Chapter 02 Thermal Properties of Matter (Pp 19-58)

CH 02 THERMAL PROPERTIES OF MATTER

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CHAPTER 02 THERMAL PROPERTIES OF MATTER 2-1 BOYLE’S LAW Problem 2-1 A perfect gas undergoes isothermal compression, which reduces its volume by 2.20 × 10 −3 m 3 . The final pressure and volume of the gas are 3.78 × 10 3 Torr and 4.65 × 10 −3 m 3 respectively. Calculate the original pressure of the gas. Solution According to Boyle’s law p1V1 = p 2V2

V    4.65 × 10 −3  p1 = p 2  2  = (3.78 × 10 3 ) −3 −3   V1   (2.20 × 10 ) + (4.65 × 10 )  p1 = 2.57 × 10 3 Torr p1 = (2.57 ×103 )(133.32) Pa = 3.43 ×105 Pa Θ 1 Torr = 133.32 Pa Problem 2-2 A sample of air occupies 1.0 litre at 25 0 C and 1.0 atm . What pressure is needed to compress it to 100 cm 3 at this temperature? Solution According to Boyle’s law p 2V2 = p1V1 −3  V1   5  1 × 10  p 2 = p1   = (1.013 × 10 ) −6   V2   100 × 10  p 2 = 1.013 × 10 4 Pa or N / m 2 = 10 atm

CH 02 THERMAL PROPERTIES OF MATTER 20 Problem 2-3 A bubble rises from the bottom of a tall open tank of water that is at uniform temperature. Just before it bursts at the surface the bubble is three times its original volume. Determine the absolute pressure at the bottom of the tank. Solution According to Boyle’s law p 2V2 = p1V1

 3 V0  V   p 2 = p1  1  = (1.013 × 10 5 )  V2   V0  p2 = 3.039 ×105 Pa or N / m 2

CH 02 THERMAL PROPERTIES OF MATTER

21

2-2 CHARLES’S LAW Problem 2-4 To what temperature must a sample of a perfect gas of volume 500 cm 3 be cooled from 35 0 C in order to reduce its volume to 150 cm 3 ? Solution According to Charles’s law Vi V f = Ti T f

Vf   150  T f = Ti   = (35 + 273)   500   Vi  T f = 92.4 K = −180.60 C Problem 2-5 A given mass of an ideal gas occupies 50 mL at 20 0 C . If its pressure is held constant, what volume does it occupy at a temperature of 50 0 C ? Solution According to Charles’s law Vi V f = Ti T f

 Tf V f = Vi   Ti

 50 + 273   = (50)  = 55.1 mL + 20 273   

CH 02 THERMAL PROPERTIES OF MATTER

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2-3 GAY-LUSSAC’S LAW Problem 2-6 A container of helium gas at STP is sealed and then raised to a temperature of 730 K . What will be its new pressure? Solution According to Gay-Lussac’s law p 2 p1 = = const. T2 T1

Now

T  p2 = p1  2   T1  p1 = 1 atm = 1.013 × 10 3 Pa T1 = 0 0 C = 273 K T2 = 730 K

Therefore  730  5 p 2 = (1.013 × 10 5 )  = 2.709 × 10 Pa  273 

Problem 2-7 An aerosol can of whipped cream is pressurized at 440 kPa when it is refrigerated at 3 0 C . The can warms against temperature in excess of 50 0 C . What is the maximum safe pressure of the can? Solution Under these conditions p 2 p1 = = const. (Gay-Lussac’s Law) T2 T1

T   50 + 273  p 2 = p1  2  = (440 × 10 3 )   3 + 273   T1  p2 = 5.149 × 105 Pa

CH 02 THERMAL PROPERTIES OF MATTER 23 Problem 2-8 One mole of oxygen gas is at a pressure of 6 atm and a temperature of 27 0 C . (a) If the gas is heated at constant volume until the pressure triples, what is the final temperature? (b) If the gas is heated until both the pressure and volume are doubled, what is the final temperature? Solution p1 p 21 (a) Now = (Gay-Lussac’s Law) T1 T2

(b) Now

p  T2 =  2  T1 = (3)(27 + 273) = 900 K  p1  p1V1 p 2V2 = (The General Gas Law) T1 T2

 p  V  T2 =  2   2  T1 = (2)(2)(27 + 273) = 1200 K  p1   V1  Problem 2-9 A sample of hydrogen gas was found to have a pressure of 125 kPa when the temperature was 23 0 C . What its pressure be expected to be when the temperature is 110 C ? Solution The relation between pressure and temperature at constant volume is given by pf p = i Tf Ti

 Tf   11 + 273  5 p f = pi   = (125 × 103 )  = 1.183 × 10 Pa  23 + 273   Ti 

CH 02 THERMAL PROPERTIES OF MATTER

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2-4 THE IDEAL GAS LAW Problem 2-10 If 3.00 m 3 of a gas initially at STP is placed under a pressure of 3.20 atm , the temperature of the gas rises to 38.0 0 C . What is the new volume? Solution According to general gas equation p1V1 p 2V2 = = const. T1 T2  pT   (1)(273 + 38)  3 V2 =  1 1  V1 =  (3.00) = 1.068 m  (3.20)(273)   p2T2 

Problem 2-11 Calculate the volume occupied by one mole of an ideal gas at STP conditions. Solution For an ideal gas p V =n R T

nRT p Now n = 1 mol , R = 8.314 J mol −1 K −1 T = 0 0 C = 273 K , p = 1 atm = 1.013 × 105 Pa Therefore (1)(8.314)(273) V = = 2.24 × 10 − 2 m 3 = 22.4 litres 5 1.013 × 10 Θ 1 m 3 = 1 × 10 3 litres Problem 2-12 A 2.5 mol sample of an ideal gas at 1.00 atm pressure has a temperature of 20 0 C . What is the volume occupied by the gas? V=

