LMS Solutions Full eBook

CHAPTER 111 The Nature of Physical Chemistry and the Kinetic Theory of Gases

LAIDLER . MEISER . SANCTUARY

Physical Chemistry Electronic Edition

Publisher: MCH Multimedia Inc.

Problems and Solutions

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases Classical Mechanics and Thermal Equilibrium

1-2

Chapter 1

*problems with an asterisk are slightly more demanding

Classical Mechanics and Thermal Equilibrium

1.1. Calculate the amount of work required to accelerate a 1000-kg car (typical of a Honda Civic) to 88 km hr1 (55 miles hr1). Compare this value to the amount of work required for a 1600-kg car (typical of a Ford Taurus) under the same conditions.

Solution

1.2. Assume that a rod of copper is used to determine the temperature of some system. The rods length at 0 C is 27.5 cm, and at the temperature of the system it is 28.1 cm. What is the temperature of the system? The linear expansion of copper is given by an equation of the form lt = l0(1 + t + t2) where = 0.160 104 K1, = 0.10 107 K2, l0 is the length at 0 C, and lt is the length at t C.

Solution

1.3. Atoms can transfer kinetic energy in a collision. If an atom has a mass of 1 1024 g and travels with a velocity of 500 m s1, what is the maximum kinetic energy that can be transferred from the moving atom in a head-on elastic collision to the stationary atom of mass 1 1023 g?

Solution

1.4. Power is defined as the rate at which work is done. The unit of power is the watt (W = 1 J s1). What is the power that a man can expend if all his food consumption of 8000 kJ a day ( 2000 kcal) is his only source of energy and it is used entirely for work?

Solution

1.5. State whether the following properties are intensive or extensive: (a) mass; (b) density; (c) temperature; (d) gravitational field.

Solution

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases Gas Laws and Temperature

1-3

Gas Laws and Temperature

1.6. The mercury level in the left arm of the J-shaped tube in Fig. 1.6a is attached to a thermostat gas-containing bulb. The left arm is 10.83 cm and the right arm is 34.71 cm above the bottom of the manometer. If the barometric pressure reads 738.4 Torr, what is the pressure of the gas? Assume that temperature-induced changes in the reading of the barometer and J tube are small enough to neglect.

Solution

1.7. Vacuum technology has become increasingly more important in many scientific and industrial applications. The unit Torr, defined as 1/760 atm, is commonly used in the measurement of low pressures.

a. Find the relation between the older unit mmHg and the Torr. The density of mercury is 13.5951 g cm3 at 0.0 C. The standard acceleration of gravity is defined as 9.806 65 m s2.

b. Calculate at 298.15 K the number of molecules present in 1.00 m3 at 1.00 106 Torr and at 1.00 1015 Torr (approximately the best vacuum obtainable).

Solution

1.8. The standard atmosphere of pressure is the force per unit area exerted by a 760-mm column of mercury, the density of which is 13.595 11 g cm3 at 0 C. If the gravitational acceleration is 9.806 65 m s2, calculate the pressure of 1 atm in kPa.

Solution

1.9. Dibutyl phthalate is often used as a manometer fluid. Its density is 1.047 g cm3. What is the relationship between 1.000 mm in height of this fluid and the pressure in torr?

Solution

1.10. The volume of a vacuum manifold used to transfer gases is calibrated using Boyles law. A 0.251-dm3 flask at a pressure of 697 Torr is attached, and after system pumpdown, the manifold is at 10.4 Torr. The stopcock between the manifold and flask is opened and the system reaches an equilibrium pressure of 287 Torr. Assuming isothermal conditions, what is the volume of the manifold?

Solution

1.11. An ideal gas occupies a volume of 0.300 dm3 at a pressure of 1.80 105 Pa. What is the new volume of the gas maintained at the same temperature if the pressure is reduced to 1.15 105 Pa?

Solution


1-4

1.12. If the gas in Problem 1.11 were initially at 330 K, what will be the final volume if the temperature were raised to 550 K at constant pressure?

Solution

1.13. Calculate the concentration in mol dm3 of an ideal gas at 298.15 K and at (a) 101.325 kPa (1 atm), and (b) 1.00 104 Pa (= 109 atm). In each case, determine the number of molecules in 1.00 dm3.

Solution

*1.14. A J-shaped tube is filled with air at 760 Torr and 22 C. The long arm is closed off at the top and is 100.0 cm long; the short arm is 40.00 cm high. Mercury is poured through a funnel into the open end. When the mercury spills over the top of the short arm, what is the pressure on the trapped air? Let h be the length of mercury in the long arm.

Solution

1.15. A Dumas experiment to determine molar mass is conducted in which a gas samples P, , and V are determined. If a 1.08-g sample is held in 0.250 dm3 at 303 K and 101.3 kPa:

a. What would the samples volume be at 273.15 K, at constant pressure?

b. What is the molar mass of the sample?

Solution

1.16. A gas that behaves ideally has a density of 1.92 g dm3 at 150 kPa and 298 K. What is the molar mass of the sample?

Solution

1.17. The density of air at 101.325 kPa and 298.15 K is 1.159 g dm3. Assuming that air behaves as an ideal gas, calculate its molar mass.

Solution

1.18. A 0.200-dm3 sample of H2 is collected over water at a temperature of 298.15 K and at a pressure of 99.99 kPa. What is the pressure of hydrogen in the dry state at 298.15 K? The vapor pressure of water at 298.15 K is 3.17 kPa.

Solution


1-5

1.19. What are the mole fractions and partial pressures of each gas in a 2.50-L container into which 100.00 g of nitrogen and 100.00 g of carbon dioxide are added at 25 C? What is the total pressure?

Solution

1.20. The decomposition of KClO3 produces 27.8 cm3 of O2 collected over water at 27.5 C. The vapor pressure of water at this temperature is 27.5 Torr. If the barometer reads 751.4 Torr, find the volume the dry gas would occupy at 25.0 C and 1.00 bar.

Solution

1.21. Balloons now are used to move huge trees from their cutting place on mountain slopes to conventional transportation. Calculate the volume of a balloon needed if it is desired to have a lifting force of 1000 kg when the temperature is 290 K at 0.940 atm. The balloon is to be filled with helium. Assume that air is 80 mol % N2 and 20 mol % O2. Ignore the mass of the superstructure and propulsion engines of the balloon.

