Physics 2 - Kinetic Theory of Gases Notes

PHYSICS

Lecturer Information

• Lecturer: Dr. DO Xuan Hoi
• Room: A1. 503
• E-mail: dxhoi@hcmiu.edu.vn

Physics 2: Fluid Mechanics and Thermal Physics

• Credits: 02 (30 periods)
• Chapters:
  • Chapter 1: Fluid Mechanics
  • Chapter 2: Heat, Temperature, and the Zeroth Law of Thermodynamics
  • Chapter 3: Heat, Work, and the First Law of Thermodynamics
  • Chapter 4: The Kinetic Theory of Gases
  • Chapter 5: Entropy and the Second Law of Thermodynamics

Chapter 4: The Kinetic Theory of Gases

Topics

• Ideal Gases, Experimental Laws, and the Equation of State
• Molecular Model of an Ideal Gas
  • The Equipartition of Energy
  • The Boltzmann Distribution Law
  • The Distribution of Molecular Speeds
  • Mean Free Path
• The Molar Specific Heats of an Ideal Gas
• Adiabatic Expansion of an Ideal Gas

1. Ideal Gases, Experimental Laws, and the Equation of State

1.1 Notions

• Properties of gases:
  • Gases do not have a fixed volume or pressure.
  • Gases expand to fill their container.
• Ideal gas:
  • A collection of randomly moving atoms or molecules.
  • Molecules exert no long-range forces on each other.
  • Molecules occupy a negligible fraction of the volume of their container.
  • Most gases at room temperature and pressure behave approximately as an ideal gas.

1.2 Moles

• The amount of gas in a given volume is expressed in terms of the number of moles, $n$.
• One mole is the amount of substance that contains as many particles as there are atoms in 12 g of carbon-12.
• Formula: $n = \frac{m}{\text{molar mass}}$

1.3 Avogadro’s Hypothesis

• "Equal volumes of gas at the same temperature and pressure contain the same numbers of molecules."
• Corollary: At standard temperature and pressure, one mole quantities of all gases contain the same number of molecules.
• This number is called $N_A$
• $N_A = 6.02 \times 10^{23} \text{ particles/mole}$
• Total number of particles: $N = nN_A$
• Mass of an individual atom: $m = \frac{\text{molar mass}}{N_A}$

Problem 1: Hope Diamond and Rosser Reeves Ruby

• The Hope diamond (44.5 carats) is almost pure carbon, and the Rosser Reeves (138 carats) is primarily aluminum oxide ($Al<em>2O</em>3$).
• One carat is equivalent to a mass of 0.200 g.
• Determine:
  • (a) the number of carbon atoms in the Hope diamond
  • (b) the number of $Al<em>2O</em>3$ molecules in the Rosser Reeves
• Solution (a):
  • Mass of the Hope diamond: $m_{\text{Hope}} = (44.5 \text{ carats}) \times (0.200 \frac{\text{g}}{\text{carat}}) = 8.90 \text{ g}$
  • Number of moles in the Hope diamond: $n_{\text{Hope}} = \frac{8.90 \text{ g}}{12.011 \frac{\text{g}}{\text{mol}}} = 0.741 \text{ mol}$
  • Number of carbon atoms in the Hope diamond: $N_H = (0.741 \text{ mol}) \times (6.022 \times 10^{23} \frac{\text{atoms}}{\text{mol}}) = 4.46 \times 10^{23} \text{ atoms}$
• Solution (b):
  • Mass of the Rosser Reeves: $m_R = (138 \text{ carats}) \times (0.200 \frac{\text{g}}{\text{carat}}) = 27.6 \text{ g}$
  • Molecular mass of $Al<em>2O</em>3$: $m_R = 2(26.9815 \text{ u}) + 3(15.9994 \text{ u}) = 101.9612 \text{ g/mol}$
  • Number of moles in the Rosser Reeves: $n_R = \frac{27.6 \text{ g}}{101.9612 \frac{\text{g}}{\text{mol}}} = 0.271 \text{ mol}$
  • Number of $Al<em>2O</em>3$ molecules in the Rosser Reeves: $N_R = (0.271 \text{ mol}) \times (6.022 \times 10^{23} \frac{\text{molecules}}{\text{mol}}) = 1.63 \times 10^{23} \text{ molecules}$

