Kinetic Theory
AS UNIT 3Oscillations and Nuclei3.3 Kinetic TheoryLearners should be able to demonstrate and apply their knowledge and understanding of: |
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| a) | The equation of state for an ideal gas expressed as [math]pV = nRT[/math] where R is the molar gas constant and [math]pV = NkT[/math] where k is the Boltzmann constant |
| b) | The assumptions of the kinetic theory of gases which includes the random distribution of energy among the molecules |
| c) | The idea that molecular movement causes the pressure exerted by a gas, and use [math]p = \frac{1}{3} \rho \overline{c^2} = \frac{1}{3} \frac{N}{V} m \overline{c^2}[/math] where N is the number of molecules |
| d) | The definition of Avogadro constant [math]N_A[/math] and hence the mole |
| e) | The idea that the molar mass M is related to the relative molecular mass [math]M_r[/math] by [math]M/kg = \frac{M_r}{1000}[/math] and that the number of moles n is given by [math]\frac{\text{total mass}}{\text{molar mass}}[/math] |
| f) | How to combine [math]\frac{1}{3} \frac{N}{V} m \overline{c^2}[/math] with [math]pV = nRT[/math] and show that the total translational kinetic energy of a mole of a monatomic gas is given by [math]\frac{3}{2} RT[/math] and the mean kinetic energy of a molecule is [math]\frac{3}{2} kT[/math] where [math]k = \frac{R}{N_A}[/math] is the Boltzmann constant, and that T is proportional to the mean kinetic energy |
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a) The equation of state for an ideal gas describes the relationship between pressure (p), volume (V), temperature (T), and the number of molecules or moles of gas present in a system. There are two common forms of this equation:
- 1. Molar Form:
- [math]pV = nRT[/math]
- Where:
- – p is the pressure of the gas
- – V is the volume,
- – n is the number of moles of gas,
- – R is the universal gas constant ( [math]R = 8.314 J/molK[/math])
- – T is the absolute temperature in kelvins.
- This form is commonly used when dealing with macroscopic amounts of gas measured in moles.
- 2. Molecular Form:
- [math]pV = NkT[/math]
- Where:
- – N is the total number of gas molecules,
- – k is the Boltzmann constant ([math]k = 1.38 × 10^{-23} J/K[/math]),
- – T is the absolute temperature.
- Figure 1 Ideal gas system
- This form relates to the microscopic behavior of gas molecules and is derived from the molar form using the relationship [math]R = N_A k,[/math], where [math]N_A[/math] is Avogadro’s number ([math]6.022 × 10^{23}[/math] molecules per mole).
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b) Assumptions of the Kinetic Theory of Gases
- The kinetic theory of gases explains the macroscopic properties of gases, such as pressure and temperature, in terms of the motion of molecules. It is based on the following key assumptions:
- Figure 2 Kinetic theory of Gas
- 1. Gas Molecules are in Constant, Random Motion
- – The molecules of a gas move in all directions randomly with different speeds.
- – They follow Newton’s laws of motion.
- 2. Collisions Between Molecules and with Walls are Perfectly Elastic
- – When molecules collide with each other or with the walls of the container, there is no loss of kinetic energy, meaning total kinetic energy is conserved.
- 3. The Volume of Gas Molecules is Negligible Compared to the Container
- – The actual volume occupied by gas molecules is much smaller than the total volume of the gas.
- – This allows gases to be compressed easily.
- 4. There are No Intermolecular Forces Except During Collisions
- – Gas molecules do not exert attractive or repulsive forces on each other except when they collide.
- – This means that between collisions, molecules travel in straight lines at constant velocity.
- 5. The Average Kinetic Energy of the Molecules is Proportional to Temperature
- – The temperature of a gas is directly related to the average kinetic energy of its molecules.
- – The average kinetic energy per molecule is given by:
- [math]KE_{avg} = \frac{3}{2} kT[/math]
- – As temperature increases, the molecules move faster.