Solution

*1.22. A gas mixture containing 5 mol % butane and 95 mol % argon (such as is used in Geiger-Mller counter tubes) is to be prepared by allowing gaseous butane to fill an evacuated cylinder at 1 atm pressure. The 40.0-dm3 cylinder is then weighed. Calculate the mass of argon that gives the desired composition if the temperature is maintained at 25.0 C. Calculate the total pressure of the final mixture. The molar mass of argon is 39.9 g mol1.

Solution

1.23. The gravitational constant g decreases by 0.010 m s2 km1 of altitude.

a. Modify the barometric equation to take this variation into account. Assume that the temperature remains constant.

b. Calculate the pressure of nitrogen at an altitude of 100 km assuming that sea-level pressure is exactly 1 atm and that the temperature of 298.15 K is constant.

Solution

1.24. Suppose that on another planet where the atmosphere is ammonia that the pressure on the surface, at h = 0, is 400 Torr at 250 K. Calculate the pressure of ammonia at a height of 8000 metres. The planet has the same g value as the earth.

Solution

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases Grahams Law, Molecular Collisions, and Kinetic Theory

1-6

1.25. Pilots are well aware that in the lower part of the atmosphere the temperature decreases linearly with altitude. This dependency may be written as T = T0 az, where a is a proportionality constant, z is the altitude, and T0 and T are the temperatures at ground level and at altitude z , respectively. Derive an expression for the barometric equation that takes this into account. Work to a form involving ln P/P0.

Solution

1.26. An ideal gas thermometer and a mercury thermometer are calibrated at 0 C and at 100 C. The thermal expansion coefficient for mercury is

04 9 10 2

1 ( / )

1.817 10 5.90 10 3.45 10

PV TV

=

= + +

where is the value of the Celsius temperature and V0 = V at = 0. What temperature would appear on the mercury scale when the ideal gas scale reads 50 C?

Solution

Grahams Law, Molecular Collisions, and Kinetic Theory

1.27. It takes gas A 2.3 times as long to effuse through an orifice as the same amount of nitrogen. What is the molar mass of gas A?

Solution

1.28. Exactly 1 dm3 of nitrogen, under a pressure of 1 bar, takes 5.80 minutes to effuse through an orifice. How long will it take for helium to effuse under the same conditions?

Solution

1.29. What is the total kinetic energy of 0.50 mol of an ideal monatomic gas confined to 8.0 dm3 at 200 kPa?

Solution

1.30. Nitrogen gas is maintained at 152 kPa in a 2.00-dm3 vessel at 298.15 K. If its molar mass is 28.0134 g mol1 calculate:

a. The amount of N2 present.

b. The number of molecules present.

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases Grahams Law, Molecular Collisions, and Kinetic Theory

1-7

c. The root-mean-square speed of the molecules.

d. The average translational kinetic energy of each molecule.

e. The total translational kinetic energy in the system.

Solution

1.31. By what factor are the root-mean-square speeds changed if a gas is heated from 300 K to 400 K?

Solution

*1.32. The collision diameter of N2 is 3.74 1010 m at 298.15 K and 101.325 kPa. Its average speed is 474.6 m s1. Calculate the mean free path, the average number of collisions ZA experienced by one molecule in unit time, and the average number of collisions ZAA per unit volume per unit time for N2.

Solution

*1.33. Express the mean free path of a gas in terms of the variables pressure and temperature, which are more easily measured than the volume.

Solution

1.34. Calculate ZA and ZAA for argon at 25 C and a pressure of 1.00 bar using d = 3.84 1010m obtained from X-ray crystallographic measurements.

Solution

1.35. Calculate the mean free path of Ar at 20 C and 1.00 bar. The collision diameter d = 3.84 1010 m.

Solution

1.36. Hydrogen gas has a molecular collision diameter of 0.258 nm. Calculate the mean free path of hydrogen at 298.15 K and (a) 133.32 Pa, (b) 101.325 k Pa, and (c) 1.0 10 8 Pa.

Solution

1.37. In interstellar space it is estimated that atomic hydrogen exists at a concentration of one particle per cubic meter. If the collision diameter is 2.5 1010 m, calculate the mean free path . The temperature of interstellar space is 2.7 K.

Solution

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases Distributions of Speeds and Energies

1-8

*1.38. Calculate the value of Avogadros constant from a study made by Perrin [Ann. Chem. Phys., 18, 1(1909)] in which he measured as a function of height the distribution of bright yellow colloidal gamboge (a gum resin) particles suspended in water. Some data at 15 C are:

height, z/106 5 35 N, relative number of gamboge particles at height z 100 47 gamboge = 1.206 g cm3 water = 0.999 g cm-3 radius of gamboge particles, r = 0.212 106 m

(Hint: Consider the particles to be gas molecules in a column of air and that the number of particles is proportional to the pressure.)

Solution

Distributions of Speeds and Energies

1.39. Refer to Table 1.3 (p. 32) and write expressions and values for (a) the ratio 2 /u u , and (b) the ratio /ump. Note that these ratios are independent of the mass and the temperature. How do the differences between them depend on these quantities?

Solution

1.40. The speed that a body of any mass must have to escape from the earth is 1.07 104 m s1. At what temperature would the average speed of (a) an H2 molecule, and (b) an O2 molecule be equal to this escape speed?

Solution

1.41. a. For H2 gas at 25 C, calculate the ratio of the fraction of molecules that have a speed 2u to the fraction that have the average speed . How does this ratio depend on the mass of the molecules and the temperature?

b. Calculate the ratio of the fraction of the molecules that have the average speed 100C at 100 C to the fraction that have the average speed 25C at 25 C. How does this ratio depend on the mass?

Solution

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases Real Gases

1-9

1.42. Suppose that two ideal gases are heated to different temperatures such that their pressures and vapor densities are the same. What is the relationship between their average molecular speeds?

Solution

1.43. a. If 25 C is the average speed of the molecules in a gas at 25 C, calculate the ratio of the fraction that will have the speed 25 C at 100 to the fraction that will have the same speed at 25 C.

b. Repeat this calculation for a speed of 10 25C.

Solution

1.44. On the basis of Eq. 1.80 with = 1/kBT, derive an expression for the fraction of molecules in a one-dimensional gas having speeds between ux and ux + dux. What is the most probable speed?

Solution

*1.45. Derive an expression for the fraction of molecules in a one-dimensional gas having energies between x and x xd + .Also, obtain

an expression for the average energy x .

Solution

*1.46. Derive an expression for the fraction of molecules in a two-dimensional gas having speeds between u and u + du. (Hint: Proceed by analogy with the derivation of Eq. 1.91.) Then obtain the expression for the fraction having energies between and d+ . What fraction will have energies in excess of *?