CH 02 THERMAL PROPERTIES OF MATTER Solution For an ideal gas p V =n R T

25

n RT p (2.5)(8.314)(20 + 273) V= = 0.060 m3 = 60 litres 1.013 × 105 Problem 2-13 Calculate the volume occupied by 5 moles of an ideal gas at 100 kPa and 300 K. Solution According to ideal gas law p V =n R T V=

V=

n R T (5)(8.314)(300) = = 0.125 m 3 3 p 100 × 10

Problem 2-14 The boiling point of helium at one atmosphere is 4.2 K . What is the volume occupied by the helium gas due to evaporation of 100 g of liquid helium at 1 atm pressure and a temperature of 300 K ? Solution According to ideal gas law p V = n R T

V=

n RT p

100 = 25 mol 4 R = 8.314 J mol −1 K −1 T = 300 K p = 1 atm = 1.013 × 105 Pa

Now

n=

Hence V =

(25)(8.314)(300) = 0.616 m 3 = 616 litres 5 1.013 × 10

CH 02 THERMAL PROPERTIES OF MATTER 26 Problem 2-15 Calculate the density of oxygen at S.T.P. using the ideal gas law. The mass of one mole of oxygen is 32 g. Solution According to ideal gas law p V =n R T

n R T (1)(8.314)(273) = = 2.241 × 10 −2 m 3 p 1.013 × 10 5 As mass of one mole of oxygen (O2 ) is equal to 2(16) = 32 g , therefore m = 32 g = 32 × 10 −3 kg = 3.2 × 10 −2 kg The desired density of oxygen at S.T.P. conditions is m 3.2 × 10 −2 = 1.428 kg / m 3 ρ= = −2 V 2.241 × 10 V=

Problem 2-16 For an ideal gas at temperature 300 K and 1 atm pressure, what are dimensions of a cube that contains 1000 particles of the gas? Solution Now pV =N k T

N k T (1000)(1.381 × 10 −23 )(300) = = 4.090 × 10 − 23 m 3 5 p 1.013 × 10 If ‘x’ is the side of the cube, then x3 = V x = V 1/ 3 = (4.090 × 10 −23 )1/ 3 = 3.445 ×10 −8 m ≅ 0.34 nm V=

Problem 2-17 What is the pressure of an ideal gas if 3.5 moles occupy 2.0 litres at a temperature of − 150 0 C ?

CH 02 THERMAL PROPERTIES OF MATTER Solution According to ideal gas law p V =n R T

27

n RT V (3.5)(8.314)(−150 + 273) p= = 1.79 ×106 Pa −3 (2.0 × 10 )

p=

Problem 2-18 A sample of 255 mg of neon occupies 3.0 litres at 122 K . Calculate the pressure of the gas. Solution Now p V =n R T

n RT V As 20.18 g of neon is equal to 1 mol, therefore 255 × 10 −3 mol = 1.264 × 10 − 2 mol = n 255 mg = 255 × 10 −3 g = 20.18 Hence (1.264 × 10 −2 )8.314)(122) p= = 4.274 × 10 3 Pa −3 3 × 10 4.274 × 103 p= atm = 4.219 × 10 − 2 atm 1.013 × 105 p=

Problem 2-19 The temperature in outer space is about 2.7 K and the matter there consists mainly of isolated hydrogen atom with a density of 0.3 atoms / m 3 . What is the pressure of this gas? Solution According to ideal gas law pV =N k T

CH 02 THERMAL PROPERTIES OF MATTER 28 N p =   k T = (0.3)(1.381× 10 − 23 )(2.7) = 1.119 × 10 − 23 Pa V  Problem 2-20 A storage tank at S.T.P. contains 18.5 kg of nitrogen ( N 2 ). (a) What is the volume of the tank? (b) What is the pressure if an additional 15.0 kg of nitrogen is added without changing the temperature? The mass of one mole of nitrogen is 28 g . Solution (a) According to ideal gas law m p V =n R T =  R T M  m RT V= p M (18.5)(8.314)(273) V= = 14.8 m3 (1.013 × 105 )(28 × 10 −3 )

m  (b) Now p V = n1 R T =  1  R T M  mRT p= 1 M V (18.5 + 15)(8.314)(273) p= = 1.835 × 105 Pa (28 ×10 −3 )(14.8)

1.835 × 10 5 p= atm = 1.811 atm 1.013 × 10 5 Problem 2-21 Estimate the mass of air in a room whose dimensions are 8 m × 5 m × 4 m high at S.T.P. The mass of one mole of air is 29 g . Solution According to ideal gas law

CH 02 THERMAL PROPERTIES OF MATTER 29 m p V =n R T =  R T M  pV M m= RT (1.013 × 10 5 )(8 × 5 × 4)(29 × 10 −3 ) m= = 207 kg (8.314)(273) Problem 2-22 Consider an ideal gas at 20 0 C and a pressure of 2.00 MPa in a 1.00 × 10 −2 m 3 tank. Determine the gram moles of gas present. Solution According to ideal gas law p V =n R T

p V (2.00 × 10 6 )(1.00 × 10 −2 ) n= = = 8.21 moles (8.314)(20 + 273) RT Problem 2-23 An industrial firm supplies compressed air cylinders of volume 0.25 m 3 filled to a pressure of 20 MPa at 17 0 C . Calculate the contents of the cylinder in (a) moles (mol) (b) kilograms. The molar mass of air is 0.029 kg mol −1 . Solution (a) p V =n R T

pV (20 × 10 6 )(0.25) = = 2.074 × 10 3 mol R T (8.314)(17 + 273) (b) n = 2.074 × 10 3 mol n = (2.074 × 103 )(0.029) = 60 kg Problem 2-24 Calculate the number of moles in one m 3 of an ideal gas at 20 0 C and atmospheric pressure. n=