1.4 Experimental Laws

• Boyle’s Law:
  • Experiment: Keeping the gas at a constant temperature.
  • Conclusion: Pressure is inversely proportional to volume.
  • Equation: $PV = \text{const}$
• Charles’ Law:
  • Experiment: Keeping the gas at constant pressure.
  • Conclusion: Temperature is directly proportional to volume.
  • Equation: $V = CT$ (C: constant)
• Gay-Lussac’s Law:
  • Experiment: Keeping the gas at constant volume.
  • Conclusion: Temperature is directly proportional to pressure.
  • Equation: $T = CP$ (C: constant)

1.5 Equation of State for an Ideal Gas

• Gay-Lussac’s law: $V = \text{constant} \rightarrow T = \text{const}$
• Boyle’s law: $PV = \text{const}$
• Charles’ law: $P = \text{const} \rightarrow V = CT$
• Number of moles $n$ of a substance of mass $m$ (g): $n = \frac{m}{M}$ (M: molar mass - g/mol)
• Equation of state for an ideal gas (Ideal gas law): $PV = nRT$
  • $T$: absolute temperature in kelvins
  • $R$: a universal constant that is the same for all gases, $R = 8.315 \text{ J/mol.K}$
• Definition of an Ideal Gas: "An ideal gas is one for which $PV/nT$ is constant at all pressures"
• Total number of molecules: $N = nN_A$
• With Boltzmann’s constant: $k<em>B = \frac{R}{N</em>A} = \frac{8.315 \text{ J/mol.K}}{6.022 \times 10^{23} \text{ mol}^{-1}} = 1.38 \times 10^{-23} \text{ J/K}$
• Ideal gas law: $PV = NkT_B$

Test Question

• An ideal gas is confined to a container with constant volume. The number of moles is constant. By what factor will the pressure change if the absolute temperature triples?
  • Answer: c. 3.0

Problem 2

• An ideal gas occupies a volume of 100 cm³ at 20°C and 100 Pa.
  • (a) Find the number of moles of gas in the container.
  • (b) How many molecules are in the container?
• Solution (a)
  • $n = \frac{PV}{RT} = \frac{(100 \text{ Pa}) \times (100 \times 10^{-6} \text{ m}^3)}{(8.315 \text{ J/mol.K}) \times (293 \text{ K})} = 4.10 \times 10^{-6} \text{ mol}$
• Solution (b)
  • $N = nN_A = (4.10 \times 10^{-6} \text{ mol}) \times (6.022 \times 10^{23} \frac{\text{molecules}}{\text{mol}}) = 2.47 \times 10^{18} \text{ molecules}$

Problem 3: Scuba Tank

• A scuba tank is designed to hold 66 ft³ of air at atmospheric pressure at 22°C. When compressed to 3,000 lb/in² and stored in a 10-L (0.35 ft³) tank, the air becomes hot.
  • (a) If the air does not cool, what is its temperature?
  • (b) What is the air temperature in degrees Celsius and in degrees Fahrenheit?
• Solution (a)
  • Since the number of moles $n$ remains constant, we use $\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2}$.
  • $T<em>2 = \frac{P</em>2V<em>2T</em>1}{P<em>1V</em>1} = \frac{(3000 \text{ lb/in}^2) \times (0.35 \text{ ft}^3) \times (295 \text{ K})}{(14.7 \text{ lb/in}^2) \times (66 \text{ ft}^3)} = 319 \text{ K}$
• Solution (b)
  • $T_C = 45.9 °C$
  • $T_F = 115 °F$