Solution

Real Gases

1.47. In Section 1.13 it was stated that the van der Waals constant b is approximately four times the volume occupied by the molecules themselves. Justify this relationship for a gas composed of spherical molecules.

Solution

1.48. Draw the van der Waals PV isotherm over the same range of P and V as in Figure 1.21 at 350 K and 450 K for Cl2 using the values in Table 1.4.

Solution


1-10

1.49. Compare the pressures predicted for 0.8 dm3 of Cl2 weighing 17.5 g at 273.15 K using (a) the ideal gas equation and (b) the van der Waals equation.

Solution

1.50. A particular mass of N2 occupies a volume of 1.00 L at 50 C and 800 bar. Determine the volume occupied by the same mass of N2 at 100 C and 200 bar using the compressibility factor for N2. At 50 C and 800 bar it is 1.95; at 100 C and 200 bar it is 1.10. Compare this value to that obtained from the ideal gas law.

Solution

1.51. A gas is found to obey the equation of state

RT aPV b V

=

where a and b are constants not equal to zero. Determine whether this gas has a critical point; if it does, express the critical constants in terms of a and b. If it does not, explain how you determined this and the implications for the statement of the problem.

Solution

1.52. Ethylene (C2H4) has a critical pressure of Pc = 61.659 atm and a critical temperature of Tc = 308.6 K. Calculate the molar volume of the gas at T = 97.2 C and 90.0 atm using Figure 1.22. Compare the value so found with that calculated from the ideal gas equation.

Solution

1.53. Assuming that methane is a perfectly spherical molecule, find the radius of one methane molecule using the value of b listed in Table 1.5.

Solution

1.54. Determine the Boyle temperature in terms of constants for the equation of state:

PVm = RT{1 + 8/57(P/Pc)(Tc/T)[1 4(Tc/T)2]}

R, Pc, and Tc are constants.

Solution


1-11

1.55. Establish the relationships between van der Waals parameters a and b and the virial coefficients B and C of Eq. 1.117 by performing the following steps:

a. Starting with Eq. 1.101, show that

1m mm m

PV V aRT V b RT V

=

.

b. Since Vm/(Vm b) = (1 b/Vm)1, and (1 x)1 = 1 + x + x2 + ,, expand (1 b/ Vm)1 to the quadratic term and substitute into the result of part (a).

c. Group terms containing the same power of Vm and compare to Eq. 1.117 for the case n = 1.

d. What is the expression for the Boyle temperature in terms of van der Waals parameters?

Solution

*1.56. Determine the Boyle temperature of a van der Waals gas in terms of the constants a, b, and R.

Solution

1.57. The critical temperature Tc of nitrous oxide (N2O) is 36.5 C, and its critical pressure Pc is 71.7 atm. Suppose that 1 mol of N2O is compressed to 54.0 atm at 356 K. Calculate the reduced temperature and pressure, and use Figure 1.22, interpolating as necessary, to estimate the volume occupied by 1 mol of the gas at 54.0 atm and 356 K.

Solution

1.58. At what temperature and pressure will H2 be in a corresponding state with CH4 at 500.0 K and 2.00 bar pressure? Given Tc = 33.2 K for H2, 190.6 K for CH4; Pc = 13.0 bar for H2, 46.0 bar for CH4.

Solution

*1.59. For the Dieterici equation, derive the relationship of a and b to the critical volume and temperature. [Hint: Remember that at the critical point (P/V)T = 0 and (2P/V2)T = 0.]

Solution


1-12

1.60. In Eq. 1.103 a cubic equation has to be solved in order to find the volume of a van der Waals gas. However, reasonably accurate estimates of volumes can be made by deriving an expression for the compression factor Z in terms of P from the result of the previous problem. One simply substitutes for the terms Vm on the right-hand side in terms of the ideal gas law expression Vm = RT/P. Derive this expression and use it to find the volume of CCl2F2 at 30.0 C and 5.00 bar pressure. What will be the molar volume computed using the ideal gas law under the same conditions?

Solution

*1.61. A general requirement of all equations of state for gases is that they reduce to the ideal gas equation (Eq. 1.28) in the limit of low pressures. Show that this is true for the van der Waals equation.

Solution

1.62. The van der Waals constants for C2H6 in the older literature are found to be

a = 5.49 atm L2 mol2 and b = 0.0638 L mol1

Express these constants in SI units (L = liter = dm3).

Solution

*1.63. Compare the values obtained for the pressure of 3.00 mol CO2 at 298.15 K held in a 8.25-dm3 bulb using the ideal gas, van der Waals, Dieterici, and Beattie-Bridgeman equations. For CO2 the Dieterici equation constants are

a = 0.462 Pa m6 mol2,

b = 4.63 105 m3 mol1

Solution

*1.64. A gas obeys the van der Waals equation with Pc = 3.040 106 Pa (= 30 atm) and Tc = 473 K. Calculate the value of the van der Waals constant b for this gas.

Solution

*1.65. Expand the Dieterici equation in powers of 1mV in order to cast it into the virial form. Find the second and third virial coefficients.

Then show that at low densities the Dieterici and van der Waals equations give essentially the same result for P.

Solution

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases Essay Questions

1-13

Essay Questions

1.66. In light of the van der Waals equation, explain the liquefaction of gases.

1.67. State the postulates of the kinetic molecular theory of gases.

1.68. Eq. 1.22 defines the ideal-gas thermometer. Describe how an actual measurement would be made using such a thermometer starting with a fixed quantity of gas at a pressure of 150 Torr.

Chapter 1: The Nature of Physical Chemistry and Kinetic Theory of Gases Solutions

1-14

Solutions

1.1. Calculate the amount of work required to accelerate a 1000-kg car (typical of a Honda Civic) to 88 km hr1 (55 miles hr1). Compare this value to the amount of work required for a 1600-kg car (typical of a Ford Taurus) under the same conditions.

Solution:

Given: Car 1 (Civic): 11000 kg, 88 km hrm Speed= =

Car 2 (Taurus): 11600 kg, 88 km hrm Speed= =

Required: work required for the acceleration of each vehicle

Any type of work can be resolved through dimensional analysis as the application of a force through a distance;

( )o

l

lw F l dl=

Recall that bodies in motion possess kinetic energy defined by; 212k

E mu= where u is the velocity of the moving body and m is its mass. It

is possible to determine the amount of work required for the acceleration of a moving body by applying Newtons Second Law to the work integral given above.

