University Physics Al No ll Kinetic Theory Class Number ame I. Choose the Correct Answer Which type of ideal gas will have the largest value for Cp-Cv? (D) (A)Monatomic (B) Diatomic (C) Polyatomic (D) The value will be the same for all Solution The difference between the molar specific heat at constant pressure and that at constant volume equal to the universal gas constant Cp-CV=R 2. What would be the most likely value for CT, the molar heat capacity at constant temperature? (A)0 (B)O<C< Cv (C)Cr<CT <CP (D)CToo Using the definition of the molar specific heat: C hen T= constant, the molar n dT heat capacity is CT=∞ 3. Which of the following speeds divides the molecules in a gas in thermal equilibrium so that half have speeds faster, and half have speeds slower? (A)Vp (C)vms (D) Non of the above Solution According to the physical meaning of v,, Vav, and vms, non of them divides the molecules in a gas so that half have speeds faster, and half have speeds slower 4. Which of the following speeds corresponds to a molecule with the average kinetic energy? (A) (C)vm (D)Non of the above Solution The average kinetic energy is KEme =)m(v2),and for vm =v v),we have KEwe=mvms Then the answer is(C)
University Physics AI No. 11 Kinetic Theory Class Number Name I．Choose the Correct Answer 1. Which type of ideal gas will have the largest value for CP-CV? （ D ） (A) Monatomic (B) Diatomic (C) Polyatomic (D) The value will be the same for all. Solution: The difference between the molar specific heat at constant pressure and that at constant volume is equal to the universal gas constant: CP-CV = R. 2. What would be the most likely value for CT, the molar heat capacity at constant temperature? （ D ） (A) 0 (B) 0< CT < CV (C) CV <CT <CP (D) CT=∞ Solution: Using the definition of the molar specific heat: T Q n Cmolar d 1 d = , so when T = constant, the molar heat capacity is CT=∞. 3. Which of the following speeds divides the molecules in a gas in thermal equilibrium so that half have speeds faster, and half have speeds slower? （ B ） (A) vp (B) vav (C) vrms (D) Non of the above. Solution: According to the physical meaning of av rms v , v , and v p , non of them divides the molecules in a gas so that half have speeds faster, and half have speeds slower. 4. Which of the following speeds corresponds to a molecule with the average kinetic energy? （ C ） (A) vp (B) vav (C) vrms (D) Non of the above Solution: The average kinetic energy is 2 2 1 KE m v ave = , and for 2 v v rms = , we have 2 2 1 KEave = mvrms , Then the answer is (C)
I. Filling the blanks The number of particles in a cubic millimeter of a gas at temperature 273 K and 1.00 atm pressure is 2.69x10. To get a feeling for the order of magnitude of this number the age of the universe in seconds assuming it is 15 billion years old is 4.37x107 ntIon The number of particles is N=nN PV=nRT .、1×1013×105×1×10°×602×103 N=nN-Pv =269×106 8.315×273 The age of the universe in seconds is 15×10°×365×24×60×60=437×10(s) 2. A sample of oxygen gas (O2)is at temperature 300 K and 1.00 atm pressure. One molecule, with a speed equal to the rms speed, makes a head-on elastic collision with your nose. Ouch! The magnitude of the impulse imparted to your schnozzle is 5.- Kg m/s Solution The speed of an oxygen molecule is 3R/y/2=3×8315×300y2=483.59m/s) 32×10 Using the Impulse-Momentum Theorem, the magnitude of the impulse imparted to the schnozzle M2×32×10-3×483.59 Ⅰ=△P=2mv=2 =514×102(Kgm/s) 6.02×10 3. When helium atoms have an rms speed equal to the escape speed from the surface of the earth (escape=11.2 km/s), the temperature is_2.01x10K Ition According to the problem vm=(.-) 50=M201240)x4=201k 3×8.315 4 The rms speed of hydrogen gas(H2)at temperature 300 K in the atmosphere is- 1.93x10-km/ Compare it with the escape speed from the Earth(11. 2 km/s). Since hydrogen is the least massive gas, hydrogen molecules will have the highest rms speeds at a given temperature. How can this calculation explain why there is essentially no hydrogen gas in the atmosphere of the earth?