CH 02 THERMAL PROPERTIES OF MATTER Solution According to ideal gas law p V =n R T

30

pV (1.013 × 10 5 )(1) = = 41.6 moles R T (8.314)(20 + 273) Problem 2-25 A sample of an ideal gas occupies a volume of 4 litres at 20 0 C with a pressure of 3 atm . How many moles are present in the sample? Solution According to ideal gas law p V =n R T n=

n=

Now

pV RT

p = 3 atm = 3 × (1.013 × 10 5 ) Pa = 3.039 × 10 5 Pa V = 4 litres = 4 × 10 −3 m 3 R = 8.314 J mol −1 K −1 T = 200 C = (20 + 273) K = 293 K

Hence n=

(3.039 ×105 )(4 × 10 −3 ) = 0.499 moles (8.314)(293)

Problem 2-26 A tyre is filled with air at 15 0 C to a gauge pressure of 220 kPa . If the tyre reaches a temperature of 38 0 C , what fraction of the original air must be removed if the original pressure of 220 kPa is to be maintained? Solution As volume and pressure remains same, therefore the fractional change in number of moles can be written as

CH 02 THERMAL PROPERTIES OF MATTER 31 ∆n ( p V / R T2 ) − ( p V / R T1 ) T1 − T2 = = n T2 ( p V / R T1 ) ∆n (15 + 273) − (38 + 273) = = −0.074 = −7.4% (38 + 273) n Hence 7.4 % of original air must be removed to maintain the same pressure i.e. 220 kPa in the tyre. Problem 2-27 A quantity of ideal gas at 12.0 0 C and a pressure of 108 kPa occupy a volume of 2.47 m 3 . (a) How many moles of the gas are present? (b) If the pressure is now raised to 316 kPa and the temperature is raised to 31.0 0 C , how much volume will the gas now occupy? Assume there are no leaks. Solution (a) According to ideal gas law p V =n R T

p V (108 × 10 3 )(2.47) = = 113 moles R T (8.314)(12 + 273) p1V1 = n R T1

n=

(b) Now

n R T1 p1 (113)(8.314)(31 + 273) V1 = = 0.904 m3 or 904 litres 316 ×103 Problem 2-28 A tank of volume 0.5 m 3 contains oxygen at an absolute pressure of 1.5 × 10 6 N / m 2 and at a temperature of 20 0 C . Assume that oxygen behaves like an ideal gas. (a) How many kilomoles of oxygen are there in the tank? (b) Find the pressure if the temperature is increased to 500 0 C . B.U. B.Sc. 2008S Solution (a) According to ideal gas law p V =n R T V1 =

CH 02 THERMAL PROPERTIES OF MATTER 32 pV (1.5 × 10 6 )(0.5) n= = R T (8.314)(20 + 273) n = 3.08 × 10 2 moles = 0.308 kilomoles p p (b) Now 2 = 1 = const. for constant V. T2 T1 T   500 + 273  6 p 2 =  2  p1 =  (1.5 × 10 )  20 + 273   T1  p 2 = 3.96 × 10 6 N / m 2

Problem 2-29 Oxygen gas having a volume of 1130 cm 3 at 42 0 C and a pressure of 101 kPa expands until its volume is 1530 cm 3 and its pressure is 106 kPa . Find (a) the number of moles of oxygen in the system and (b) its final temperature. Solution (a) For an ideal gas piVi = n R Ti

piVi (101× 103 )(1130 × 10 −6 ) n= = = 4.36 × 10− 2 mol R Ti (8.314)(42 + 273) pfVf p i Vi = (b) Now Ti Tf  pf   T f =  p  i 

Vf   Vi

  Ti 

 106   1530  Tf =    (42 + 273) = 447.6 K  101   1130 

Problem 2-30

CH 02 THERMAL PROPERTIES OF MATTER 33 Calculate the value of gas constant if one mole of an ideal gas at S.T.P. conditions occupies 22.41 litres . Solution According to ideal gas law p V =n R T

R=

p V (1.013 × 10 5 )(22.41× 10 −3 ) = = 8.316 J mol −1 K −1 ( 1 )( 0 + 273 ) nT

Problem 2-31 Write the ideal gas law in terms of the density of the gas. Solution The ideal gas law is m p V =n R T =  R T M  where ‘m’ and ‘M’ are mass and molecular weight of the gas respectively. The above equation can be rewritten as m R T ρ R T p=  = M V  M

Problem 2-32 Calculate the density of oxygen at S.T.P. using the ideal gas. The molecular weight of oxygen is 32 g / mol . Solution According to ideal gas law ρ RT p= M pM ρ= RT (1.013 × 10 5 )(32 × 10 −3 ) ρ= = 1.428 kg / m 3 (8.314)(273)

Problem 2-33

CH 02 THERMAL PROPERTIES OF MATTER 34 How many molecules are in ideal gas sample at 350 K that occupies 8.5 litres when the pressure is 180 kPa ? Solution According to gas law pV =N k T

pV k T (180 × 10 3 )(8.5 × 10 −3 ) N= = 3.165 × 10 23 molecules − 23 (1.381 × 10 )(350) N=