- 6. The Distribution of Molecular Speeds is Given by Maxwell-Boltzmann Statistics
- – Not all molecules have the same speed; rather, their speeds follow a statistical distribution, known as the Maxwell-Boltzmann distribution.
- – This distribution shows that some molecules move very fast, some very slow, but most have an intermediate speed.
- ⇒ Random Distribution of Energy Among Molecules
- Due to constant collisions, the energy of gas molecules is randomly distributed at a given temperature.
- The distribution of kinetic energy among molecules follows the Maxwell-Boltzmann distribution, meaning some molecules will have higher or lower energies than others.
- Even at a fixed temperature, molecules have a wide range of speeds and kinetic energies.
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c) Molecular Movement and Gas Pressure
- The pressure exerted by a gas on the walls of its container arises from the continuous motion and collisions of gas molecules.
- Each molecule moves randomly in all directions and collides with other molecules and the container walls. These collisions transfer momentum to the walls, creating a force per unit area, which is observed as pressure (p).
- To derive the relationship between pressure and molecular motion, consider a container with a gas of molecules, each having mass mmm and velocity c.
- Figure 3 The kinetic-molecular theory
- ⇒ Derivation of Pressure Expression
- 1. Momentum Change During Collision
- – Consider a single gas molecule moving in the x-direction with velocity cx .
- – When it collides elastically with a wall perpendicular to the x-axis, it reverses its velocity, changing from cx to −cx
- – The change in momentum due to this collision is:
- [math]Δp = m(-c_x ) – m(c_x ) = -2mc_x[/math]
- Figure 4 Conservation of momentum
- 2. Time Between Successive Collisions with the Same Wall
- – The molecule moves to the opposite wall at distance L, reflects, and returns, covering 2L in time:
- [math]t = \frac{2L}{c_x}[/math]
- – The number of times the molecule collides with the wall per unit time is:
- [math]\frac{c_x}{2L}[/math]
- 3. Force Exerted on the Wall by One Molecule
- – The force is the rate of momentum change:
- [math]F = \frac{\Delta p}{t} \\
F = \frac{-2m c_x}{\frac{2L}{c_x}} \\
F = \frac{-2m c_x^2}{L}[/math] - 4. Total Pressure from All Molecules
- – Considering N molecules in the volume [math]V = L^3[/math], the total force exerted by all molecules is:
- [math]F_{\text{total}} = \sum \frac{m c_x^2}{L}[/math]
- – The pressure (p) is force per unit area [math]A = L^2[/math]):
- [math]p = \frac{F_{\text{total}}}{L^2} \\
p = \frac{1}{V} \sum \frac{m c_x^2}{}[/math] - 5. Accounting for Motion in Three Dimensions
- – Since molecules move in three directions (x, y, and z), the total velocity is:
- [math]c^2 = c_x^2 + c_y^2 + c_z^2[/math]
- – On average, energy is equally distributed among directions, so:
- [math]\langle c_x^2 \rangle = \langle c_y^2 \rangle = \langle c_z^2 \rangle = \frac{1}{3} \langle c^2 \rangle[/math]
- – Therefore, substituting in the pressure equation:
- [math]p = \frac{1}{V} \sum \frac{m}{3} c^2 \\
p = \frac{1}{3} \rho \overline{c^2}[/math] - Where [math]ρ = \frac{Nm}{V}[/math] is the mass density of the gas.
- 6. Final Expression for Pressure
- – Using [math]ρ = \frac{Nm}{V}[/math], we obtain:
- [math]p = \frac{1}{3} \frac{N}{V} m \overline{c^2}[/math]
- – This equation shows that gas pressure depends on the number density ( [math]N/V[/math]), molecular mass (m), and the mean squared speed ([math]\overline{c^2}[/math] ) of gas molecules.
