Lecture 3 Notes
Maxwell's Molecular Velocity Distribution
Introduction
Lecture: This section represents the third and final part of the gas chapter in general chemistry, focusing on the microscopic behavior of gas particles and their velocities. It provides a statistical understanding of how molecules in a gas distribute their speeds at a given temperature, moving beyond macroscopic gas laws.
James Clark Maxwell
Mid-19th century physicist (1800s), renowned for his seminal work in unifying electricity and magnetism into Maxwell's electromagnetism. He also made significant contributions to the kinetic theory of gases, particularly developing the Maxwell-Boltzmann distribution which describes the distribution of molecular speeds in an ideal gas.
Maxwell's Velocity Distribution Equation
Concept: This equation is fundamental to understanding the varying distribution of molecular velocities within a gas sample. It shows that not all molecules in a gas move at the same speed; instead, there is a range of velocities, with some being more probable than others.
Students do not need to memorize the complex full equation but should focus on comprehending its interpretation and the factors that influence the distribution, such as temperature and molecular mass.
Equation:
The general form of the distribution for a component of velocity often resembles a Boltzmannian decay, indicating an exponential decrease in the probability of finding very high energy (and thus high velocity) particles.
It represents exponential decay with respect to energy (related to velocity squared) for a generic probability distribution function F(V), where V could represent velocity or speed:
f(v)=Ke^{\frac{-E}{k_{B}T}}
Where:
E = kinetic energy of a molecule, typically E = \frac{1}{2}mv^2
T = absolute temperature in Kelvin
k_B = Boltzmann constant (1.380649 \times 10^{-23} J/K), which relates the average kinetic energy of particles in a gas to its temperature.
Velocity as a Vector
Velocity Components
Velocity is a vector quantity, possessing both magnitude (speed) and direction. In a three-dimensional space, velocity has components along the x, y, and z axesf\left(^{v}x\right)f\left(^{v}y\right)f\left(^{v}z\right)\Rightarrow f\left(^{v}x\right)=\frac13f\left(v\right). These components are independent of each other.
For simplification in some derivations, consider one axis (e.g., the x-axis). The probability distribution function for velocity along the x-axis, f(V_x), is typically described by a Gaussian function, reflecting that positive and negative velocities are equally probable:
f(Vx) = K' e^{-\frac{mVx^2}{2k_B T}}
Here, K' is a normalization constant that ensures the total probability is 1. The original note's expression F(Vx)=\frac{3}{F\left(V\right)}=\frac{K^3e}{-\frac{mv^2}{2kB^{T}}}^{} is a simplified representation where K^3 relates to the 3D velocity distribution constant and the \frac13 might represent partitioning of energy into one degree of freedom, but typically would be derived independently for the 1D case.
Assumptions and Integrations
Assumptions about Velocity
The variable Vx (and similarly Vy, V_(-\infty, +\infty)). must be a finite measurable quantity, meaning its value can range continuously from negative infinity to positive infinity
Integral Representation
The integral of the probability distribution function over all possible values of V_x must equal 1 (100%), which is a fundamental requirement for any probability distribution. This normalization ensures that there is a 100% chance of a molecule having some velocity along the x-axis.
\int{-\infty}^{+\infty} f(Vx) dV_x = 1
Replacing F(V_x)
To find the normalization constant K'
We substitute f(Vx) = K' e^{-\frac{mVx^2}{2k_B T}} into the integral equation:
\int{-\infty}^{+\infty} K' e^{-\frac{mVx^2}{2kB T}} dVx = 1
Applying Gaussian Integral
Template from Gaussian Integral
The definite integral of a Gaussian function is well-known and crucial here. The general form states:
\int_{-\infty}^{+\infty} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}}
Comparing this to our integral, we can identify:
a = \frac{m}{2kB T} and x = Vx
Equation Manipulation
Applying the Gaussian integral identity to the normalization equation:
K' \sqrt{\frac{2\pi k_B T}{m}} = 1
Thus, the normalization constant for the one-dimensional velocity component is:
K' = \left(\frac{m}{2\pi k_B T}\right)^{1/2}
The full Maxwell-Boltzmann speed distribution (for the magnitude of velocity, V) is derived by combining the three independent velocity components and converting to spherical coordinates. The normalization constant for the 3D speed distribution is related to (K')^3 . The constant K mentioned in the note, K=\left(\frac{m}{2\pi k_B T}\right)^{\frac32}, is indeed the normalization constant for the 3D velocity distribution (without the V^2 term).
Final Equation and Proportions
Mass and Temperature Relationships
When considering the distribution of molecular speeds (magnitudes of velocity), we move from a linear (1D) velocity axis to a three-dimensional velocity space. The factor 4\pi V^2 arises from integrating over all possible angular directions in velocity space at a constant speed V, representing the surface area of a spherical shell. The complete Maxwell-Boltzmann distribution for molecular speeds, f(V), is:
f(V) = 4\pi \left(\frac{m}{2\pi kB T}\right)^{\frac32} V^2 e^{-\frac{mV^2}{2kB T}} This standard equation shows essential proportionalities:
The probability density of a certain speed f(V) is directly proportional to (\frac{m}{T})^{3/2} and to V^2, while the exponential decay term is also critically dependent on the ratio of molecular mass to temperature, \frac{m}{T} . This implies that at a given temperature, heavier molecules move slower on average.