II. Filling the Blanks 1. The number of particles in a cubic millimeter of a gas at temperature 273 K and 1.00 atm pressure is 2.69×1016 . To get a feeling for the order of magnitude of this number, the age of the universe in seconds assuming it is 15 billion years old is 4.37×1017 s . Solution: The number of particles is A N = nN Q PV = nRT 16 5 9 23 2.69 10 8.315 273 1 1.013 10 1 10 6.02 10 = × × × × × × × × ∴ = = = − A NA RT PV N nN The age of the universe in seconds is 15 10 365 24 60 60 4.37 10 (s) 9 17 × × × × × = × 2. A sample of oxygen gas (O2) is at temperature 300 K and 1.00 atm pressure. One molecule, with a speed equal to the rms speed, makes a head-on elastic collision with your nose. Ouch! The magnitude of the impulse imparted to your schnozzle is 5.14×10-23 Kg m/s . Solution: The speed of an oxygen molecule is ) 483.59(m/s) 32 10 3 8.315 300 ) ( 3 ( 1/ 2 3 1/ 2 = × × × = = = − M RT v vrms Using the Impulse-Momentum Theorem, the magnitude of the impulse imparted to the schnozzle is 5.14 10 (Kg m/s) 6.02 10 2 32 10 483.59 2 2 23 23 3 = × ⋅ × × × × = ∆ = = = − − − v N M I P mv A 3. When helium atoms have an rms speed equal to the escape speed from the surface of the Earth (vescape = 11.2 km/s), the temperature is 2.01×104 K . Solution: According to the problem 1/ 2 ) 3 ( M RT vrms = So the temperature is 2.01 10 (K) 3 8.315 (11.2 10 ) 4 10 3 4 2 3 2 3 = × × × × × = = − R v M T rms 4 The rms speed of hydrogen gas (H2) at temperature 300 K in the atmosphere is 1.93×103 km/s . Compare it with the escape speed from the Earth (11.2 km/s). Since hydrogen is the least massive gas, hydrogen molecules will have the highest rms speeds at a given temperature. How can this calculation explain why there is essentially no hydrogen gas in the atmosphere of the Earth?
Solution The rms speed of hydrogen gas is 3RT√23×8.315×300 12=193×10°(m/s) 2×10 The rms speed of hydrogen gas is less than the escape speed from the Earth(11.2 km/s). Since hydrogen is the least massive gas, hydrogen molecules have the highest rms speeds at a given temperature. Hydrogen have highest escape rate in the atmosphere of the Earth, so there is very few hydrogen gas in the atmosphere of the Earth 5. The average kinetic energy of a particle in a gas at temperature T is given by Equation 14.13 KE=二kT.or kT. The special theory of relativity states that there is an upper limit on the speed of an particle: the speed of light c=3.00 x 10 m/s. For a gas of hydrogen atoms, the immediately preceding equation implies an upper limit on the temperature. The absolute temperature such that(v") for a gas of hydrogen atoms is equal to the square of the speed of light is 4.35x10 K In fact, there is no upper limit on the temperature, so the classical expression for the kinetic energy cannot be valid for speeds approaching the speed of light Solution According to the problem (v*)=c,so T 2×10-3×(3×103)2 3×1.38×10-23=4.35×1036(K) I. Give the solutions of the following problems Consider helium gas at temperature 300 K near the surface of the earth (a) Calculate the average kinetic energy of one of the helium atoms (b) Calculate the gravitational potential energy of a single helium atom near the surface of the Earth Choose the zero of gravitational potential energy to be infinitely far away from the earth (c)What is the absolute value of the ratio of the gravitational potential energy of the helium atom to its average kinetic energy? Is it justifiable to neglect the gravitational potential energy in theory? Why or why not (a) The average kinetic energy of one of the helium atoms KE 38×10-2×300=6.21×1021(J) (b) The gravitational potential energy of a single helium atom near the surface of the Earth is 981×6374×10°=-416×10-(J) 6.02×10
