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PROGRAM OF “PHYSICS” Lecturer: Dr. DO Xuan Hoi Room

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PHYSICS 2 (FLUID MECHANICS AND THERMAL PHYSICS) 02 credits (30 periods) Chapter 1 Fluid Mechanics Chapter 2 Heat, Temperature and the Zero th 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

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References : Halliday D., Resnick R. and Walker, J. (2005), Fundamentals of Physics, Extended seventh edition. John Willey and Sons, Inc. Alonso M. and Finn E.J. (1992). Physics, Addison-Wesley Publishing Company Hecht, E. (2000). Physics. Calculus, Second Edition. Brooks/Cole. Faughn/Serway (2006), Serway’s College Physics, Brooks/Cole. Roger Muncaster (1994), A-Level Physics, Stanley Thornes.

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astr.gsu.edu/hbase/HFrame.html ml

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CHAPTER 4 The Kinetic Theory of Gases 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

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1. Ideal Gases, Experimental Laws and the Equation of State 1.1 Notions ► ► Properties of gases A gas does not have a fixed volume or pressure In a container, the gas expands to fill the container ► ► Ideal gas Collection of atoms or molecules that move randomly Molecules exert no long-range force on one another Molecules occupy a negligible fraction of the volume of their container ► ► Most gases at room temperature and pressure behave approximately as an ideal gas

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1.2 Moles ► ► It’s convenient to express the amount of gas in a given volume in terms of the number of moles, n ► One mole is the amount of the substance that contains as many particles as there are atoms in 12 g of carbon-12

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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 NANA Can also look at the total number of particles: The number of particles in a mole is called Avogadro’s Number N A =6.02 x particles / mole The mass of an individual atom :

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The Hope diamond (44.5 carats) is almost pure carbon and the Rosser Reeves (138 carats) is primarily aluminum oxide (Al 2 O 3 ). One carat is equivalent to a mass of g. Determine (a) the number of carbon atoms in the Hope diamond and (b) the number of Al 2 O 3 molecules in the ruby Rosser Reeves. SOLUTION The mass of the Hope diamond : (a) The number of moles in the Hope diamond : The number of carbon atoms in the Hope diamond : PROBLEM 1

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The Hope diamond (44.5 carats) is almost pure carbon and the Rosser Reeves (138 carats) is primarily aluminum oxide (Al 2 O 3 ). One carat is equivalent to a mass of g. Determine (a) the number of carbon atoms in the Hope diamond and (b) the number of Al 2 O 3 molecules in the ruby Rosser Reeves. SOLUTION The mass of the Rosser Reeves : (b) Molecular mass : The number of moles in the Rosser Reeves : PROBLEM 1

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(b) The number of Al 2 O 3 molecules in the Rosser Reeves : The Hope diamond (44.5 carats) is almost pure carbon and the Rosser Reeves (138 carats) is primarily aluminum oxide (Al 2 O 3 ). One carat is equivalent to a mass of g. Determine (a) the number of carbon atoms in the Hope diamond and (b) the number of Al 2 O 3 molecules in the ruby Rosser Reeves. SOLUTION PROBLEM 1

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1.4 Experimental Laws Boyle’s Law Experiment : Conclusion : When the gas is kept at a constant temperature, its pressure is inversely proportional to its volume (Boyle’s law)

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Charles’ Law Experiment : Conclusion : At a constant pressure, the temperature is directly proportional to the volume (Charles’ law) ( C : constant )

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Gay-Lussac’s Law Experiment : Conclusion : At a constant volume, the temperature is directly proportional to the pressure (Gay-Lussac’ law) ( C : constant )

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1.5 Equation of State for an Ideal Gas Gay-Lussac’ law :V = constant T = const Boyle’s law : Charles’ law : P = const The number of moles n of a substance of mass m (g) : (M : molar mass-g/mol) Equation of state for an ideal gas : (Ideal gas law) T : absolute temperature in kelvins R : a universal constant that is the same for all gases R =8.315 J/mol.K