Effects of Temperature on Velocity Distribution
Different Temperatures for the Same Gas
When the Maxwell-Boltzmann distribution (plotted as f(V) against V) is analyzed for the same gas at varying temperatures, specific trends in population density are observed:
As temperature increases, the distribution curve shifts to higher velocities, and its peak decreases in height while becoming broader.
Observations:
Increase in temperature leads to a decrease in the height of the peak, because the molecules' velocities are more spread out over a wider range.
The peak represents the most probable velocity (V_{mp}). A higher peak (at lower temperature) corresponds to a lower most probable velocity and a narrower distribution. A shorter, wider curve (at higher temperature) indicates a higher most probable velocity and a broader range of velocities for the molecules.
Same Temperature for Different Gases
Comparison Across Gases
At the same absolute temperature, different gases exhibit distinct speed profiles based on their molecular mass. Lighter gases, having smaller masses, will have higher average speeds compared to heavier gases.
Example: At 298 K, helium (He, 4 amu) molecules will move faster on average than neon (Ne, 20 amu) molecules, which in turn move faster than xenon (Xe, 131 amu) molecules.
Spherical Distribution Implications
When comparing gases at the same temperature, heavier gases will have distribution curves with taller and narrower peaks, shifted towards lower velocities. This reflects a less spread-out distribution of speeds because their higher mass results in lower velocities for the same kinetic energy. Lighter gases will have shorter, broader curves, reflecting a wider range of higher speeds.
Kinetic Energy Associated with Temperature
Kinetic Energy and Temperature Graph
When the distribution is plotted as a function of kinetic energy (f(KE) vs. KE), the average kinetic energy of gas molecules is directly proportional to the absolute Kelvin temperature.
Average translational kinetic energy: KE{avg} = \frac{1}{2} m \overline{V^2} = \frac{3}{2} kB T
This equation underscores that temperature is a direct measure of the average translational kinetic energy of the gas molecules. Importantly, at the same temperature, all ideal gas molecules have the same average kinetic energy, regardless of their mass.
Effect of Varying Temperature
Increasing the temperature results in a broader population distribution of kinetic energies, meaning a larger fraction of molecules possesse higher kinetic energies.
Kinetic Molecular Theory (KMT)
Fundamental Concepts
The Kinetic Molecular Theory provides a model for understanding the behavior of ideal gases based on several postulates:
Gas particles are in constant, random motion, moving in straight lines until they collide with other particles or the container walls (random vectors).
Collisions between gas particles and with the container walls are perfectly elastic, meaning that the total kinetic energy of the system is conserved during these collisions.
The average kinetic energy of the gas particles is directly proportional to the absolute temperature of the gas.
Assumptions of KMT
The KMT makes simplifying assumptions to describe ideal gas behavior:
The constituent particles (atoms or molecules) are considered to have negligible volume compared to the total volume of the gas. This simplifies gas equations by allowing us to ignore the excluded volume of the particles themselves.
There are no significant attractive or repulsive forces (inter-particle interactions) between gas particles. This assumption means that the internal energy of the gas is purely kinetic, with no potential energy contributions from intermolecular forces.
Activation Energy Concepts
Activation Energy
Activation energy (E_a) is the minimum amount of energy that reactant molecules must possess for a chemical reaction to occur. It represents an energy barrier that must be overcome for bonds to break and new ones to form.
Temperature Influence on Activation Energy
The Maxwell-Boltzmann distribution shows that only a fraction of molecules at any given temperature possess enough kinetic energy to exceed the activation energy threshold. Higher temperatures significantly increase this fraction, as the distribution curve broadens and shifts to higher energies, leading to a greater number of molecules with sufficient energy for reactions. This is why reaction rates generally increase with temperature.
Graham's Law of Effusion
Definitions
Effusion: The process by which a gas escapes from a container through a very small hole or orifice into a vacuum (or region of lower pressure).
Diffusion: The process by which gas molecules spread out to uniformly fill a space, moving from an area of higher concentration to an area of lower concentration. This involves collisions between molecules.
Derivation of Graham's Law
Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This law is derived from the average kinetic energy equation (KE_{avg} = \frac{1}{2}m\overline{v^2}).
Since all gases at the same temperature have the same average kinetic energy:
\frac{1}{2} mA \overline{vA^2} = \frac{1}{2} mB \overline{vB^2}
Which simplifies to: \frac{\overline{vA^2}}{\overline{vB^2}} = \frac{mB}{mA}
Taking the square root and noting that effusion rates are proportional to average molecular velocities:
\frac{\text{Rate}A}{\text{Rate}B} = \frac{vA}{vB} = \sqrt{\frac{mB}{mA}}
This law can be presented in terms of velocity or time of travel (time is inversely proportional to rate).