Solution: The rms speed of hydrogen gas is ) 1.93 10 (m/s) 2 10 3 8.315 300 ) ( 3 ( 1/ 2 3 3 1/ 2 = × × × × = = − M RT vrms The rms speed of hydrogen gas is less than the escape speed from the Earth (11.2 km/s). Since hydrogen is the least massive gas, hydrogen molecules have the highest rms speeds at a given temperature. Hydrogen have highest escape rate in the atmosphere of the Earth, so there is very few hydrogen gas in the atmosphere of the Earth. 5. The average kinetic energy of a particle in a gas at temperature T is given by Equation 14.13: KEave kT 2 3 = , or m v kT 2 3 2 1 2 = . The special theory of relativity states that there is an upper limit on the speed of an particle: the speed of light c = 3.00 × 108 m/s. For a gas of hydrogen atoms, the immediately preceding equation implies an upper limit on the temperature. The absolute temperature such that 〈v 2 〉 for a gas of hydrogen atoms is equal to the square of the speed of light is 4.35×1036 K . In fact, there is no upper limit on the temperature; so the classical expression for the kinetic energy cannot be valid for speeds approaching the speed of light. Solution: According to the problem 〈v 2 〉 = c 2 , so 4.35 10 (K) 3 1.38 10 2 10 (3 10 ) 3 3 36 23 2 3 8 2 2 = × × × × × × = = = − − k mc k m v T III. Give the Solutions of the Following Problems 1. Consider helium gas at temperature 300 K near the surface of the Earth. (a) Calculate the average kinetic energy of one of the helium atoms. (b) Calculate the gravitational potential energy of a single helium atom near the surface of the Earth. Choose the zero of gravitational potential energy to be infinitely far away from the Earth. (c) What is the absolute value of the ratio of the gravitational potential energy of the helium atom to its average kinetic energy? Is it justifiable to neglect the gravitational potential energy in theory? Why or why not? Solution: (a) The average kinetic energy of one of the helium atoms is 1.38 10 300 6.21 10 (J) 2 3 2 3 −23 −21 KEave = kT = × × × = × (b) The gravitational potential energy of a single helium atom near the surface of the Earth is 9.81 6.374 10 4.16 10 (J) 6.02 10 4 10 6 19 23 3 − − × × × = − × × × PEgrav = −mgREarth = −
(c) The absolute value of the ratio is PE gm=4.16×10 621×1067 It is justifiable to neglect the gravitational potential energy in theory. As the average kinetic energy is a statistical value for the whole system regardless the gravitational potential energy 2. A well-insulated 4.00 liter box contains a partition dividing the box into two equal volumes as shown in Figure 1. Initially, 2.00 g of molecular hydrogen gas(H2) at 300 K is confined to the left-hand side of the partition, and the other half is a vacuum (a)What is the rms speed of the particles in the gas? Insulation Thin partitio b)What is the initial pressure of the gas? (c)The partition is removed or broken suddenly, so that the gas now is contained throughout the entire box. Assume that the gas is ideal Gas Vacuum Does the temperature of the gas change? What is the change in the internal energy of the system? Fig. l (d)When the gas reaches equilibrium, what is the final pressure? eed of th 3RT 3×8315×300y2=193×10°(m) 2×10 (b)Using the equation of state for an ideal gas: PV=nRT=RT So the initial pressure of the gas is mRT2×10-3×8.315×300 =1.25×10°(P (c) The temperature is not changed. The internal energy of the system is not changed too (d) Because it is so fast that the gas reaches equilibrium. So it is an adiabatic process. For 1=300K and PV=PV, So the final pressure is 0 4×10 3. A gas initially at temperature Ti, pressure Pi and volume Vi has its pressure reduced to a final value P/via one of the following types of processes: (1)isochoric; (2) isothermal; (3)adiabatic (a)Sketch each process schematically on a P-V diagram b)In which process is the work done by the gas zero? (c)In which process is the work done by the gas greatest?