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Definition of an Ideal Gas : “An ideal gas is one for which PV/nT is constant at all pressures” Total number of molecules : With Boltzmann’s constant : Ideal gas law :

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Ideal gas law for an quantity of gas:

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Test 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? a. 1/9 b. 1/3 c. 3.0 d. 9.0

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An ideal gas occupies a volume of 100cm 3 at 20°C and 100 Pa. (a) Find the number of moles of gas in the container SOLUTION The number of moles of gas : (a) PROBLEM 2 (b) How many molecules are in the container? The number molecules in the container : (b)

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A certain scuba tank is designed to hold 66 ft 3 of air when it is at atmospheric pressure at 22°C. When this volume of air is compressed to an absolute pressure of lb/in. 2 and stored in a 10-L (0.35-ft 3 ) tank, the air becomes so hot that the tank must be allowed to cool before it can be used. (a) If the air does not cool, what is its temperature? (Assume that the air behaves like an ideal gas.) PROBLEM 3 ) SCUBA (Self-Contained Underwater Breathing Apparatus) The number of moles n remains constant : (a)

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A certain scuba tank is designed to hold 66 ft 3 of air when it is at atmospheric pressure at 22°C. When this volume of air is compressed to an absolute pressure of lb/in. 2 and stored in a 10-L (0.35-ft 3 ) tank, the air becomes so hot that the tank must be allowed to cool before it can be used. (b) What is the air temperature in degrees Celsius and in degrees Fahrenheit? PROBLEM °C; 115°F. (b)

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A sculpa consists of a m3 m3 tank filled with compressed air at a pressure of 2.02 10 7 Pa. Assume that air is consumed at a rate of m3 m3 per minute and that the temperature is the same at all depths, determine how long the diver can stay under seawater at a depth of (a) 10.0 m and (b) 30.0 m The density of seawater is = 1025 kg/m 3. PROBLEM 4 SOLUTION (a) The volume available for breathing :

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A sculpa consists of a m 3 tank filled with compressed air at a pressure of 2.02 10 7 Pa. Assume that air is consumed at a rate of m 3 per minute and that the temperature is the same at all depths, determine how long the diver can stay under seawater at a depth of (a) 10.0 m and (b) 30.0 m The density of seawater is = 1025 kg/m 3. PROBLEM 4 SOLUTION (a) The compressed air will last for : (b) The deeper dive must have a shorter duration

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A spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having a volume of 125 cm 3 is at 22°C. It is then tossed into an open fire. When the temperature of the gas in the can reaches 195°C, what is the pressure inside the can? Assume any change in the volume of the can is negligible. PROBLEM 5 The number of moles n remains constant : SOLUTION Because the initial and final volumes of the gas are assumed to be equal :

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An ideal gas at 20.0 O C at a pressure of 1.50 10 5 Pa when has a number of moles of 6.16 mol. SOLUTION The volume : (a) PROBLEM 6 (a) Find the volume of the gas.

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An ideal gas at 20.0 O C at a pressure of 1.50 10 5 Pa when has a number of moles of 6.16 mol. SOLUTION The volume : (a) PROBLEM 6 (b) The gas expands to twice its original volume, while the pressure falls to atmospheric pressure. Find the final temperature. (b)

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A beachcomber finds a corked bottle containing a message. The air in the bottle is at the atmospheric pressure and a temperature of 30.0 O C. The cork has the cross-sectional area of 2.30 cm 3. The beachcomber places the bottle over a fire, figuring the increased pressure will pushout the cork. At a temperature of 99 o C the cork is ejected from the bottle PROBLEM 7 (a) What was the pressure in the bottle just before the cork left it ? (a) SOLUTION Message in a bottle found 24 years later - Yahoo!7

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A beachcomber finds a corked bottle containing a message. The air in the bottle is at the atmospheric pressure and a temperature of 30.0 O C. The cork has the cross-sectional area of 2.30 cm 3. The beachcomber places the bottle over a fire, figuring the increased pressure will pushout the cork. At a temperature of 99 o C the cork is ejected from the bottle PROBLEM 7 (b) What force of friction held the cork in place? (b) SOLUTION