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d) Definition of Avogadro Constant [math]N_A[/math] and the Mole
- ⇒ Avogadro’s Constant [math]N_A[/math]
- The Avogadro constant ([math]N_A[/math] ) is the number of atoms, ions, or molecules in one mole of a substance. It has a precise value:
- [math]N_A = 6.022 × 10^{23} mol^{-1}[/math]
- This number represents the count of discrete particles (molecules, atoms, or ions) in one mole of a substance.
- ⇒ Definition of the Mole
- A mole is a fundamental SI unit used to measure the amount of a substance. It is defined as:
- [math]1 mole = 6.022 × 10^{23} \text{particles}[/math]
- This means:
- 1 mole of oxygen ( [math]O_2[/math]) contains [math]6.022 × 10^{23}[/math] oxygen molecules.
- 1 mole of carbon (C) contains [math]6.022 × 10^{23}[/math] carbon atoms.
- 1 mole of electrons contains [math]6.022 × 10^{23}[/math] electrons.
- ⇒ Relationship Between Avogadro’s Number and the Ideal Gas Law
- Using the molecular form of the ideal gas law:
- [math]pV = NkT[/math]
- Where N is the number of molecules and k is the Boltzmann constant:
- [math]k = \frac{R}{N_A} \\
k = 1.38 \times 10^{-23} \text{ J/K}[/math] - Since [math]N = nN_A[/math] (where n is the number of moles), substituting this into the ideal gas law gives:
- [math]pV = nN_A kT[/math]
- Which simplifies to the molar form of the ideal gas equation:
- [math]pV = nRT[/math]
- This shows that the ideal gas law applies both at the molecular level (using Boltzmann’s constant) and at the macroscopic level (using the molar gas constant R).
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e) Relationship Between Molar Mass M and Relative Molecular Mass [math]M_r[/math]
- The molar mass (M) of a substance is the mass of one mole of its particles (atoms, molecules, or ions) expressed in kilograms per mole ([math]kg/mol[/math] ) or grams per mole ([math]kg/mol[/math] ).
- The relative molecular mass ([math]M_r[/math] ) (also called molecular weight) is a dimensionless quantity that represents the mass of a molecule relative to 1/12th the mass of a carbon-12 atom.
- – The two quantities are related by:
- [math]\frac{M}{\text{kg/mol}} = \frac{M_r}{1000}[/math]
- or equivalently,
- [math]M = \frac{M_r}{1000} \ \text{kg/mol} \\
M = M_r \ \text{g/mol}[/math] - ⇒ Explanation of the Relationship
- 1. Relative molecular mass [math]M_r[/math] is a ratio
- – It has no unit because it is the ratio of the mass of a molecule to 1/12th the mass of a carbon-12 atom.
- – Example: The relative molecular mass of water (H2O) is:
- [math]M_r = (2 × 1.008) + 16.0 \\ M_r = 18.016[/math]
- 2. Molar mass M has units (g/mol or kg/mol)
- – The mass of one mole of a substance is given by:
- [math]M = M_r \ \text{g/mol} \\ M = \frac{M_r}{1000} \ \text{kg/mol}[/math]
- – For water (H2O):
- [math]M = 18.016 \ \text{g/mol} \\
M = 0.018016 \ \text{kg/mol}[/math] - Thus, molar mass and relative molecular mass differ only by a factor of 1000 because molar mass is expressed in grams per mole while mass in physics is often expressed in kilograms.
- ⇒ Number of Moles and Its Calculation
- The number of moles (n) of a substance is given by:
- [math]n = \frac{\text{total mass of the substance}}{\text{molar mass} \ (M)}
[/math] - or:
- [math]n = \frac{m}{M}[/math]
- Where:
- – n = number of moles (mol),
- – m = total mass of the substance (g or kg),
- – M = molar mass (g/mol or kg/mol).