(c) The absolute value of the ratio is 67 6.21 10 4.16 10 21 19 = × × = − − ave grav KE PE It is justifiable to neglect the gravitational potential energy in theory. As the average kinetic energy is a statistical value for the whole system regardless the gravitational potential energy. 2. A well-insulated 4.00 liter box contains a partition dividing the box into two equal volumes as shown in Figure 1. Initially, 2.00 g of molecular hydrogen gas (H2) at 300 K is confined to the left-hand side of the partition, and the other half is a vacuum. (a) What is the rms speed of the particles in the gas? (b) What is the initial pressure of the gas? (c) The partition is removed or broken suddenly, so that the gas now is contained throughout the entire box. Assume that the gas is ideal. Does the temperature of the gas change? What is the change in the internal energy of the system? (d) When the gas reaches equilibrium, what is the final pressure? Solution: (a) The rms speed of the particles in the gas is ) 1.93 10 (m/s) 2 10 3 8.315 300 ) ( 3 ( 1/ 2 3 3 1/ 2 = × × × × = = − M RT vrms (b) Using the equation of state for an ideal gas: RT M m PV = nRT = So the initial pressure of the gas is 1.25 10 (Pa) 2 10 2 10 2 10 8.315 300 6 3 3 3 = × × × × × × × = = − − − V RT M m P (c) The temperature is not changed. The internal energy of the system is not changed too. (d) Because it is so fast that the gas reaches equilibrium. So it is an adiabatic process. For T=300K and P1 V1 = P2V2 , So the final pressure is 6.25 10 (Pa) 4 10 1.25 10 2 10 5 3 6 3 2 1 1 2 = × × × × × = = − − V PV P 3. A gas initially at temperature Ti, pressure Pi and volume Vi has its pressure reduced to a final value Pf via one of the following types of processes: (1) isochoric; (2) isothermal; (3) adiabatic. (a) Sketch each process schematically on a P-V diagram. (b) In which process is the work done by the gas zero? (c) In which process is the work done by the gas greatest? Gas Vacuum Insulation Thin partition Fig.1
(d)Show that the ratio of the absolute magnitude of the heat transfer to the gas during the isothermal process to that during the isochoric process is RP P choric (e)In which process is the absolute magnitude of the change in the internal energy of the gas the greatest? Explain your reasoning Solution Isothermal process Adiabatic process P (b)See the area under the curve of the graph, the work done by the gas is zero in isochoric process (c) See the area under the curve of the graph, the work done by the gas is greatest in isothermal (d) For isothermal process: Pv=P/ So the heat transfer is O.m, =nRT In=nRT In P. For isochoric process P T P T P-P T Thus the ratio of the absolute magnitude of the heat transfer to the gas during the isothermal process to that during the isochoric process is
(d) Show that the ratio of the absolute magnitude of the heat transfer to the gas during the isothermal process to that during the isochoric process is f i V i isochoric isothermal P P c P RP Q Q ln ∆ = (e) In which process is the absolute magnitude of the change in the internal energy of the gas the greatest? Explain your reasoning. Solution: (a) (b) See the area under the curve of the graph, the work done by the gas is zero in isochoric process. (c) See the area under the curve of the graph, the work done by the gas is greatest in isothermal process. (d) For isothermal process: Pi Vi = PfVf So the heat transfer is f i i i f isothermal i P P nRT V V Q = nRT ln = ln For isochoric process: f i f i T T P P = So the heat transfer is ( ) Qisochoric = nCV Tf −Ti f i f i T T P P Q = f i i f i i T T T P P P − = − ∴ i i f i isochoric V T P P P Q nC ⋅ − ∴ = ⋅ Thus the ratio of the absolute magnitude of the heat transfer to the gas during the isothermal process to that during the isochoric process is Pi Pf P V Isothermal process Adiabatic process Isochoric process