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A room of volume 60.0 m3 m3 contains air having an equivalent molar mass of 29.0 g/mol. If the temperature of the room is raised from 17.0°C to 37.0°C, what mass of air (in kilograms) will leave the room? Assume that the air pressure in the room is maintained at 101 kPa. PROBLEM 8 SOLUTION

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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.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. A particle having a brownian motion inside a polymer like network Brownian motion

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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.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 Its x component of momentum is reversed, while its y component remains unchanged : The average force exerted on the molecule : The average force exerted by the molecule on the wall :

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T he total force F exerted by all the molecules on the wall : The average value of the square of the velocity in the x direction for N molecules : The total pressure exerted on the wall:

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The equation of state for an ideal gas : Temperature is a direct measure of average molecular kinetic energy The average translational kinetic energy per molecule is Each degree of freedom contributes to the energy of a system: (the theorem of equipartition of energy)

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The total translational kinetic energy of N molecules of gas : The number of moles of gas : Boltzmann’s constant Assume: Ideal gas is a monatomic gas (which has individual atoms rather than molecules: helium, neon, or argon) and the internal energy E int of ideal gas is simply the sum of the translational kinetic energies of its atoms

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The root-mean-square (rms) speed of the molecules : M is the molar mass in kilograms per mole : M = mN A

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Five gas molecules chosen at random are found to have speeds of 500, 600,700, 800, and 900 m/s. Find the rms speed. Is it the same as the average speed? SOLUTION PROBLEM 9 In general, v rms and v av are not the same.

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A tank used for filling helium balloons has a volume of m3 m3 and contains 2.00 mol of helium gas at 20.0°C. Assuming that the helium behaves like an ideal gas, (a) what is the total translational kinetic energy of the molecules of the gas? SOLUTION PROBLEM 10 (a)

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A tank used for filling helium balloons has a volume of m3 m3 and contains 2.00 mol of helium gas at 20.0°C. Assuming that the helium behaves like an ideal gas, (b) What is the average kinetic energy per molecule? (c) Using the fact that the molar mass of helium is 4.00 10 3 kg/mol, determine the rms speed of the atoms at 20.0°C. SOLUTION PROBLEM 10 (b) (c)

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(a) What is the average translational kinetic energy of a molecule of an ideal gas at a temperature of 27°C ? (b) What is the total random translational kinetic energy of the molecules in 1 mole of this gas? (c) What is the root-mean-square speed of oxygen molecules at this temperature ? SOLUTION PROBLEM 11 (a) (b)

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(a) What is the average translational kinetic energy of a molecule of an ideal gas at a temperature of 27°C ? (b) What is the total random translational kinetic energy of the molecules in 1 mole of this gas? (c) What is the root-mean-square speed of oxygen molecules at this temperature ? SOLUTION PROBLEM 11 (c)

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(a) A deuteron, 2 1 H, is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a nuclear fusion reactor must be heated to about 300 million K. What is the rms speed of the deuterons? Is this a significant fraction of the speed of light (c = 3.0 x 10 8 m/s) ? (b) What would the temperature of the plasma be if the deuterons had an rms speed equal to 0.10c ? SOLUTION PROBLEM 12

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2.3 The Boltzmann Distribution Law The Maxwell–Boltzmann distribution function Consider the distribution of molecules in our atmosphere : Determine how the number of molecules per unit volume varies with altitude Consider an atmospheric layer of thickness dy and cross-sectional area A, having N particles. The air is in static equilibrium : where nV nV is the number density. Law of Exponential Atmospheres From the equation of state :

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The Boltzmann distribution law : the probability of finding the molecules in a particular energy state varies exponentially as the negative of the energy divided by kBT.kBT.