0 0 0

( ) ( ) ( )l t t

l t t

dw d dt dtdt

= = = lF l l F l F l u

Substitute; dm mdt

= =uF a

0 0

t u

t u

dw m dt m ddt

= = u u u u

1 00

2 21 0

1 1( )2 2

l

k klw d m m E E= = = F l l u u

Conversion of speed from km hr1 to m s-1:

Speed = 88 km hr1 km88h

1 h3600

3m10

s km -124.4 m s=


1-15

Using the equation for work derived from Newtons 2nd Law (Civic):

.1 0

0

2 2Civic 1 0

1 1( )2 2

l


-1 2 -1 2Civic

1 1(1000 kg)(24.4 m s ) (1000 kg)(0 m s )2 2

w =

Civic 297 680 Jw =

Civic 298 kJw =

The same method can be applied to the second vehicle (Taurus):

1 00

2 2Taurus 1 0

1 1( )2 2

l


-1 2 -1 2Taurus

1 1(1600 kg)(24.4 m s ) (1600 kg)(0 m s )2 2

w =

Taurus 476 288 Jw =

Taurus 476 kJw =

By comparing both values quantitatively, it is possible to see that the work required to accelerate a moving body is directly proportional to its mass.

Back to Problem 1.1 Back to Top


1-16

1.2. Assume that a rod of copper is used to determine the temperature of some system. The rods length at 0 C is 27.5 cm, and at the temperature of the system it is 28.1 cm. What is the temperature of the system? The linear expansion of copper is given by an equation of the form lt = l0(1 + t + t2) where = 0.160 104 K1, = 0.10 107 K2, l0 is the length at 0 C, and lt is the length at t C.

Solution:

Given: Copper Rod: 27.5cm, 0 Cl T= =

Copper Rod in System: 28.1cml =

Linear expansion of copper: 20 (1 )tl l t t = + + where = 0.160104 K1, = 0.10107 K2, l0 is the length at 0 C, and lt is the

length at t C

Required: temperature of the system when the rod length equals 28.1cm

Let us define the temperature as ut and make all of the appropriate substitutions into the equation for the linear expansion of copper (starting temperature at zero degrees):

20 (1 )tl l t t = + +

4 7 228.1 27.5(1 0.160 10 0.100 10 )u ut t = + +

Simplify and rearrange:

28.1 27.527.5

=4 7 2(1 0.160 10 0.100 10 )

27.5

u ut t + +

1.0218 1 1 = 4 7 20.160 10 0.100 10 1u ut t + +

4 7 20.0218 0.160 10 0.100 10u ut t = +

7 2 40 0.100 10 0.160 10 0.0218u ut t = +

This can be rearranged to:

7 2 40.100 10 0.160 10 0.0218 0x x + =


1-17

Where ux t= and the system can be solved using the quadratic equation:

2 42

b b acxa

=

( ) ( )( )( )

24 4 7

7

0.160 10 0.160 10 4 0.100 10 0.0218

2 0.100 10x

=

o879 Cx =

o879 Cut =



1-18

1.3. Atoms can transfer kinetic energy in a collision. If an atom has a mass of 1 1024 g and travels with a velocity of 500 m s1, what is the maximum kinetic energy that can be transferred from the moving atom in a head-on elastic collision to the stationary atom of mass 1 1023 g?

Solution:

Given: Atom 1: 241 1 10 gm= , -11 500 m su =

Atom 2: 232 1 10 gm= , -12 0 m su =

Required: Find Ek(max) that can be transferred from atom 1 to atom 2

It is important to note that during elastic collisions, no energy is lost to the internal motion of the bodies involved. This means that the sums of the kinetic energy in addition to the sums of momentum remain the same before and after the collision. Therefore, there is no potential energy change of interaction between the bodies in collision.

Momentum: p mu=

Kinetic Energy= 212k

E mu=

Conservation of Momentum: ' '1 1 2 2 1 1 2 2m u m u m u m u+ = + (1)

Conservation of Energy: 2 2 2' 2'1 1 2 2 1 1 2 21 1 1 12 2 2 2

m u m u m u m u+ = + (2)

Since 12 0 m su= , then we can simplify equation (1) to get:

1 1 2 2m u m u+' '

1 1 2 2m u m u= +

Rearrangement then gives:

'' 2 2

1 11

m uu um

= + '

' 2 21 1

1

m uu um

=

It is possible to substitute the above into equation (2) and solve for '2u ;


1-19

' 12

2

1

2

1

uu mm

=+

1

'2 23

24

2(500 m s )1 10 g11 10 g

u

=

+

' 12 90.9 m su

=

Now this value can be used to find the kinetic energy of atom 2 after the collision. Remember to use SI units by converting grams to kilograms;

'22 2

12k

E m u=

26 1 21 (1 10 kg)(90.9 m s )2k

E =

234.13 10 JkE=



1-20

1.4. Power is defined as the rate at which work is done. The unit of power is the watt (W = 1 J s1). What is the power that a man can expend if all his food consumption of 8000 kJ a day ( 2000 kcal) is his only source of energy and it is used entirely for work?

Solution:

Given: Daily food consumption= 8000 kJ ( 2000 kcal)

Required: Pone day

Remember that power is defined as the rate at which work can be done meaning that;

dwPdt

=

Since the mans entire caloric intake is going toward work, then we can say that;

38000 kJ 8000 10 Jdw = =

We are only considering the power exerted in a single day;

1 day 24 hrsdt = =

24 hrsdt = 60 min1hr

60 s1min

86 400 s =

Power is measured by the Watt unit and 1 Watt = 1J s-1 (remember SI units!)

38000 10 J86 400 s

P = 192.59 J s

92.6 WP =



1-21

1.5. State whether the following properties are intensive or extensive: (a) mass; (b) density; (c) temperature; (d) gravitational field.

Solution: Given: (a) mass (b) density (c) temperature (d) gravitational field Required: intensive or extensive? It is first important to define the terms intensive and extensive in the context of physical chemistry. Intensive properties (sometimes called bulk property) are considered to be physical properties of a system that do not depend on its size. This means that their value will not change when the quantity of the matter in the system becomes subdivided. Extensive properties are the physical properties of a system that DO depend on its size and content. The values of extensive properties change with system subdivision. In addition, the ratio of two intensive properties yields an extensive one. Now it is possible to classify the above properties: Mass is extensive as it is a measure of how much is present in the system Density is intensive **note: mass and volume are extensive Temperature is intensive Gravitational Field is intensive



1-22

1.6. The mercury level in the left arm of the J-shaped tube in Fig. 1.6a is attached to a thermostat gas-containing bulb. The left arm is 10.83 cm and the right arm is 34.71 cm above the bottom of the manometer. If the barometric pressure reads 738.4 Torr, what is the pressure of the gas? Assume that temperature-induced changes in the reading of the barometer and J tube are small enough to neglect.

