Question

A crude model for the molecular distribution of atmospheric gases above Earth's surface (denoted by height \(h\) ) can be obtained by considering the potential energy due to gravity: $$P(h)=e^{-m g h} / k T$$ In this expression \(m\) is the per-particle mass of the gas, \(g\) is the acceleration due to gravity, \(k\) is a constant equal to \(1.38 \times 10^{-23} \mathrm{JK}^{-1},\) and \(T\) is temperature. Determine \(\langle h\rangle\) for methane \(\left(\mathrm{CH}_{4}\right)\) using this distribution function.

Short answer

The average height, ⟨h⟩, for methane using the distribution function \(P(h) = e^{-mhg} / kT\), can be calculated through a series of steps which include finding the values for the constants m, g, k, and T; integrating the expression for average height; and solving for ⟨h⟩. After performing these steps, we find that the average height, ⟨h⟩, for methane according to this crude model, is approximately 0.00091174 meters.

Step-by-step solution

1. Understand the distribution function

The given distribution function represents the probability of finding the methane molecule at a height h: \(P(h) = e^{-mhg} / kT\)

2. Write down the expression for average height, ⟨h⟩

The average height can be calculated using the following formula: \(\langle h\rangle = \int_0^\infty hP(h)dh\)

3. Determine the values of constants m, g, k, and T for methane

m(CH4) = 16 atomic mass units = 16 × 1.66054 x 10^(-27) kg g = 9.81 m/s^2 (acceleration due to gravity) k = 1.38 x 10^(-23) J/K (Boltzmann constant) T = we can assume the temperature remains constant at a given height, let's choose room temperature 298 K

4. Integrate the expression to find the value of ⟨h⟩

We will substitute the distribution function P(h) and then integrate: \(\langle h \rangle = \int_0^\infty h\frac{e^{-mgh/kT}}{kT}dh\) Integration by parts: Let u = h, dv = e^(-mgh/kT)dh Then, du = dh, and v = -\(\frac{kT}{mg}\)e^(-mgh/kT) Therefore, \(\langle h \rangle = -\frac{kT}{mg}\left[\int_0^\infty -hv'e^{-v'}dv'\right]\) (change variables, with v' = mgh/kT) \(\langle h \rangle = \frac{kT}{mg}\int_0^\infty hv'e^{-v'}dv'\) Now, we apply integration by parts again: Let u' = v', dv' = e^(-v')dv' Then, du' = dv', and v' = -e^(-v') So, \(\langle h \rangle = \frac{kT}{mg}\left[-v'e^{-v'}\Big|_0^\infty + \int_0^\infty e^{-v'}dv'\right]\) \(\langle h \rangle = \frac{kT}{mg}\left[-v'e^{-v'}\Big|_0^\infty - [e^{-v'}\Big|_0^\infty]\right]\)

5. Solve for ⟨h⟩ using the values found in step 3

Substitute the values we found for m, g, k, and T: \(\langle h \rangle = \frac{1.38 \times 10^{-23}\mathrm{J/K} \times 298\mathrm{K}}{16 \times 1.66054 \times 10^{-27}\mathrm{kg} \times 9.81\mathrm{m/s^2}}\left[-v'e^{-v'}\Big|_0^\infty - [e^{-v'}\Big|_0^\infty]\right]\) \(\langle h \rangle = \frac{1.38 \times 10^{-23}\mathrm{J/K} \times 298\mathrm{K}}{16 \times 1.66054 \times 10^{-27}\mathrm{kg} \times 9.81\mathrm{m/s^2}}(-1)\) Calculate the result: \(\langle h \rangle \approx 0.00091174 m\) The average height, ⟨h⟩, for methane according to this crude model, is approximately 0.00091174 meters.

Key concepts

Boltzmann constant

The Boltzmann constant, represented by the symbol \( k \), is a fundamental physical constant that relates the average kinetic energy of particles in a gas to the temperature of the gas.
This constant is essential in the field of statistical mechanics and thermodynamics, providing a bridge between macroscopic and microscopic physics.
Its value is \( 1.38 \times 10^{-23} \text{J/K} \) (joules per kelvin). This small number might seem peculiar, but it's crucial for understanding phenomena at the molecular level.
With this constant, you can calculate the energy associated with thermal motion (kinetic energy) using the formula:

• \( E = kT \), where \( E \) is the energy, and \( T \) is the temperature in kelvins.

In the context of atmospheric pressure and gas distributions, the Boltzmann constant helps explain how temperature affects the distribution of gas molecules at different heights.
In our exercise, it plays a role in determining the average height of methane molecules in Earth's atmosphere.

Probability Distribution

Probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes.
In this exercise, the probability distribution represents the likelihood of finding a molecule at any given height above Earth's surface.
The function \( P(h) = \frac{e^{-mgh/kT}}{kT} \) describes how these probabilities change with height \( h \).
This model assumes that the probability of finding a molecule decreases exponentially with height due to the gravitational potential energy component \( mgh \). Here, \( m \) is the molecular mass, \( g \) is the acceleration due to gravity, and \( T \) is the temperature.
Key aspects of probability distributions include:

• Normalization: The total area under a probability distribution curve equals 1, ensuring that all probabilities combined equal certainty.
• Exponential Decay: As height increases, the probability decreases exponentially.

Understanding the probability distribution allows scientists to predict how gases will behave under various conditions and aids in calculating average properties, such as the average height \( \langle h \rangle \).

Molecular Weight

Molecular weight is the mass of a molecule and is sometimes referred to as molecular mass. It is typically expressed in atomic mass units (amu).
For methane (\( \text{CH}_4 \)), the molecular weight is 16 amu.
To use this value in calculations involving gravitational potential energy and atmospheric pressure, it's standard to convert it to kilograms.
This conversion uses the conversion factor that 1 amu = \( 1.66054 \times 10^{-27} \text{kg} \). Thus, methane's molecular weight is converted as follows:

• \( m(\text{CH}_4) = 16 \times 1.66054 \times 10^{-27} \text{kg} \)

This converted mass is then used in equations like \( P(h) = e^{-mgh/kT} \) as it allows scientists to calculate the effects of gravity on the molecule and, subsequently, its probability distribution with height in the atmosphere.

Integration by Parts

Integration by parts is a powerful technique used in calculus to integrate products of functions. It is particularly helpful when dealing with integrals that contain an exponential function multiplied by some polynomial, which is what we encounter in this exercise.
The integration by parts formula is derived from the product rule for differentiation and is given by:

• \( \int u \, dv = uv - \int v \, du \)

Here, \( u \) and \( v \) are functions of a variable, and \( du \) and \( dv \) are their respective differentials.
In our exercise, integration by parts helps in solving the integral \( \int_0^\infty h \frac{e^{-mgh/kT}}{kT} \, dh \) to find the average height \( \langle h \rangle \).
Through strategic choice of \( u \) and \( dv \), we repeated the process to simplify this complex exponential expression, leading us to calculate the final average height at a specific temperature and gravitational setting.
This method shows how calculus can be applied to real-world physics scenarios, helping scientists resolve intricate integrals that describe how gases distribute in an atmosphere.