- ⇒ Example Calculations
- 1. Example 1: Calculating Moles of Oxygen Gas (O2)
- – Given: 32 g of O2, with [math]M_r (O_2)[/math]= 32.
- – Molar mass:
- [math]M = 32 \ \text{g/mol} = 0.032 \ \text{kg/mol}[/math]
- – Number of moles:
- [math]n = \frac{32 \ \text{g}}{32 \ \text{g/mol}} \\
n = 1 \ \text{mol}[/math] - 2. Example 2: Finding Mass from Moles of Water
- – Given: n=2 moles of H2O.
- – Molar mass of water:
- [math]M = 18.016 g/mol \\ M = 0.018016 kg/mol[/math]
- – Total mass:
- [math]m = nM[/math]
- 1. Molar mass M and relative molecular mass are related by:
- [math]M = \frac{M_r}{1000} \ \text{kg/mol} \\
M = M_r \ \text{g/mol}[/math] - Number of moles n is calculated as:
- [math]n = \frac{m}{M}[/math]
- – More mass (m) means more moles.
- – A higher molar mass (M) means fewer moles for a given mass.
- This relationship is crucial in chemical calculations, stoichiometry, and thermodynamics, linking the microscopic world of atoms and molecules to macroscopic measurements like grams and kilograms.
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f) The total translational kinetic energy:
- The total translational kinetic energy of a mole of gas and the mean kinetic energy per molecule by combining the kinetic theory equation:
- [math]pV = \frac{1}{3} m N \overline{c^2}[/math]
- with the ideal gas law:
- [math]pV = nRT[/math]
- ⇒ Step 1: Equating the Two Expressions for [math][/math]
- From kinetic theory:
- [math]pV = \frac{1}{3} m N \overline{c^2}[/math]
- From the ideal gas law:
- [math]pV = nRT[/math]
- Since both equations represent the same physical quantity, we set them equal:
- [math]\frac{1}{3} m N \overline{c^2} = nRT[/math]
- ⇒ Step 2: Expressing N in Terms of n
- We know that the total number of molecules is:
- [math]N = nN_A[/math]
- Where:
- – n = number of moles,
- – [math]N_A[/math] = Avogadro’s number ([math]6.022 × 10^{23}[/math] molecules/mol).
- Substituting [math]N = nN_A[/math] into the equation:
- [math]\frac{1}{3} mnN_A = nRT[/math]
- Canceling n from both sides:
- [math]\frac{1}{3} mN_A = RT[/math]
- Multiplying both sides by [math]\frac{3}{2}[/math]:
- [math]\frac{3}{2} \left(\frac{1}{3} m N \overline{c^2} \right) = \frac{3}{2} RT \\
\frac{1}{2} m N \overline{c^2} = \frac{3}{2} RT[/math] - This equation represents the total translational kinetic energy of one mole of a monatomic gas:
- [math]E_{translational} = \frac{3}{2} RT[/math]
- ⇒ Step 3: Finding the Mean Kinetic Energy per Molecule
- The mean kinetic energy per molecule is given by:
- [math]KE_{mean} = \frac{1}{N} E_{total}[/math]
- Since [math]N = nN_A[/math], dividing both sides by [math]N_A[/math]:
- [math]\text{KE}_{\text{mean}} = \frac{1}{N_A} \times \frac{3}{2} RT[/math]
- Using the relation [math]k = \frac{R}{N_A}[/math](Boltzmann’s constant):
- [math]\text{KE}_{\text{mean}} = \frac{3}{2} \left(\frac{R}{N_A}\right) T \\
\text{KE}_{\text{mean}} = \frac{3}{2} k T[/math] - ⇒ Step 4: Showing that T is Proportional to Mean Kinetic Energy
- From the final equation:
- [math]\text{KE}_{\text{mean}} = \frac{3}{2} k T[/math]
- The temperature T is directly proportional to the mean kinetic energy per molecule. This confirms that temperature is a measure of the average kinetic energy of gas molecules.