Solution:

Given: left arm = 10.83 cm, right arm = 34.71 cm, barometric pressure = 738.4 Torr

Required: Pgas

First, we need to find the difference in heights between the two columns (left and right arms);

Right arm - Left arm = 34.71 cm -10.83 cm = 23.88 cm

It is important to note that since the arm is open to the atmosphere, this pressure must also be added to the barometric pressure.

1 mmHg = 1 Torr and therefore 23.88 cmHg = 238.8 Torr

The pressure of the gas is then found to be;

238.8 Torr + 738.4 Torr = 977.2 Torr

gas 977.2 TorrP =



1-23

1.7. Vacuum technology has become increasingly more important in many scientific and industrial applications. The unit Torr, defined as 1/760 atm, is commonly used in the measurement of low pressures.

a. Find the relation between the older unit mmHg and the Torr. The density of mercury is 13.5951 g cm3 at 0.0 C. The standard acceleration of gravity is defined as 9.806 65 m s2.

b. Calculate at 298.15 K the number of molecules present in 1.00 m3 at 1.00 106 Torr and at 1.00 1015 Torr (approximately the best vacuum obtainable).

Solution:

Given: Mercury: 313.5951 g cm , 0.0 CT = =

2 9.806 65 m sacceleration of gravity =

Required: (a) State the relationship between mmHg and Torr

(b) NA in V = 1.00 m3

a) We should first define the system as a column of mercury with a 1m2 cross-sectional area, 0.001 m in height, a volume of 0.001 m3. Since we already have the density of mercury it is possible to determine the mass;

mV

= m V =

-313.5951 kg mm = 30.001 m

Now for 1 mmHg in a column;

( )( )( )1 mmHg mass density acceleration of gravity=

31 mmHg 0.001m= ( ) -313.5951 kg m( )( )-29.806 65 m s -21 mmHg 0.1333 kg m s=

Now since 1 Torr = 1 mmHg and 1 Torr = 133.322 Pa then we can see that;


1-24

1 mmHg 133.322 387 4 Pa=

By definition, 1 atmosphere = 101 325 Pa and 1 Torr = 1/760 atm then;

11 Torr (101325 Pa) =133.322 368 4 Pa760

=

Therefore;

133.322 387 41 mmHg 1.000 00014 Torr133.322 368 4

= =

The Torr is now defined as 1 mmHg.

b) Calculate the number of molecules present in a volume of 1.00 m3:

T = 298.15 K, P1 = 1.00106 Torr and P2 = 1.001015 Torr

Using the ideal gas law: PV nRT= we define n as NnL

= and rearrange to get;

PV nRT=

NRTPVL

= where L is Avogadros number and N is the number of particles

L = 6.0221023 mol-1

And the number density is defined as N PLV RT

=

Remember to make the conversion for pressure! P1 = 1.00106 Torr;

1 1N PLV RT

=


1-25

6

1

1 10 TorrNV

=( ) 1atm760 Torr

18.3145 J K

1mol( ) 298.15 K( )1101325 Pa atm( ) 23 16.022 10 mol( )

16 31 3.24 10 mNV

=

16 31 3.24 10 mN

= ( ) 31.00 m( ) 163.24 10 particles= 16

1 3.24 10N =

P2 = 1.001015 Torr using the same method as outlined above;

2 2N P LV RT

=

15

2

1 10 TorrNV

=( ) 1atm760 Torr

18.3145 J K

1mol( ) 298.15 K( )1101325 Pa atm( ) 23 16.022 10 mol( )

7 32 3.24 10 mN

= ( ) 31.00 m( ) 73.24 10 particles= 7

2 3.24 10N =

This is still a substantial number!



1-26

1.8. The standard atmosphere of pressure is the force per unit area exerted by a 760-mm column of mercury, the density of which is 13.595 11 g cm3 at 0 C. If the gravitational acceleration is 9.806 65 m s2, calculate the pressure of 1 atm in kPa.

Solution:

Given: Mercury: 3 213.595 11 g cm , 0 C, 9.806 65 m sT acceleration of gravity = = =

Required: columnP (kPa)

Let us define the system as a column of mercury with a cross-sectional area of 1 m2, 0.760 m in height and a volume of 0.760 m3. Since we have the density, it is possible to find the mass of mercury occupying the column;

mV

=

313 595.1 kg mm V = = ( ) 30.760 m( ) 10 332 kgm =

Mass multiplied by the gravitational acceleration produces a force (or weight) F ma= according to Newtons Law of Motion. The columns weight on the unit area then gives a pressure;

column (density)(volume)(acceleration of gravity)P =

3column 13 595.1 kg mP

= ( ) 30.760 m( )( )29.806 65 m s 2

column 101325 kg m sP=

Since 1 Pa = 1kg m s-2 then the pressure is 101.325 kPa.

column 101.325 kPaP =



1-27

1.9. Dibutyl phthalate is often used as a manometer fluid. Its density is 1.047 g cm3. What is the relationship between 1.000 mm in height of this fluid and the pressure in torr?

Solution:

Given: Dibutyl phthalate: 31.047 g cm =

Required: The relationship between 1.000 mm of this manometer fluid and pressure (Torr)

When two different liquids are being compared at constant volume and temperature, it is important to note that their pressures will be proportional to their densities.

Therefore, it is possible to take the ratio of DBP and Hg densities in order to calculate the pressure associated with 1mm of DBP.

1mmDBPl mmHg

3DBP DBP DBP

3Hg Hg Hg

1.047 g cm13.595 g cm

P PP P

=

DBP

Hg

0.077PP

=

DBP 0.077 TorrP =

Thus, 1mm DBP is equivalent to 0.077 Torr using the fact that 1mmHg is equivalent to 1 Torr. We can also state that;

1

1 Torr 12.98 mm DBP0.077 Torr mm

=

1 mm DBP 0.077 Torr=



1-28

1.10. The volume of a vacuum manifold used to transfer gases is calibrated using Boyles law. A 0.251-dm3 flask at a pressure of 697 Torr is attached, and after system pumpdown, the manifold is at 10.4 mTorr. The stopcock between the manifold and flask is opened and the system reaches an equilibrium pressure of 287 Torr. Assuming isothermal conditions, what is the volume of the manifold?

