Problem 72 - University Physics with Modern Physics

Question

(a) Compute the increase in gravitational potential energy for a nitrogen molecule (molar mass 28.0 \(\mathrm{g} / \mathrm{mol} )\) for an increase in elevation of 400 \(\mathrm{m}\) near the earth's surface. (b) At what temperature is this equal to the average kinetic energy of a nitrogen molecule? (c) Is it possible that a nitrogen molecule near sea level where \(T=15.0^{\circ} \mathrm{C}\) could rise to an altitude of 400 \(\mathrm{m} ?\) Is it likely that it could do so without hitting any other molecules along the way? Explain.

Step-by-Step Solution

Answer

(a) 1.83 × 10^-23 J; (b) 0.883 K; (c) Possible energetically, but unlikely without collisions.

Step 1: Convert mass to kilograms

First, convert the molar mass of the nitrogen molecule from grams per mole to kilograms per molecule. The molar mass is given as 28.0 g/mol. Thus, \( \text{molar mass} = 28.0 \times 10^{-3} \text{ kg/mol} \). To find the mass of a single nitrogen molecule, divide by Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol):\[ m = \frac{28.0 \times 10^{-3}}{6.022 \times 10^{23}} = 4.65 \times 10^{-26} \text{ kg} \].

Step 2: Calculate gravitational potential energy

The formula for gravitational potential energy is \( U = mgh \), where \( m \) is the mass, \( g \) is the gravitational acceleration (9.81 m/s^2), and \( h \) is the height change (400 m). Substituting the values:\[ U = 4.65 \times 10^{-26} \times 9.81 \times 400 = 1.83 \times 10^{-23} \text{ J} \].

Step 3: Relate potential energy to kinetic energy

The average kinetic energy (K.E.) of a molecule at temperature \( T \) is given by \( K.E. = \frac{3}{2}kT \), where \( k \) is the Boltzmann constant (\(1.38 \times 10^{-23} \text{ J/K}\)). Set this equal to the calculated potential energy:\[ \frac{3}{2}kT = 1.83 \times 10^{-23} \].

Step 4: Solve for temperature

Solving for \( T \), we have:\[ T = \frac{2}{3} \frac{1.83 \times 10^{-23}}{1.38 \times 10^{-23}} \approx 0.883 \text{ K} \].

Step 5: Assess possibility of ascent without collisions

The typical kinetic energy of a nitrogen molecule at sea level and \( T = 15.0^{\circ} \text{C} \) (288.15 K) is significantly higher than at the required temperature (0.883 K). It is, therefore, energetically possible for a molecule to reach the 400 m altitude. However, given the high density of air, it is unlikely that a nitrogen molecule could rise this distance without colliding with other molecules.

Key Concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics, describing the energy an object possesses due to its motion. For molecules, this energy can be calculated using the formula \( K.E. = \frac{3}{2}kT \), where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin.
This relationship shows how kinetic energy is directly proportional to temperature; higher temperatures lead to greater kinetic energies for molecules.
In the context of molecular motion, understanding kinetic energy helps explain how temperature influences the speed and movement of molecules in a gas.

• Molecules at higher temperatures exhibit greater speeds and hence higher kinetic energy.
• The kinetic energy transformation into potential energy plays a crucial role in processes like phase changes and molecular collisions.

For nitrogen molecules at a temperature of 0.883 K, as calculated in the exercise, the energy is just enough to match their potential energy when rising 400 m, but under typical sea level conditions, this temperature is unnaturally low.

Molecular Mass Conversion

Converting molecular mass into a usable form is essential when dealing with molecular calculations. Molecular mass, often given in grams per mole, must be converted into kilograms per molecule to perform accurate energy calculations.
To achieve this, we use Avogadro's number, \(6.022 \times 10^{23}\) molecules per mole, a constant that reflects the number of particles in one mole of a substance.
The conversion steps are straightforward yet pivotal for precise outcomes in physical chemistry.

• Convert grams to kilograms by multiplying by \(10^{-3}\) (since 1 g = \(10^{-3}\) kg).
• Divide by Avogadro's number to find the mass of a single molecule.

In the solution, the molar mass of nitrogen (28.0 g/mol) was converted, yielding \(4.65 \times 10^{-26}\) kg for a single nitrogen molecule, allowing further calculations of energy changes related to gravitational influence.

Boltzmann Constant

The Boltzmann constant, denoted as \( k \), is a key factor in the kinetic theory of gases, linking temperature and energy at a molecular level. This constant is expressed as \( 1.38 \times 10^{-23} \text{ J/K} \), providing a bridge between macroscopic and microscopic physics.
It appears in various thermodynamic equations, crucial for calculating the average kinetic energy of particles.
Boltzmann constant's role in energy calculations is illustrated through:

• Determining how much energy is associated with each molecule at a given temperature.
• Forming the backbone of the equation \( K.E. = \frac{3}{2}kT \) to compute kinetic energy in thermodynamics.

The constant ensures we can smoothly transition from individual particle considerations to system-wide thermal properties, thus helping us relate microscopic molecular behavior with real-world temperature measurements.

Molecular Collisions

Molecular collisions refer to the interactions between molecules as they move and transfer energy in a substance. These collisions are frequent and varied, playing a significant role in determining the properties and behavior of gases.
In a gaseous environment, molecules are in constant motion and can collide with each other and with walls of their container, leading to various outcomes based on speed and energy.
Collisions are influenced by several factors:

• The speed of the molecules, which increases with temperature.
• The density of the gas, which affects the likelihood of collision.

For a nitrogen molecule to ascend 400 m without a collision might be theoretically possible if its kinetic energy is sufficient. However, given the density of air at sea level, the probability of traveling such a distance without encountering another molecule is extremely low, reinforcing the dynamic and ever-interacting nature of gaseous molecules.