nRT. In RP In P sothernnal isochoric (P-P) Cr(P-P)Cr.A (e) For isothermal process: AU,=0. Using the first law of thermodynamics, for an adiabatic AU2=W=nC, (T-T) For isochoric process: W=0 AU3=9=nCy(T-T) T>T U2|>U So the absolute magnitude of the change in the internal energy of the gas is the greatest in adiabatic 4. An ideal gas experiences an adiabatic compression from P=122kpa, /=10.7m, T--230C to P=1450kpa, V=1. 36m.(a) Calculate the value of y(b)Find the final temperature(c) How many moles of gas are present?(d) What is the total translational kinetic energy per mole before and after the compression?(e)Calculate the ratio of the rms speed before to that after the compression Solution (a)For adiabatic process: P=PV 1450 Hence"y==(107y=1450 1.36 122 2≈1.2 1.36 (b)Using the equation of state for an ideal gas: PV=nRT, we have i'= ev So the final temperatur2=1450×10×136×(273-23)=376K 122×103×107 (c)Using the equation of state for an ideal gas: Pv=nRT, we have
f i V i V f i f i i i i V f i f i i isochoric isothermal P P C P RP C P P P P RP P T nC P P P P nRT Q Q ln ( ) ln ( ) ln ⋅ ∆ = ⋅ − = ⋅ − = (e) For isothermal process: 0 ∆U1 = . Using the first law of thermodynamics, for an adiabatic process: Q = 0 ( ) 2 V Tf Ti ∆U = W = nC − For isochoric process: W = 0 ( ) 3 V Tf Ti ∆U = Q = nC ′ − Tf Tf Q > ′ ∴ ∆U2 > ∆U3 So the absolute magnitude of the change in the internal energy of the gas is the greatest in adiabatic process. 4. An ideal gas experiences an adiabatic compression from P=122kpa, V=10.7m3 , T=-23.0°C to P=1450kpa, V=1.36m3 . (a) Calculate the value of γ. (b) Find the final temperature. (c) How many moles of gas are present? (d) What is the total translational kinetic energy per mole before and after the compression? (e) Calculate the ratio of the rms speed before to that after the compression. Solution: (a) For adiabatic process: γ γ Pi Vi = Pf Vf , Hence 1.2 1.36 10.7 ln 122 1450 ln 122 1450 ) 1.36 10.7 ( ) = ⇒ ( = ⇒ γ = = γ γ i f f i P P V V , (b) Using the equation of state for an ideal gas: PV = nRT , we have f f f i i i T P V T PV = So the final temperature is 377.66(K) 122 10 10.7 1450 10 1.36 (273 23) 3 3 = × × × × × − = = i i f f i f PV P V T T (c) Using the equation of state for an ideal gas: PV = nRT , we have
P122×103×10.7 RT8.315×(273-23) 62797(mole) d)The total translational kinetic energy per mole before and after the compression is KE=nR7=62797××8.31×250=1.96×10°(J) KEa=nR7=62797××8.31×37766=296×10°(J (e)The rms speed is v BRT/ Hence the ratio of the rms speed before to that after the compression =081 377.66
627.97 (mole) 8.315 (273 23) 122 10 10.7 3 = × − × × = = RT PV n (d)The total translational kinetic energy per mole before and after the compression is 8.31 377.66 2.96 10 (J) 2 3 627.97 2 3 8.31 250 1.96 10 (J) 2 3 627.97 2 3 6 after 6 before = = × × × = × = = × × × = × f i KE n RT KE n RT (e) The rms speed is 1/ 2 ) 3 ( M RT vrms = Hence the ratio of the rms speed before to that after the compression is ) 0.81 377.66 273 23 ( ) ( 1/ 2 1/ 2 = − = = f i rms f rms i T T v v