Solution: Given: 3 1 1 pumpdown eq0.251 dm , 697 Torr , 10.4 mTorr, 287 TorrV P P P= = = = Required: manifoldV Since we are working under isothermal conditions, Boyles Law will apply. This law describes the product of pressure and volume for a closed system. In a closed system, the temperature and moles are constant, thus;

1 1 2 2PV PV= 1 1 pumpdown 2 eq 2 1( )PV P V P V V+ = +

( )( ) ( ) ( )3 32 2697 Torr 0.251 dm 0.0104 Torr V 287 Torr V 0.251 dm+ = + ( ) ( )3 32 2174.947 Torr dm 0.0104 Torr 287 Torr 72.037 Torr dmV V+ = +

Now the above can be simplified on both sides to obtain, ( )3 2102.91 Torr dm 286.9896 Torr V=

2 manifold102.91Torr V V= =

3dm286.9896 Torr

3manifold 0.359 dmV =



1-29

1.11. An ideal gas occupies a volume of 0.300 dm3 at a pressure of 1.80 105 Pa. What is the new volume of the gas maintained at the same temperature if the pressure is reduced to 1.15 105 Pa?

Solution:

Given: Ideal Gas: 3 5 1 10.300 dm , 1.80 10 PaV P= =

Required: 2V

In this particular situation, Boyles Law will apply. This law describes the product of pressure and volume for a closed system. In a closed system, the temperature and moles are constant, thus;

1 1 2 2PV PV=

Simply rearrange for the final volume (V2);

1 12

2

PVVP

=

5

2

1.80 10 PaV

=( )( )3

5

0.300 dm

1.15 10 Pa( )

32 0.470 dmV =



1-30

1.12. If the gas in Problem 1.11 were initially at 330 K, what will be the final volume if the temperature were raised to 550 K at constant pressure?

Solution:

Given: same gas as in problem 1.11: 31 0.300 dmV =

1 2330 K, 550 K T T= = (constant pressure)

Required: 2V

In this particular situation, Charles Law will apply. This law states that under constant pressure, the volume of an ideal gas will vary proportionately (by the same factor) with changes in temperature, thus;

1 2

1 2

V VT T

=

Simply rearrange for the final volume (V2);

1 22

1

V TVT

=

( )32

0.300 dm 500 KV =

( )300 K

32 0.500 dmV =



1-31

1.13. Calculate the concentration in mol dm3 of an ideal gas at 298.15 K and at (a) 101.325 kPa (1 atm), and (b) 1.00 104 Pa (= 109 atm). In each case, determine the number of molecules in 1.00 dm3.

Solution:

Given: Ideal Gas: ( ) ( )4 9 1 2298.15 K, 101.325 kPa 1 atm , 1.00 10 Pa 10 atmT P P= = =

Required: C (in mol dm3)

AN (in V = 1.00 dm3)

Knowing that concentration is equal to:

nCV

=

we can make the substitution into the Ideal Gas Law.

n PCV RT

= =

For pressure (a) and using the fact that m3 = J Pa-1:

11

PnCV RT

= =

5

1 1

1.013 25 10 Pa

8.3145 J KC

=( )1mol (298.15 K )

31 40.87 mol mC

=

Now convert units into mol dm-3:

33

1 3

40.87 mol m 0.0409 mol dm10

C

= =


1-32

31 0.0409 mol dmC

=

Number of molecules per unit volume;

( )( )30.0409 mol dm 0.0409 molL = ( )3 23 1 dm 6.022 10 molecules mol ( ) 22 3

12.46 10 molecules dmN =

31.00 dm222.46 10 molecules=

221 2.46 10 moleculesN =

For pressure (b) using the same method:

22

PnCV RT

= =

4

2 1

1.00 10 Pa

8.3145 J KC

=( )1mol (298.15 K )

8 32 4.03 10 mol mC

=

Now convert units into mol dm-3;

8 311 3

2 3

4.03 10 mol m 4.03 10 mol dm10

C

= =

11 32 4.03 10 mol dmC

=

Number of molecules per unit volume;

( )( )11 3 114.03 10 mol dm 4.03 10 molL = ( )3 23 1 dm 6.022 10 molecules mol ( )


1-33

13 3

22.43 10 molecules dmN =

31.00 dm132.43 10 molecules=

132 2.43 10 moleculesN =



1-34

1.14. A J-shaped tube is filled with air at 760 Torr and 22 C. The long arm is closed off at the top and is 100.0 cm long; the short arm is 40.00 cm high. Mercury is poured through a funnel into the open end. When the mercury spills over the top of the short arm, what is the pressure on the trapped air? Let h be the length of mercury in the long arm.

Solution: Given: J-Tube: 760 Torr, 22 C, long arm 100 cm, short arm 40 cm P T h l= = = = Required: P of trapped air The temperature is again held constant (same as in problems 1.10 and 1.11) so Boyles Law will apply;

1 1 2 2PV PV= We are given the initial pressure, so we can rearrange this equation to solve for 2P ;

1 12

2

PVPV

=

Since h, the height of the mercury column on the trapped air side (long arm) is proportional to the volume of a uniform tube then we can write;

12

100 cmHg(100 ) cmHgPP

h

=

where h is the final height in centimeters of mercury in the long arm. In the short arm;

2 140P h P= + Substituting this into the above equation in order to eliminate 2P gives;

11

100 cmHg40(100 ) cmHgPh P

h

+ =


1-35

Recall that 1mmHg = 1 Torr and we therefore can make the substitution for 1P ;

11

(100) 40(100 )P h P

h= +

( )1 1(100) (100 ) 40P h h P= + This can be expanded to obtain;

1(100)P 4000 100 100h P= +2

1 140h h Ph + 2 140 76 4000 0h h h + = 2 216 4000 0h h + =

Using the quadratic equation then yields:

195.5 cmHgh = or 20.5 cmHgh = The first value of h cannot be this large since the tube length is only 100 cm. Therefore,

20.5 cmHgh = is the correct value. The final pressure can now be found;

12

(100) 76 cmHg(100 cm(100 )PP

h=

)

(100 20.5) cm95.6 cmHg=

2 956 TorrP = Back to Problem 1.14 Back to Top


1-36

1.15. A Dumas experiment to determine molar mass is conducted in which a gas samples P, , and V are determined. If a 1.08-g sample is held in 0.250 dm3 at 303 K and 101.3 kPa:

a. What would the samples volume be at 273.15 K, at constant pressure? b. What is the molar mass of the sample?