Problem 2-34 An auditorium has dimensions 10.0 m × 20 m × 30 m . How many molecules of air fill the auditorium at 20.0 0 C and a pressure of 1 atm ? Solution According to ideal gas law N N p V =n R T = RT Θ n= NA NA

p V NA RT (1.013 × 10 5 )(10.0 × 20.0 × 30.0)(6.022 × 10 23 ) N= (8.314)(20 + 273) N = 1.50 × 10 29 molecules N=

Problem 2-35 A sealed flask of volume 80 cm 3 contains argon gas at a pressure of 10 kPa and a temperature of 27 0 C . Calculate the number of molecules of argon gas in the vessel. Solution According to ideal gas law pV =N k T

CH 02 THERMAL PROPERTIES OF MATTER 35 pV N= k T (10 × 10 3 )(80 × 10 −6 ) N= = 1.93 × 10 20 molecules − 23 (1.381 × 10 )(27 + 273) Problem 2-36 A certain vacuum pump is capable of reading the gas pressure in a sealed container is 0.001 × standard pressure if the temperature is maintained at 300 K . Calculate the number of molecules per m 3 in the container at this pressure and temperature. Solution According to ideal gas law pV =N k T

N p 0.001 × (1.013 × 10 5 ) = = V k T (1.381 × 10 − 23 )(300) N = 2.45 × 10 22 molecules / m3 V Problem 2-37 Calculate the number of molecules / m3 in an ideal gas at S.T.P. Solution According to ideal gas law pV =N k T

N p (1.013 × 10 5 ) = = = 2.687 × 10 25 molecules / m 3 V k T (1.381 × 10 − 23 )(300) Problem 2-38 The best vacuum attainable in the laboratory is about 5.0 × 10 −18 Pa at 293 K . How many molecules are there per m 3 in such a vacuum? Solution N N pV =n R T = RT Θ n= NA NA

CH 02 THERMAL PROPERTIES OF MATTER 36 N p NA = V RT N (5.0 ×10 −18 )(6.022 ×10 23 ) = = 1236 molecules / m 3 V (8.314)(293) Problem 2-39 One mole of helium gas is at room temperature ( 300 K ) and one atmospheric pressure. Calculate the number of helium atoms per unit volume. Solution For an ideal gas N N pV =n R T = RT Θ n= NA NA

N p NA = V RT N (1.013 × 10 5 )(6.022 × 10 23 ) = = 2.45 × 10 25 atoms / m 3 V (8.314)(300) Problem 2-40 A measured amount of heat is added to 2 × 10 −3 mol of a particular ideal gas to change its volume from 63.0 cm 3 to 113.0 cm 3 at constant pressure of 1 atmosphere . Calculate the change in temperature. P.U. B.Sc. 2006 Solution For an ideal gas at constant pressure p V1 = n R T1 (1) p V2 = n R T2 (2) Subtract Eq.(1) from Eq.(2) p (V2 − V1 ) = n R (T2 − T1 ) p (V2 − V1 ) n R 5 (1× 1.013 × 10 ){(113 − 63) × 10 −6 } ∆T = = 305 K (2 × 10−3 )(8.314) ∆T = T2 − T1 =

CH 02 THERMAL PROPERTIES OF MATTER

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2- 5 THE INTERNAL ENERGY OF AN IDEAL GAS Problem 2-41 k T = 1.00 eV holds? At what temperature the relation 2 Solution k T Now = 1.00 eV = 1.602 × 10 −19 J 2 Θ 1.00 eV = 1.602 × 10 −19 J (1.381 × 10 −23 )T = 1.602 × 10 −19 2 2(1.602 × 10 −19 ) T= = 2.32 × 10 4 K − 23 1.381 × 10 Problem 2-42 What is the average kinetic energy of a molecule of a gas at 300 K ? Solution The average kinetic energy of a molecule is given by 3 = k T 2 3 < K > = (1.381 × 10 − 23 )(300) = 6.215 × 10 − 21 J 2 Problem 2-43 What is the total random kinetic energy (in Joules) of the molecules in one mole of a gas at temperature 27 0 C . Solution The total random kinetic energy of one mole of a monoatomic gas is given by 3 E= RT 2 3 E = (8.314)(27 + 273) = 3741 J 2

CH 02 THERMAL PROPERTIES OF MATTER 38 Problem 2-44 Find the average translational kinetic energy of individual nitrogen molecule at 1600 K , in electron volts. P.U. B.Sc. 2004 Solution The average translational kinetic energy of a nitrogen molecule is given by 3 = k T 2 3 < K > = (1.381 × 10 − 23 )(1600) = 3.3144 × 10 − 20 J 2 3.3144 × 10 −20 = = 0.207 eV 1.602 × 10 −19 Θ 1 eV = 1.602 × 10 −19 J Problem 2-45 What is the total translational kinetic energy of 2 mol of O2 molecule at 20 0 C ? Solution The average translational kinetic energy of a molecule of a gas is given by 3 = k T 2 3 < K > = (1.381 × 10 − 23 )(20 + 273) = 6.069 × 10 − 21 J 2 As there are 2 N A molecules in 2 mol of O2 , therefore total translational kinetic energy is K TOTAL = 2 N A < K > = 2(6.022 × 10 23 )(6.069 × 10 −21 ) K TOTAL = 7310 J Problem 2-46 The mean kinetic energy of hydrogen molecule at 0 0 C is 5.62 × 10 −21 J . Calculate the value of Avogadro’s number.

