\) (Section 7.5). Assume that for a given type of gas to eventually escape the Earth's atmosphere, its average molecular speed must be about \(10 \%\) of the escape speed. (a) Which gas would be more likely to escape the Earth: (1) oxygen, (2) nitrogen, or (3) helium? (b) Assuming a temperature of \(-40^{\circ} \mathrm{F}\) in the upper atmosphere, determine the rms speed of a molecule of oxygen. Is it enough to escape the Earth? (Data: The mass of an oxygen molecule is \(5.34 \times 10^{-26} \mathrm{~kg}\), that of a nitrogen molecule is \(4.68 \times 10^{-26} \mathrm{~kg},\) and that of a helium molecule is \(6.68 \times 10^{-27} \mathrm{~kg}\).

Step-by-Step Solution

Answer

Helium is more likely to escape Earth. The rms speed of oxygen is 461.9 m/s, not enough to escape Earth's atmosphere.

Step 1: Calculate 10% of Escape Speed

First, calculate 10% of Earth's escape speed. The escape speed is 11000 m/s, so 10% of this speed is \(0.1 \times 11000\, \text{m/s} = 1100\, \text{m/s}\). This is the molecular speed required for a gas to escape Earth's atmosphere.

Step 2: Use the Formula for RMS Speed

The root-mean-square speed \(v_{rms}\) of a gas is calculated using the formula \(v_{rms} = \sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant \(1.38 \times 10^{-23}\, \text{J/K}\), \(T\) is the temperature in Kelvin, and \(m\) is the molar mass of the gas.

Step 3: Convert Temperature to Kelvin

First, convert the temperature from Fahrenheit to Kelvin. The given temperature is \(-40^{\circ}F\). Using the formula \(T(K) = \frac{5}{9}(T(\degree F) - 32) + 273.15\), solve for \(-40^{\circ}F\):\[-40^{\circ}F = \frac{5}{9}(-40 - 32) + 273.15 = \frac{5}{9}(-72) + 273.15\] \[= -40 + 273.15 = 233.15\, K\].

Step 4: Calculate rms Speed for Each Gas

Calculate the rms speed for each type of gas using the formula from Step 2.For Oxygen: \[v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \cdot 233.15}{5.34 \times 10^{-26}}} = 461.9\, \text{m/s}\]For Nitrogen: \[v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \cdot 233.15}{4.68 \times 10^{-26}}} = 502.1\, \text{m/s}\]For Helium: \[v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \cdot 233.15}{6.68 \times 10^{-27}}} = 1327.2\, \text{m/s}\]

Step 5: Determine Likelihood of Escape

Compare the rms speeds to 1100 m/s. Helium has an rms speed (1327.2 m/s) greater than 1100 m/s, making it more likely to escape the Earth's atmosphere. Oxygen and Nitrogen have rms speeds much lower than 1100 m/s.

Key Concepts

Root-Mean-Square Speed

Understanding the concept of root-mean-square (RMS) speed is essential when analyzing molecular motions in gases. RMS speed is a type of average speed that considers the molecular speed distribution in a gas. It's calculated using: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \] where \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the mass of a single molecule of the gas.

• Boltzmann constant: This represents the relation between temperature and energy at the particle level.
• Temperature: Higher temperatures typically lead to higher RMS speeds because molecules move faster when they have more energy.
• Molecular mass: Heavier molecules (higher mass \( m \)) have lower RMS speeds, as the mass inversely affects the motion speed.

In essence, RMS speed helps to determine how quickly a gas molecule can move through and potentially escape an environment, such as a planet's atmosphere.

Boltzmann Constant

The Boltzmann constant \( k \) is a fundamental physical constant that links the average kinetic energy of particles in a gas with the temperature of the gas. It is valued at \( 1.38 \times 10^{-23} \, \text{J/K} \). This constant is a bridge between macroscopic and microscopic descriptions of temperature.

• Energy per Particle: The constant helps determine the energy that each particle possesses on average due to thermal motion.
• Unit Explanation: Expressed in Joules per Kelvin, it reflects the energy change per temperature unit that affects the RM speeds of gas molecules.
• Application: The Boltzmann constant plays a vital role in statistical mechanics and thermodynamics, providing deep insights into atomic behavior at various temperature scales.

Understanding the Boltzmann constant is crucial in predicting how quickly different gases within Earth's atmosphere might reach speeds necessary to escape into space. It helps compute kinetic energy shared among gas molecules which drives these assessments.

Molecular Speed

Molecular speed refers to the rate at which molecules within a gas move. This speed is variable and follows a distribution because not all molecules travel at the same pace at a given temperature. The Maxwell-Boltzmann distribution describes this, highlighting that most molecules travel around a mean speed, while some are much faster or slower.

• Speed Range: Molecules within a gas can move at speeds from very slow to very fast, but there is an average speed known as the mean speed and an RMS speed.
• RMS Speed: As mentioned, this provides a measure closer to the speed of most molecules due to its weighting by the square of speeds.
• Influence of Temperature and Mass: Warmer temperatures increase molecular speed by providing more kinetic energy, while heavier molecules move slower due to greater inertia.

Molecular speed concepts help us understand phenomena like gas diffusion, pressure, and in this specific Earth escape problem, the speeds necessary for gases to overcome gravitational forces.

Temperature Conversion

Converting temperatures from one unit to another is crucial when engaging in physics problems. In this exercise, Celsius, Fahrenheit, and Kelvin conversions play key roles.

• Kelvin Conversion: Useful because many physical formulas, including those for RMS speed, require temperatures in Kelvin.
• Fahrenheit to Kelvin: The conversion formula \( T(K) = \frac{5}{9}(T(\degree F) - 32) + 273.15 \) helps translate temperature to absolute scale.
• Significance: Using Kelvin ensures calculations remain consistent as it is an absolute temperature scale without negative values.

These conversions are indispensable for proper context in scientific problems as they align temperature scales with the formulas' requirements, thus producing accurate and meaningful results.