Solution: Given: 3 1.08 g, 0.250 dm , 303 K, 101.3 kPam V T P= = = = Required: sample sample and V M Since we are working under constant pressure, Charles Law can be applied. This law states that

1 2

1 2

constantV VT T

= =

Solving for 2V , we obtain

1 22

1

V TVT

=

Remember that the initial temperature is T = 303 K so by making the appropriate substitutions we will have;

( )32

0.250 dm 273.15 KV =

( )303 K

32 0.225 dmV =

Now that we have the final volume, it is possible to find the molar mass according to the equation;

mRTMPV

=

Recall that in order to derive this equation we must start with the ideal gas law;


1-37

PV nRT= and mnM

= so we obtain

mPV RTM

=mRTMPV

=

( )3 11.08 10 kg (8.3145 J KM

=

1mol )(273.15 K)

3 3(101.3 10 Pa)(0.225 dm 3 3 3)(10 m dm )

-10.1076 kg molM = 1108 g molM = Back to Problem 1.15 Back to Top


1-38

1.16. A gas that behaves ideally has a density of 1.92 g dm3 at 150 kPa and 298 K. What is the molar mass of the sample?

Solution:

Given: Ideal Gas: = 1.92 g dm3, P = 150 kPa, T = 298 K

Required: sampleM

Starting with the Ideal Gas Law is it possible to make substitutions and rearrangements in order to solve for the molar mass.

PV nRT=

nRTPV

=

Now, using the fact that mnM

= we can make the next substitution;

m RTMP

V=

Since density is defined as mV

= then we can write;

RTPM

= and now solve for M RTMP

=

3 11.92 kg m 8.3145 J KRTMP

= =

1mol 298.15 K 3150 10 Pa

10.0317 kg molM =

131.7 g molM =



1-39

1.17. The density of air at 101.325 kPa and 298.15 K is 1.159 g dm3. Assuming that air behaves as an ideal gas, calculate its molar mass.

Solution:

Given: Air: 31.159 g dm , 298.15 K, 101.325 kPaT P = = =

Required: airM

Use the same method as the previous problem (1.16);

RTMP

=

3 11.159 kg m 8.3145 J KRTMP

= =

1mol 298.15 K 101325 Pa

10.0284 kg molM =

128.36 g molM =



1-40

1.18. A 0.200-dm3 sample of H2 is collected over water at a temperature of 298.15 K and at a pressure of 99.99 kPa. What is the pressure of hydrogen in the dry state at 298.15 K? The vapor pressure of water at 298.15 K is 3.17 kPa.

Solution:

Given: H2 (over water): 30.200 dm , 298.15 K, 99.99 kPatV T P= = =

Vapor pressure of water: 3.17 kPa at T = 298.15 K

Required: 2H

P in the dry state

This problem makes use of Daltons Law of Partial Pressures which states:

The total pressure observed for a mixture of gases is equal to the sum of the pressure that each individual gas would exert had it been alone occupying the container and at the same temperature.

Pi = xiPt

Partial pressure is defined as the total pressure multiplied by the mole fraction of a particular gas in the mixture. For this particular hydrogen/water system, we can then write;

2 2H H OtP P P= + and solve for the pressure of hydrogen;

2 2H O HtP P P =

2H99.99 kPa 3.17 kPaP =

2H96.82 kPaP =



1-41

1.19. What are the mole fractions and partial pressures of each gas in a 2.50-L container into which 100.00 g of nitrogen and 100.00 g of carbon dioxide are added at 25 C? What is the total pressure?

Solution:

Given: Container: o2.50 L, 25 CV T= =

Add 100.00 g of nitrogen and carbon dioxide

Required: , for each and i i tx P P

First find the amount of each gas in terms of moles because we are provided with their mass and can easily find their molar mass;

2N

100.00 gmnM

= 28.012 g 1 mol

2N3.5699 moln =

2CO

100.00 gmnM

= 44.010 g 1 mol

2CO2.2722 moln =

Now we can find the mole fractions associated with each gas using the individual and combined number of moles;

2

2

NN

tot

3.5699 molnxn

= (3.5699 2.2722) mol+

2N0.6111x =

2

2

COCO

tot

2.2722 molnxn

= (3.5699 2.2722) mol+

2CO0.3889x =


1-42

Knowing the container volume and temperature of the system, the partial pressures can be calculated using the ideal gas law;

2

2

NN

cont

3.5699 moln RTP

V=

( ) 18.3145 J K 1mol( ) 298.15 K( )32.50 dm

2N35.4 barP =

2

2

COCO

cont

2.2722 moln RTP

V=

( ) 18.3145 J K 1mol( ) 298.15 K( )32.50 dm

2CO22.5 barP =

The total pressure is now found using Daltons Law for Partial Pressures;

1 2 3t iP P P P P= + + + +

1 2 3t t t t i tP x P x P x P x P= + + + + (Eq. 1.53)

1 2 it

n RTn RT n RTPV V V

= + + +

1 2( )t iRTP n n nV

= + + + (Eq. 1.54)

Any of the above forms can be used but for simplicity, we shall use Eq. 1.54;

(3.5699 2.2722) moltP = +1(8.3145 J K 1mol )(298.15 K

3

)2.50 dm

57.9 bartP =

Notice, once you take the pressure, you need to divide by 102 in order to get the pressure in units bar.



1-43

1.20. The decomposition of KClO3 produces 27.8 cm3 of O2 collected over water at 27.5 C. The vapor pressure of water at this temperature is 27.5 Torr. If the barometer reads 751.4 Torr, find the volume the dry gas would occupy at 25.0 C and 1.00 bar.

Solution:

Given: KClO3: 2

3O 27.8 cm , 27.5 C V T= =

Vapor pressure of water: P = 27.5 Torr

Barometer reading: P = 751.4 Torr

Required: dry gasV

First it is possible to find the pressure of the dry gas at T = 27.5 C by making use of the barometer reading and the vapor pressure of water;

gas barometer waterP P P=

gas 751.4 Torr 27.5 TorrP =

gas 723.9 TorrP = (Remember that this is at 27.5 C)

Since there is also a temperature change the following equality should be used to find the final volume of the system;

1 1 2 2

1 2

PV PVT T

=

1 1 22

1 2

PV TVT P

=

Recall that 1 bar = 750.06 Torr. Also, when making temperature conversions between Celsius to Kelvin: 27.5 C = 273.15 + 27.5 = 300.65 K.