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What is the number density of air at an altitude of 11.0 km (the cruising altitude of a commercial jetliner) compared with its number density at sea level? Assume that the air temperature at this height is the same as that at the ground, 20°C. SOLUTION PROBLEM 13 The Boltzmann distribution law : Assume an average molecular mass of :

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Density of the number of molecules with speeds between v and dv : The Maxwell–Boltzmann distribution function With :

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Poisson's Integral Formula: Density of the number of molecules with speeds between v and dv is

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Density of the number of molecules with speeds between v and dv is The rms speed : The average speed: The most probable speed:

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Definition: The average value of v n : PROOF: The average speed: The mean square speed: The most probable speed:

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For diatomic carbon dioxide gas ( CO 2, molar mass 44.0 g/mol) at T = 300 K, calculate (a) the most probable speed v mp ; (b) the average speed v av ; (c) the root-mean-square speed v rms. SOLUTION PROBLEM 14 The rms speed : The average speed: The most probable speed:

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At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at C? SOLUTION PROBLEM 15 The rms speed : A N 2 molecule has more mass so N 2 gas must be at a higher temperature to have the same v rms.

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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.

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The mean free path for a gas molecule Consider N spherical molecules with radius r in a volume V.V. Suppose only one molecule is moving. When it collides with another molecule, the distance between centers is 2r.2r. In a short time dt a molecule with speed v travels a distance vdt ; during this time it collides with any molecule that is in the cylindrical volume of radius 2r and length vdt. The volume of the cylinder : The number of the molecules with centers in this cylinder : The number of collisions per unit time : When all the molecules move at once :

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The average time between collisions (the mean free time) The mean free path (the average distance traveled between collisions) is For the ideal-gas :

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Approximate the air around you as a collection of nitrogen molecules, each of which has a diameter of 2.00 m. How far does a typical molecule move before it collides with another molecule? SOLUTION PROBLEM 16 Assume that the gas is ideal: The mean free path:

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A cubical cage 1.25 m on each side contains 2500 angry bees, each flying randomly at 1.10 m/s. We can model these insects as spheres 1.50 cm in diameter. On the average, (a) how far does a typical bee travel between collisions, (b) what is the average time between collisions, and (c) how many collisions per second does a bee make? SOLUTION PROBLEM 17

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3. The Molar Specific Heats of an ldeal 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:

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C : molar specific heat of Various Gases Gas constant: R = J/mol.K

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C : molar specific heat of Various Gases monatomic molecules: diatomic molecules: (not vibration) polyatomic molecules: f : degree of freedom (the number of independent coordinates to specify the motion of a molecule)

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V = const dW = 0 If the heat capacity is measured under constant- volume conditions: the molar heat capacity CV CV at constant volume First law dU = dQ = nC V dT By definition : (Ideal gas : PV = nRT) First law : dQ = dU + dW Relating Cp Cp and Cv Cv for an Ideal Gas

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The total work done by the gas as its volume changes from V 1 to Vf Vf : Ideal gas : Isothermal process: Work done by an ideal gas at constant temperature

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Also : When a system expands : work is positive. When a system is compressed, its volume decreases and it does negative work on its surroundings

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Work done by an ideal gas at constant volume Work done by an ideal gas at constant pressure

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PROBLEM 18 A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase of C at constant pressure. As a result, the bubble expands. The helium is monatomic and ideal. a) How much energy is added to the helium as heat during the increase and expansion? SOLUTION

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PROBLEM 18 A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase of C at constant pressure. As a result, the bubble expands. The helium is monatomic and ideal. a) How much energy is added to the helium as heat during the increase and expansion? (b) What is the change in the internal energy of the helium during the temperature increase? SOLUTION

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PROBLEM 18 A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase of C at constant pressure. As a result, the bubble expands. The helium is monatomic and ideal. a) How much energy is added to the helium as heat during the increase and expansion? (b) What is the change in the internal energy of the helium during the temperature increase? (c) How much work is done by the helium as it expands against the pressure of the surrounding water during the temperature increase? SOLUTION