CH 02 THERMAL PROPERTIES OF MATTER Solution The mean kinetic energy is given by

39

3 3 R   T k T =  2 2  N A  3 RT 3(8.314)(273) = = 6.058 ×10 23 NA = − 21 2 < K > 2(5.62 × 10 )

=

Problem 2-47 Calculate the total kinetic energy of the molecules of one gram of helium gas at 0 0 C . Solution The mean kinetic energy of a gas molecule is given by 3 = k T 2 As molar mass of helium is 4 g / mole, therefore the number of molecules in one gram of helium gas will be N N= A 4 Hence the total kinetic energy of molecules of the given gas sample will be N 3  K TOTAL = N < K > = A  k T  4 2  3 3 K TOTAL = (k N A ) T = R T 8 8 3 K TOTAL = (8.314)(273) = 851 J 8 Problem 2-48 Calculate the total rotational kinetic energy of all the molecules in one mole of air at 25 0 C . Solution The internal energy of air molecules has rotational, vibrational and translational kinetic energies, therefore on average the

CH 02 THERMAL PROPERTIES OF MATTER 40 1 E INTERNAL at a given rotational kinetic energy will be 3 temperature. Hence for polyatomic gas 1 ( K .E.) ROTATIONAL = E INTERNAL 3 1 ( K .E.) ROTATIONAL = (8.314)(25 + 273) = 2478 J 3 Problem 2-49 Calculate the internal energy of one mole of an ideal gas at 250 0 C . Solution The internal energy of monoatomic gas is given by 3 3 E INTERNAL = N k T = n R T 2 2 3 E INTERNAL = (1)(8.314)(250 + 273) = 6522 J 2 Problem 2-50 Calculate the internal energy of an ideal gas of volume 3.4 × 10 −4 m 3 when its pressure is 100 kPa . The internal energy of monoatomic gas is given by 3 3 E INTERNAL = N k T = n R T 2 2 3 E INTERNAL = pV Θ p V =n R T 2 3 E INTERNAL = (100 × 103 )(3.4 × 10 − 4 ) = 51 J 2

CH 02 THERMAL PROPERTIES OF MATTER 41 Problem 2-51 What is the internal energy of 4.50 mol of an ideal diatomic gas at 600 K , assuming that all degrees of freedom are active? Solution A diatomic molecule free to translate, rotate and vibrate will have seven degrees of freedom. The internal energy of such diatomic gas will be 7 E INTERNAL = n R T 2 7 E INTERNAL = (4.50)(8.314)(600) = 7.857 × 10 4 J 2 Problem 2-52 Certain excited state of hydrogen atom is found to have energy of 1.632 × 10 −18 J above the lowest (ground) state. At what temperature would the average translational kinetic energy be equal to the energy of the excited state? (Given that k = 8.6 × 10 −5 eV / K ) P.U. B.Sc. 2007 Solution The average translational kinetic energy is given by 3 = k T 2 −18 Now K = 1.632 × 10 J k = 8.6 × 10 −5 eV / K = (8.6 × 10 −5 ) × (1.602 × 10 −19 ) J / K k = 1.378 × 10 −23 J / K Hence 3 1.632 × 10 −18 = (1.378 × 10 − 23 ) T 2 2(1.632 × 10 −18 ) T= = 7.9 × 10 4 K − 23 3(1.378 × 10 )

CH 02 THERMAL PROPERTIES OF MATTER 42 Example 2-53 A mole of an ideal at 300 K is subjected to a pressure of 10 4 Pa and its volume is 0.025 cm 3 . Calculate (a) the molar gas constant R (b) the Boltzmann constant k and (c) the average translational kinetic energy of a molecule of the gas. Solution (a) According to ideal gas law p V =n R T

p V (10 5 )(0.025) 25 = = = 8.33 J mol −1 K −1 (1)300) 3 nT (b) The Boltzmann constant is related to gas constant as R = k NA R (25 / 3) k= = = 1.38 × 10 − 23 J K −1 N A 6.022 × 10 23 (c) The average translational kinetic energy of gas molecules is given by 3 < K >= k T 2 3 < K >= (1.38 × 10 − 23 )(300) = 6.21×10 − 21 J 2 R=

CH 02 THERMAL PROPERTIES OF MATTER

43

2-6 THE KINETIC THEORY OF GASES Problem 2-54 Twelve molecules have the following speeds, given in arbitrary units: 6,2,4,6,0,4,1,8,5,3,7 and 8. Calculate (a) the mean speed and (b) the r.m.s. speed. Solution ∑v = 6 + 2 + 4 + 6 + 0 + 4 +1+ 8 + 5 + 3 + 7 + 8 (a) v mean = N 12 54 = 4 .5 m / s vmean = 12

(b) v rms = v rms =

∑v

2

N

6 2 + 2 2 + 4 2 + 6 2 + 0 2 + 4 2 + 12 + 8 2 + 5 2 + 3 2 + 7 2 + 8 2 12 v rms =

320 = 5.16 m / s 12

Problem 2-55 The speeds of ten particles in m/s are 0, 1.0, 2.0, 3.0, 3.0, 3.0, 4.0, 4.0, 5.0 and 6.0 Find (a) the average speed (b) the root-mean-square speed and (c) the most probable speed of these particles. K.U. B.Sc. 1999, 2006 Solution (a) The average speed is calculated as ∑v v average = N 0 + 1 .0 + 2 .0 + 3 .0 + 3 .0 + 3 .0 + 4 .0 + 4 .0 + 5 .0 + 6 .0 v average = 10