Example Problem Using Graham's Law
Practical Calculation
Consider a scenario where you are given the effusion rate of nitrogen gas and asked to calculate the relative effusion rate of hydrogen gas at the same temperature.
Given: Molar mass of nitrogen (N_2) = 2 \times 14.007 amu = 28.014 amu.
Molar mass of hydrogen (H_2) = 2 \times 1.008 amu = 2.016 amu.
Using Graham's Law:
\frac{\text{Rate}{H2}}{\text{Rate}{N2}} = \sqrt{\frac{m{N2}}{m{H2}}} = \sqrt{\frac{28.014 \text{ amu}}{2.016 \text{ amu}}} \approx \sqrt{13.9} \approx 3.73
This estimate indicates that hydrogen gas effuses about 3.73 times faster than nitrogen gas under the same conditions.
Ideal vs Real Gas Equation Application
Comparison
The ideal gas law, PV = nRT, provides a useful approximation for gas behavior, especially at high temperatures and low pressures. For example, it can be used to calculate a specific property (like pressure) for CO_2 assuming ideal conditions.
Note: Temperature (T) must always be converted to Kelvin when using gas law equations, as it represents absolute temperature and appears in the denominator of many relationships.
Real Gas Corrections
For gases under non-ideal conditions (e.g., high pressure, low temperature) where molecular volume and intermolecular attractions become significant, the van der Waals equation of state offers greater accuracy:
\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT
Where:
The term \frac{an^2}{V^2} corrects for the attractive forces between molecules, which reduce the effective pressure exerted on the container walls.
The term nb corrects for the finite volume occupied by the gas molecules themselves, reducing the available free volume for the gas to move in.
Final Results
A comparison between pressures calculated for a real gas (like CO_2) using the ideal gas law versus the van der Waals equation often reveals:
The ideal gas equation significantly (and often incorrectly) overestimates the pressure compared to the real gas behavior. This overestimation occurs because the ideal gas law neglects the attractive forces (which reduce pressure) and assumes negligible molecular volume (which would slightly increase pressure if ignored, but attractions are usually more dominant in leading to overestimation of pressure). Real gases exhibit lower pressures than predicted by the ideal gas law due to intermolecular attractions.
Conclusion
Remember to refer to additional resources on Canvas for practical problems, examples, and further exploration related to the various gas.
Certainly! Let's consider an example applying the Ideal Gas Law and then how the van der Waals equation (for real gases) would differ, drawing from the explanation in the notes.
Example: Calculating Pressure of Carbon Dioxide Gas
Scenario: You have 1.0 mole of carbon dioxide (CO_2) gas in a 2.0 L container at a temperature of 300 K.
1. Using the Ideal Gas Law (PV = nRT):
Given:
Number of moles (n) = 1.0 mol
Volume (V) = 2.0 L
Temperature (T) = 300 K
Ideal gas constant (R) = 0.08206{ L}\frac{Latm}{molK}
Rearrange the equation to solve for Pressure (P):
P = \frac{nRT}{V}Substitute values:
P=\frac{\left(1.0mol_{})\left(\right.0.08206_{\frac{Latm}{molK}}\right)(300_{K})}{2.0_{L}}Calculate:
P=12.309{ atm}So, under ideal conditions, the pressure of 1.0 mol of CO_2 in a 2.0 L container at 300 K would be approximately 12.3 atm.
2. Using the van der Waals Equation (for Real Gases):
To account for real gas behavior (molecular volume and intermolecular forces), we use the van der Waals equation: \left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT
Given (same as above, plus van der Waals constants for CO_2):
n = 1.0 mol
V = 2.0 L
T = 300 K
R = 0.08206\frac{{L}atm}{molK}
van der Waals constant 'a' for CO_2 (corrects for attractive forces) = 3.59{atm}{L}^2mol^2
van der Waals constant 'b' for CO_2 (corrects for molecular volume) = 0.0427{ L}{mol}
Rearrange to solve for P:
P = \frac{nRT}{(V - nb)} - \frac{an^2}{V^2}Calculate the individual correction terms first:
Volume correction term (V - nb):
(2.0{ L}-(1.0{ mol})(0.0427{ L/mol}))=2.0{ L}-0.0427{ L}=1.9573{ L}Attractive force correction term (\frac{an^2}{V^2}):
\frac{(3.59{ atm L}^2/{mol}^2)(1.0{ mol})^2}{(2.0{ L})^2}=\frac{3.59{ atm L}^2}{4.0{ L}^2}=0.8975{ atm}
Substitute these and other values into the rearranged equation:
P=\frac{(1.0{mol})(0.08206{ L atm/mol K})(300{ K})}{1.9573{ L}}-0.8975{ atm}
P=\frac{24.618{ L atm}}{1.9573{ L}}-0.8975{ atm}
P = 12.577atm is the final calculation for the total pressure of the gas in the given system.