Problem 4: Sculpa Tank

• A sculpa tank has a volume of 0.0150 m³ filled with compressed air at a pressure of 2.02 × 10⁷ Pa. Air is consumed at a rate of 0.0300 m³/minute. Determine how long the diver can stay under seawater at a depth of:
  • (a) 10.0 m
  • (b) 30.0 m
• The density of seawater is $\rho = 1025 \text{ kg/m}^3$.
• Solution (a)
  • Pressure at 10.0 m: $P<em>2 = P</em>1 + \rho gh = 1.01 \times 10^5 \text{ Pa} + (1025 \text{ kg/m}^3) \times (9.80 \text{ m/s}^2) \times (10.0 \text{ m}) = 2.01 \times 10^5 \text{ Pa}$
  • Volume available for breathing: $\frac{P<em>1V</em>1}{P_2} = V \Rightarrow V = \frac{(2.02 \times 10^7 \text{ Pa}) \times (0.0150 \text{ m}^3)}{1.01 \times 10^5 \text{ Pa}} = 3.01 m^3$
  • The compressed air will last for:
  • $t=\frac{1.50 m^3}{0.0300 m^3/min} = 50.0 min$
• Solution (b)
  • The deeper dive must have a shorter duration. $t = 24.6 min$

Problem 5: Spray Can

• A spray can contains propellant gas at twice atmospheric pressure (202 kPa) and has a volume of 125 cm³ at 22°C. It is tossed into an open fire. When the temperature reaches 195°C, what is the pressure inside the can? Assume any change in the volume of the can is negligible.
• Solution
  • Since the number of moles $n$ remains constant and the initial and final volumes are assumed to be equal: $\frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2}$
  • $P<em>2 = \frac{T</em>2}{T<em>1}P</em>1 = \frac{(468 \text{ K})}{(295 \text{ K})} \times (202 \text{ kPa}) = 320 \text{ kPa}$

Problem 6

• An ideal gas at 20.0°C at a pressure of 1.50 × 10⁵ Pa has a number of moles of 6.16 × 10⁻² mol.

  • (a) Find the volume of the gas.
  • (b) The gas expands to twice its original volume, while the pressure falls to atmospheric pressure. Find the final temperature.

• Solution (a)

  • $V = \frac{nRT}{P} = \frac{(6.16 \times 10^{-2} \text{ mol}) \times (8.315 \text{ J/mol.K}) \times (293 \text{ K})}{1.50 \times 10^5 \text{ Pa}} = 1.00 \times 10^{-3} \text{ m}^3 = 1.00 \text{ L}$

• (b)

  • Since the number of moles $n$ remains constant, we use $\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2}$
  • $T<em>2 = \frac{P</em>2V<em>2T</em>1}{P<em>1V</em>1} = \frac{(1.01 \times 10^5 \text{ Pa}) \times (2.00 \text{ L}) \times (293 \text{ K})}{(1.50 \times 10^5 \text{ Pa}) \times (1.00 \text{ L})} = 395 \text{ K}$

Problem 7: Message in a Bottle

• A beachcomber finds a corked bottle containing a message. The air in the bottle is at atmospheric pressure and a temperature of 30.0°C. The cork has a cross-sectional area of 2.30 cm². The beachcomber places the bottle over a fire, figuring the increased pressure will push out the cork. At a temperature of 99°C the cork is ejected from the bottle.
  • (a) What was the pressure in the bottle just before the cork left it?
  • (b) What force of friction held the cork in place?
• Solution (a)
  • Since the number of moles $n$ and Volume $V$ remains constant, so $\frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2}$
  • $P_2 = \frac{(372 \text{ K})}{(303 \text{ K})} \times (1.01 \times 10^5 \text{ Pa}) = 1.24 \times 10^5 \text{ Pa}$
• Solution (b)
  • $\sum F = 0$
  • $F<em>{\text{fric}} = (P</em>{\text{in}} - P_{\text{out}})A = (1.24 \times 10^5 \text{ Pa} - 1.01 \times 10^5 \text{ Pa}) \times (2.30 \times 10^{-4} \text{ m}^2) = 5.29 \text{ N}$

Problem 8

• A room of volume 60.0 m³ contains air with an equivalent molar mass of 29.0 g/mol. If the temperature is raised from 17.0°C to 37.0°C, what mass of air will leave the room? Assume that the air pressure is maintained at 101 kPa.
• Solution
  • $m = \frac{\mu PV}{RT} = (29.0 \times 10^{-3} \text{ kg/mol}) \times (1.01 \times 10^5 \text{ Pa}) \frac{60.0}{8.315 \text{ J/mol K}} (\frac{1}{290 \text{ K}} - \frac{1}{310 \text{ K}}) = 4.70 \text{ kg}$