CH 02 THERMAL PROPERTIES OF MATTER 31 v average = = 3 .1 m / s 10 (b) The mean square speed is defined as

v rms =

∑v

44

2

N 1/ 2

 1 (0) 2 + (1.0) 2 + (2.0) 2 + (3.0) 2 + (3.0) 2 + (3.0) 2  v rms =    2 2 2 2  10 + (4.0) + (4.0) + (5.0) + (6.0) 125 v rms = = 3.54 m / s 10 (c) As the entry 3.0 has the highest frequency in the given data, therefore the most probable speed is v prob = 3.0 m / s Problem 2-56 Calculate the root-mean-square speed of hydrogen molecule at 0 0 C and 1 atm pressure assuming hydrogen to be an ideal gas. Under these conditions for hydrogen ρ = 8.99 × 10 −2 kg / m 3 . P.U. B.Sc. 2000, 2009 Solution The root-mean square speed is defined as v rms =

3 p

ρ

p = 1 atm = 1.013 × 10 5 N / m 2 or Pa ρ = 8.99 ×10 −2 kg / m 3 Therefore

Now

v rms =

3(1.013 × 10 5 ) = 1839 m / s 8.99 × 10 − 2

Problem 2-57 At 44.0 0 C and 1.23 × 10 −2 atm the density of a gas is 1.32 × 10 −5 g / cm 3 . Find v rms for the gas molecule. P.U. B.Sc. 2001, 2008

CH 02 THERMAL PROPERTIES OF MATTER Solution The root-mean-square speed is defined as

v rms = Now

45

3 p

ρ

−2

p = 1.23 × 10 atm

p = (1.23 × 10 −2 )(1.013 × 105 ) N / m 2 or Pa = 1.24599 × 103 Pa ρ = 1.32 × 10 −5 g / cm 3

ρ=

(1.32 ×10 −5 )(10 −3 ) kg / m3 = 1.32 × 10 − 2 kg / m3 (10 − 2 ) 3

Hence 3(1.24599 × 10 3 ) = 532 m / s 1.32 × 10 − 2

v rms =

Problem 2-58 Calculate the root mean square velocity of Nitrogen at 0 0 C . Given that the density of Nitrogen at N.T.P. is 1.25 kg / m 3 . B.U. B.Sc. 2002A Solution The v rms is given by

v rms =

3 p

ρ

=

3(1.013 × 10 5 ) = 493 m / s 1.25

Problem 2-59 A cylindrical container of length 56.0 cm and diameter 12.5 cm holds 0.350 mol of nitrogen gas at a pressure of 2.05 atm . Find the r.m.s. speed of nitrogen molecule. Solution The root-mean-square speed of nitrogen molecule is defined as

v rms = Now

3 p

ρ

p = 2.05 atm = (2.05)(1.013 × 10 5 ) Pa

(1)

CH 02 THERMAL PROPERTIES OF MATTER 46 p = 2.077 × 105 Pa For calculation of density we proceed as under. 1 mol of nitrogen gas has mass = 28 g = 28 × 10 −3 kg

0.350 mol of nitrogen gas has mass = (0.350)(28 × 10 −3 ) kg m = 9.8 × 10 −3 kg The volume of the cylinder is given by V = π r 2 λ = π (D / 2) 2 λ 2

 12.5 × 10 −2   (56.0 × 10 − 2 ) m 3 = 6.872 × 10 −3 m 3 V = π  2   The density is defined as m 9.8 × 10 −3 ρ= = = 1.426 kg / m 3 −3 V 6.872 × 10 Substitute the values of p and ρ in Eq.(1)

v rms =

3(2.077 × 10 5 ) = 661 m / s 1.426

Problem 2-60 What is the average speed and root-mean-square speed of oxygen molecule at 300 K ? Molar mass of oxygen = 0.032 kg / mole . K.U. B.Sc. 2007 Solution The average speed is defined as

8 RT π M 8(8.314)(300) vaverage = = 445.5 m / s π (0.032) The root mean square velocity is defined as v average =

v rms =

3 RT M

CH 02 THERMAL PROPERTIES OF MATTER 47 3(8.314)(300) v rms = = 483.6 m / s 0.032 Problem 2-61 Root mean square velocity of the gas molecules (of certain density and pressure) if found to be 531 m s −1 at a temperature of 44 0 C . Find the molecular mass of the gas. P.U. B.Sc. 2005 Solution The root mean square velocity is defined as

3 RT M

v rms =

3 RT M 3 RT M= 2 v rms

2 v rms =

M =

3(8.314)(44 + 273) (531) 2

M = 2.80 × 10 −2 kg / mol = 28 g / mol Problem 2-62 The temperature in interstellar space is 2.7 K . Find the root-mean-square speed of hydrogen molecule at this temperature. The molar mass of hydrogen is 2 g . Solution The root-mean-square speed is defined as

3 RT M 3(8.314)(2.7) = = 183.5 m / s 2 × 10 −3

v rms = v rms

Problem 2-63

CH 02 THERMAL PROPERTIES OF MATTER 48 At what temperature do helium atoms have an r.m.s. speed equal to the escape speed from the surface of Earth ( v ESCAPE = 11.2 km / s )? Solution The root-mean-square speed is defined as

v rms = 2 = v rms

Now

3 RT M 3 RT M

2 v rms M T= 3 R v rms = v ESCAPE = 11.2 km / s

vrms = 11.2 ×103 m / s = 1.12 ×10 4 m / s

M = 4 g / mol = 4 × 10 −3 kg / mol R = 8.314 J mol −1 K −1 Therefore T=

(1.12 × 10 4 ) 2 (4 × 10 −3 ) = 2.01 × 10 4 K 3(8.314)