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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 : The Ratio of Heat Capacities 4 Adiabatic Expansion of an Ideal Gas

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For ideal gas : From : R = CP CP - C V : Divide by PV :

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For ideal gas :

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PROBLEM 19 One mole of oxygen (assume it to be an ideal gas) expands at a constant temperature of 310 K from an initial volume 12 L to a final volume of 19 L. a/ How much work is done by the gas during the expansion? SOLUTION

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PROBLEM 19 One mole of oxygen (assume it to be an ideal gas) expands at a constant temperature of 310 K from an initial volume 12 L to a final volume of 19 L. a/ How much work is done by the gas during the expansion? b/ What would be the final temperature if the gas had expanded adiabatically to this same final volume? Oxygen (O 2 is diatomic and here has rotation but not oscillation.) SOLUTION

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PROBLEM 19 One mole of oxygen (assume it to be an ideal gas) expands at a constant temperature of 310 K from an initial volume 12 L to a final volume of 19 L. a/ How much work is done by the gas during the expansion? b/ What would be the final temperature if the gas had expanded adiabatically to this same final volume? Oxygen (O 2 is diatomic and here has rotation but not oscillation.) c/ What would be the final temperature and pressure if, instead, the gas had expanded freely to the new volume, from an initial pressure of.2.0 Pa? SOLUTION The temperature does not change in a free expansion:

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PROBLEM 20 Air at 20.0°C in the cylinder of a diesel engine is compressed from an initial pressure of 1.00 atm and volume of cm 3 to a volume of 60.0 cm 3. Assume that air behaves as an ideal gas with = 1.40 and that the compression is adiabatic. Find the final pressure and temperature of the air. SOLUTION

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PROBLEM 21 A typical dorm room or bedroom contains about 2500 moles of air. Find the change in the internal energy of this much air when it is cooled from 23.9°C to 11.6°C at a constant pressure of 1.00 atm. Treat the air as an ideal gas with = SOLUTION

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PROBLEM 22 The compression ratio of a diesel engine is 15 to 1; this means that air in the cylinders is compressed to 1/15 of its initial volume (Fig). If the initial pressure is 1.01 10 5 Pa and the initial temperature is 27°C (300 K), (a) find the final pressure and the temperature after compression. Air is mostly a mixture of diatomic oxygen and nitrogen; treat it as an ideal gas with = SOLUTION (a)

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The compression ratio of a diesel engine is 15 to 1; this means that air in the cylinders is compressed to 1/15 of its initial volume (Fig). If the initial pressure is 1.01 10 5 Pa and the initial temperature is 27°C (300 K),(b) how much work does the gas do during the compression if the initial volume of the cylinder is 1.00 L? Assume that CV CV for air is 20.8 J/mol.K and = SOLUTION (b) PROBLEM 22

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Two moles of carbon monoxide (CO) start at a pressure of 1.2 atm and a volume of 30 liters. The gas is then compressed adiabatically to 1/3 this volume. Assume that the gas may be treated as ideal. What is the change in the internal energy of the gas? Does the internal energy increase or decrease? Does the temperature of the gas increase or decrease during this process? Explain. SOLUTION PROBLEM 23

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Two moles of carbon monoxide (CO) start at a pressure of 1.2 atm and a volume of 30 liters. The gas is then compressed adiabatically to 1/3 this volume. Assume that the gas may be treated as ideal. What is the change in the internal energy of the gas? Does the internal energy increase or decrease? Does the temperature of the gas increase or decrease during this process? Explain. SOLUTION The internal energy increases because work is done on the gas (ΔU > 0) and Q = 0. The temperature increases because the internal energy has increased. PROBLEM 23

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On a warm summer day, a large mass of air (atmospheric pressure 1.01 10 5 Pa) is heated by the ground to a temperature of 26.0°C and then begins to rise through the cooler surrounding air. (This can be treated as an adiabatic process). Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 10 5 Pa. Assume that air is an ideal gas, with = SOLUTION PROBLEM 24

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