2 Molecular Model of an Ideal Gas

2.1 Assumptions of the molecular model of an ideal gas

• A container with volume $V$ contains a very large number $N$ of identical molecules, each with mass $m$.
• The molecules behave as point particles; their size is small in comparison to the average distance between particles and to the dimensions of the container.
• The molecules are in constant motion; they obey Newton's laws of motion. Each molecule collides occasionally with a wall of the container. These collisions are perfectly elastic.
• The container walls are perfectly rigid and infinitely massive and do not move.

2.2 Collisions and Gas Pressure

• Consider a cubical box with sides of length $d$ containing an ideal gas. The molecule shown moves with velocity $v$.
• Consider the collision of one molecule moving with a velocity $v$ toward the right-hand face of the box.
• Elastic collision with the wall implies that the x component of momentum is reversed, while its y component remains unchanged:
  • $\Delta p<em>x = -mv</em>x - (mv<em>x) = -2mv</em>x$
• The average force exerted on the molecule:
  • $F<em>x = \frac{\Delta p</em>x}{\Delta t} = \frac{-2mv<em>x}{\frac{2d}{v</em>x}} = -\frac{mv_x^2}{d}$
• The average force exerted by the molecule on the wall:
  • $F<em>x = \frac{mv</em>x^2}{d}$
• The total force $F$ exerted by all the molecules on the wall:
  • $F = \frac{m}{d}(v<em>{x1}^2 + v</em>{x2}^2 + …)$
• The average value of the square of the velocity in the x direction for $N$ molecules:
  • $v<em>x^2 = \frac{v</em>{x1}^2 + v<em>{x2}^2 + … + v</em>{xN}^2}{N}$
• The total pressure exerted on the wall:
  • $F = \frac{Nm}{d}v_x^2$
  • $v^2 = v<em>x^2 + v</em>y^2 + v<em>z^2$ , $v</em>x^2=v<em>y^2=v</em>z^2$
  • $P = \frac{F}{A} = \frac{Nmv^2}{3V}$
• The equation of state for an ideal gas:
  • $PV = NkT$
• Temperature is a direct measure of average molecular kinetic energy
  • $\frac{1}{2}mv^2 = \frac{3}{2}kT$
• The average translational kinetic energy per molecule is
  • Each degree of freedom contributes to the energy of a system: $\frac{1}{2}kT$ (the theorem of equipartition of energy)

The total translational kinetic energy of N molecules of gas

• $\text{E}_{\text{trans}} = \frac{3}{2}NkT = \frac{3}{2}nRT$

2.3 The Boltzmann Distribution Law

• The Maxwell–Boltzmann distribution function
• Consider the distribution of molecules in our atmosphere

2.4 The mean free path

• A molecule moving through a gas collides with other molecules in a random fashion. Notion of the mean free path
• Between collisions, the molecules move with constant speed along straight lines. The average distance between collisions is called the mean free path.

3. The Molar Specific Heats of an Ideal Gas

• Constant volume: $C_V$ : the molar specific heat at constant volume
• Constant pressure: $C_P$ : the molar specific heat at constant pressure
• First law of thermodynamics: $\Delta E_{\text{int}} = Q - W$
• $Q = nC_V\Delta T$
• $Q = nC_P\Delta T$
• $C<em>P = C</em>V + R$

4 Adiabatic Expansion of an Ideal Gas

• For adiabatic process : no energy is transferred by heat between the gas and its surroundings: dQ = 0
• dU = dQ – dW = -dW
• Definition of the Ratio of Heat Capacities :
  • $\gamma = C<em>P / C</em>V$
• For ideal gas :
  • Equation of state: PV = nRT
  • Adiabatic process: $PV^\gamma = \text{const.}$
  • Adiabatic process: $TV^{\gamma-1} = \text{const.}$