Problem 2-64 At what temperature, pressure remaining unchanged, will the speed of hydrogen molecules be double of its value at S.T.P.? Solution For a given mass of gas v rms ∝ T Let v1 and v 2 be the r.m.s. speeds of hydrogen molecules at T1 and T2 respectively, then

v1 ∝

T1

v 2 ∝ T2 Divide second relation by first relation

CH 02 THERMAL PROPERTIES OF MATTER v2 T2 = v1 T1 Substituting the given values in above equation 2 v1 T2 = 273 v1

49

T2 273 T 4= 2 273 T2 = 4(273) = 1092 K 2=

Problem 2-65 Calculate the temperature of oxygen molecules to have the same root mean square speed as that of hydrogen molecules at 100 0 C . Solution The root-mean-square speed is defined as

v rms =

3 RT M

Now v rms of oxygen molecule at T1 = v rms of oxygen molecule at T2 3 R T1 = M1

3 R T2 M2

3 R T1 3 R T2 = M1 M2 M  T1 =  1  T2  M2  As the molecular weights of oxygen and hydrogen are 32 and 2 respectively, therefore M 1 32 = = 16 M2 2

CH 02 THERMAL PROPERTIES OF MATTER 50 and T2 = −100 0 C = (−100 + 273) K = 173 K Hence T1 = (16)(173) = 2768 K Problem 2-66 At what temperature do atoms of helium have an r.m.s. speed equal to 1.00 % of the speed of light? Solution The root-mean-square speed is defined as

v rms = 2 = v rms

3 RT M 3 RT M

2 v rms M T= 3 R

Now v rms = 1.00% of c = 0.01 c v rms = (0.01)(2.998 × 10 8 ) m / s = 2.998 × 10 6 m / s

M = 4 g / mol = 4 × 10 −3 kg / mol , R = 8.314 J mol −1 K −1 Therefore (2.998 × 10 6 ) 2 (4 × 10 −3 ) T= = 1.441 × 10 9 K 3(8.314) Problem 2-67 To increase the r.m.s. speed of a gas by 1 %, by what percentage must the temperature increase? Solution The root-mean-square speed is defined as 3 RT (1) M Let T + ∆T be the temperature at which the r.m.s. speed is v rms + 1% of v rms v rms =

= vrms + 0.01 vrms = 1.01 vrms

CH 02 THERMAL PROPERTIES OF MATTER Hence

51

3 R (T + ∆T ) (2) M Divide Eq.(2) by Eq.(1) T + ∆T 1.01 = T Square both sides of above equation T + ∆T ∆T 1.0201 = = 1+ T T ∆T = 1.0201 − 1 = 0.0201 T ∆T × 100 = 0.0201× 100 = 2.01 T The desired percentage increase in temperature is 2.01 %. Problem 2-68 At what temperature do the atoms of helium gas have same root mean square speed as molecules of hydrogen gas at 27 0 C ? The molar mass of helium is double that of hydrogen. K.U. B.Sc. 2002 Solution The root-mean-square speed is defined as 1.01 v rms =

3 RT M where M is the molar mass. Now v rms of helium gas atom = v rms of hydrogen gas molecule v rms =

3 R T1 = M1

T1 = M1

3 R T2 M2

T2 M2

T1 T = 2 M1 M 2

CH 02 THERMAL PROPERTIES OF MATTER M  T1 = T2  1   M2 

52

 2M 2   = 600 K = 327 0 C T1 = (27 + 273)   M2  Problem 2-69 Calculate the root-mean-square speed of smoke particles of mass 5.2 × 10 −14 g in air at 14 0 C and 1 atm pressure. Solution The root-mean-square speed is given by 3 k T m −23 k = 1.381 × 10 J / K T = 14 0 C = (14 + 273) K = 287 K m = 5.2 × 10−14 g = 5.2 × 10 −17 kg v rms =

Now

Therefore v rms =

3(1.381 × 10 −23 )(287) = 1.521 × 10 − 2 m / s −17 5.2 × 10

Problem 2-70 Calculate the root mean square speed of hydrogen at 127 0 C . The mass of hydrogen molecule is 3.34 × 10 −27 kg . Solution The root-mean-square speed is given by

3 k T m −23 k = 1.381 × 10 J / K T = 127 0 C = (127 + 273) K = 400 K m = 3.34 × 10 −27 kg v rms =

Now

Therefore v rms =

3(1.381 × 10 −23 )(400) = 2.23 × 10 3 m / s − 27 3.34 × 10

CH 02 THERMAL PROPERTIES OF MATTER

53

2-7 THE MEAN FREE PATH Problem 2-71 What is the average distance between nitrogen molecules at S.T.P.? Solution The ideal gas law can be used to calculate the molecular density (i.e. number of molecules per m3) for nitrogen at S.T.P. as under. pV =N k T N p = V k T N 1.013 × 10 5 = = 2.687 × 10 25 molecules / m 3 − 23 V (1.381 × 10 )(273) Assuming that nitrogen molecule has a spherical shape of diameter‘d’, then 3

V 4 4 d  π d3 3 = π r = π   = N 3 3 2 6 6V 6 d3 = = π N π (N /V ) 1/ 3

  6 d =  π ( N / V ) 

1/ 3

  6 = 4.142 × 10 −9 m = 4.142 nm d = 25  π (2.687 × 10 )  the desired distance between nitrogen molecules. Problem 2-72 The mean free path of CO2 molecules at S.T.P. is measured

to be about 5.6 × 10 −8 . Estimate the diameter of CO2 molecule. Solution The mean free path of a gas molecule is given by

CH 02 THERMAL PROPERTIES OF MATTER kT λ= 2 π d2p

d2 =

54

k T 2 π p λ 1/ 2

 k T  d =   2 π p λ 

1/ 2

  (1.381 × 10 − 23 )(273) −10 d =  = 3.868 × 10 m 5 −8  2 π (1.013 × 10 )(5.6 × 10 )  the desired diameter of a CO2 molecule. Problem 2-73 The molecular diameter of different kinds of gas molecules can be found experimentally by measuring the rates at which different diffuse into each other. For nitrogen, d = 3.15 × 10 −10 m has been reported. What are the mean free path and average rate of collision for nitrogen at room temperature ( 300 K ) and at atmospheric pressure? P.U. B.Sc. 2000, 2008 Solution The number of molecules per unit volume is given by N N p ρn = = = Θ pV=N k T V ( N k T / p) k T

1.013 × 10 5 = 2.445 × 10 25 molecules / m 3 (1.381 × 10 − 23 )(300) The mean free path is given by 1 λ= 2 π d 2 ρn 1 λ= = 9.278 × 10 −8 m −10 2 25 2 π (3.15 ×10 ) (2.445 × 10 ) The average rate of collision is given by

ρn =

CH 02 THERMAL PROPERTIES OF MATTER 55 v 1 3 RT Rate of collision = rms = λ λ M 1 3(8.314)(300) = −8 0.028 9.278 × 10 9 −1 = 5.572 ×10 s Problem 2-74 Calculate the mean free path of a gas if its diameter at S.T.P. is 2 Å. Solution The mean free path of a gas molecule is given by 1 λ= 2 π d 2 ρn

λ= λ=

kT

p k T

Θ ρn =

2

2π d p

(1.381 × 10 −23 )(273) 2 π (2 × 10

−10 2

5

) (1.013 × 10 )

= 2.09 × 10 −7 m

Problem 2-75 Calculate the diameter of benzene molecule if there are 2.79 × 10 25 molecules / m 3 and mean free path for benzene is 2.2 × 10 −8 m . Solution The mean free path is given by 1 λ= 2 π d 2 ρn 1 d2 = 2 π ρnλ 1 d= ( 2 π ρ n λ )1 / 2

CH 02 THERMAL PROPERTIES OF MATTER 56 1 d= 1/ 2 2 π (2.79 × 10 25 )(2.2 × 10 −8 ) d = 6.06 × 10 −10 m = 6.06 Å the desired diameter of benzene molecule. Problem 2-76 A cubic box 1.20 m on a side is evacuated so the pressure of air inside is 10 −6 torr . Estimate how many molecular collisions there are per each collision with a wall ( 0 0 C ). The diameter of air molecule is 3 × 10 −10 m . Solution The mean free path of a gas molecule is given by kT λ= 2 π d2p

{

λ=

}

(1.381 × 10 −23 )(273) 2 π (3 × 10

−10 2

−6

2

= 70.7 m

) (10 × 1.333 × 10 ) Θ 1 torr = 1.333 × 10 2 N / m 2 The desired number of molecular collisions is a 1.20 = = 1.697 × 10 − 2 m −1 λ 70.7 Problem 2-77 At what frequency would the wavelength of sound be of the order of mean free path in oxygen at S.T.P.? Take the diameter of oxygen molecule to be 3 × 10 −10 m and speed of sound in oxygen equal to 330 m / s . Solution The number of molecules per unit volume is given by N N p = ρn = = Θ pV=N k T V ( N k T / p) k T 1.013 × 10 5 = 2.687 × 10 25 molecules / m 3 (1.381 × 10 − 23 )(273) The mean free path of a gas molecule is given by

ρn =

CH 02 THERMAL PROPERTIES OF MATTER 57 1 λ= 2 π d 2 ρn 1 = 9.305 × 10 −8 m λ= −10 2 25 2 π (3 × 10 ) (2.687 × 10 ) The desired frequency is v 330 f = = = 3.546 × 10 9 Hz λ 9.307 × 10 −8 Problem 2-78 At standard temperature and pressure (0 0C and 1.00 atm) the mean free path in helium gas is 285 nm. Determine (a) the number of molecules per cubic metre and (b) the effective diameter of the helium atoms. Solution (c) The number of molecules per unit volume is given by N N p = ρn = = Θ pV=N k T V ( N k T / p) k T

1.013 × 15 5 = 2.687 × 10 25 molecules / m 3 − 23 (1.381 × 10 )(273) (d) The mean free path is given by 1 λ= 2 π d 2 ρn 1 d2 = 2 π ρnλ 1 d= ( 2 π ρ n λ )1 / 2 1 d= 1/ 2 2 π (2.687 × 10 25 )(285 × 10 −9 ) d = 1.741 × 10 −10 m ≅ 0.174 nm

ρn =

{

}

CH 02 THERMAL PROPERTIES OF MATTER 58 ADDITIONAL PROBLEMS (1) Calculate the root-mean-square (r.m.s.) velocity of nitrogen molecule at S.T.P. conditions. The density of nitrogen at S.T.P. is 1.25 kg m −3 . B.U. B.Sc. 2002A (2) Calculate the root mean square speed of oxygen molecule at 27 0 C . The density of oxygen at S.T.P. is 1.43 kg m −3 . (3) Calculate the r.m.s. speed of carbon dioxide molecule at 27 0 C . The molar mass of carbon dioxide is 44 g . (4) Calculate the r.m.s. speed of the molecules of nitrogen gas at 10 0 C . The molar mass of nitrogen molecule is 0.028 kg . (5) Calculate the number of molecules in a flask of volume 500 cm 3 containing oxygen at a pressure of 200 kPa and a temperature of 27 0 C .

ANSWERS (1) 493 m / s (2) 483.5 m / s (3) 412 m / s (4) 502 m / s (5) 2.42 × 10 22 molecules