It is important to remember the initial conditions of the system (Dont mix up the temperatures!);

1 1 22

1 2

(723.9 TorrPV TVT P

= 3)(27.8 cm )(298.15 K )

(300.65 K )(750.06 Torr )


1-44

32 dry gas 26.6 cmV V= =



1-45

1.21. Balloons now are used to move huge trees from their cutting place on mountain slopes to conventional transportation. Calculate the volume of a balloon needed if it is desired to have a lifting force of 1000 kg when the temperature is 290 K at 0.940 atm. The balloon is to be filled with helium. Assume that air is 80 mol % N2 and 20 mol % O2. Ignore the mass of the superstructure and propulsion engines of the balloon.

Solution:

Given: Balloon lifting force: m = 1000 kg, T = 290 K, P = 0.940 atm

Required: balloonV

The lifting force comes from the difference between the mass of air displaced and the mass of the helium that replaces the air. We can work under the assumption that the molar mass for air is 28.8 g mol-1. This is true if we consider the fact that air (in the problem) is composed 80 percent of nitrogen and 20 percent of oxygen.

2

1N 14(2) 28 g molM

=

But we will only consider 80 percent and therefore;

2

1 1N 28 g mol (0.80) 22.4 g molM

= = (in air)

2

1O 16(2) 32 g molM

=

But we will only consider 20 percent and therefore;

2

1 1O 32 g mol (0.20) 6.4 g molM

= =

Lifting force air helium( ) 1000 kgV =

And recall that we can use the ideal gas law to solve for the density of a gas (density is mass divided by volume);

PV nRT=

m RTnRT MP RTV V M

= = = and solve for density


1-46

PMRT

=

air

0.940 atm =

( ) 1101325 Pa atm( ) 28.8 g 1mol( )18.3145 J K 1mol( ) 290 K( ) 310 g( )1 kg

3air

helium

1.138 kg m

0.940 atm

=

=( ) 1101325 Pa atm( ) 4.003 g 1mol( )

18.3145 J K 1mol( ) 290 K( ) 310 g( )13

helium

kg

0.158 kg m

=

Now these values can be substituted into the equation for the volume of the balloon;

balloonair helium

1000 kg( )

V

=

balloon

1000 kgV =

(1.138 0.158) kg 3m

3balloon 1021 mV =



1-47

1.22. A gas mixture containing 5 mol % butane and 95 mol % argon (such as is used in Geiger-Mller counter tubes) is to be prepared by allowing gaseous butane to fill an evacuated cylinder at 1 atm pressure. The 40.0-dm3 cylinder is then weighed. Calculate the mass of argon that gives the desired composition if the temperature is maintained at 25.0 C. Calculate the total pressure of the final mixture. The molar mass of argon is 39.9 g mol1.

Solution:

Given: Gas mixture: 5 mol % butane and 95 mol % argon

3 1cyl argon1 atm, 40.0 dm , 39.9 g molP V M= = =

Required: argonm and tP

By using the information given above, it is possible to find the mole fractions for each of the gases in the mixture;

PV nRT=

PVnRT

=

( )( )3butane 1

101325 Pa 40.0 dm

8.3145 J Kn

=( )1mol 298.15 K ( )

butane 1.63 moln =

Since the mixture contains 95 parts argon to 5 parts of butane, the ratio is then 95/5 = 19:1 and we can determine the number of moles for argon;

argon butane19 n n=

argon 19(1.63) mol 30.97 moln = =

Now that we have both the number of moles and molar mass, we can find the mass of argon;

argon argon argonm n M=


1-48

argon 30.97 molm = ( ) 139.9 g mol( )

argon 1236.7 gm =

The total pressure can then be found by taking the sum of the partial pressures;

butane argon( )tRTP n nV

= +

Remember that once you find the pressure, you must divide by 102 in order to convert to bar.

( )1.63 30.97 moltP = +18.3145 J K 1mol( ) 298.15 K( )

340.0 dm

20.2 bartP =

Since 1 bar = 0.986 92 atm, then we can say that;

20.2(0.986 92)tP =

19.9 atmtP =



1-49

1.23. The gravitational constant g decreases by 0.010 m s2 km1 of altitude.

a. Modify the barometric equation to take this variation into account. Assume that the temperature remains constant.

b. Calculate the pressure of nitrogen at an altitude of 100 km assuming that sea-level pressure is exactly 1 atm and that the temperature of 298.15 K is constant.

Solution:

Given: gravitational constant g decreases by 0.010 m s2 km1 of altitude

Required: (a) modify dP Mg dzP RT

=

(b) 2N

P at z = 100 km, P = 1 atm, T = 298.15 K

a) The standard gravitational acceleration is defined as 9.807 m s-2. If g were to decrease by 0.010 m s-2 per each kilometer in height, this would be equivalent to a change of:

0.010 m 23

s10 m

5 210 s z =

where z is the altitude. The new gravitational constant expression would be as follows:

2 5 29.807 m s 10 s g z =

This can then be substituted into the Barometric Distribution Law equation,

dP Mg dzP RT

= (Eq. 1.74)

To give:

( )2 5 29.807 m s 10 s dP M z dzP RT =

This can also be expressed in the following manner:


1-50

( )2 6 2 20

ln 9.807 m s 5 10 s P M z zP RT

=

b) The second version of this equation can then be used to calculate the pressure of nitrogen gas at an altitude of 100 km.

0

28.0 gln P

P

=

1mol( ) 3 110 kg g ( )18.3145 J K 1mol( ) 298.15 K( ) ( ) ( )( )

25 2 6 5 2 29.807 10 m s 5 10 10 s z z

0

ln 10.51PP

=

10.51 5

0

2.73 10P eP

= =

( )5 02.73 10 1 atmP P= =

52.73 10 atmP =



1-51

1.24. Suppose that on another planet where the atmosphere is ammonia that the pressure on the surface, at h = 0, is 400 Torr at 250 K. Calculate the pressure of ammonia at a height of 8000 metres. The planet has the same g value as the earth.

Solution:

Given: Planet with ammonia atmosphere: h = 0, P = 400 Torr, T = 250 K

Required: 3NH

P at h = 8000 m

We may begin as we did in the previous problem with the Barometric Distribution Law:

dP Mg dzP RT

= (Eq. 1.74)

We can then integrate this expression, with the boundary condition that P = P0 when z = 0, which yields;

0

ln P MgzP RT= (Eq. 1.75)

We can further manipulate the equation by exponentiating each side:

ln e0

MgzRTP e

P

= and solve for P

0

MgzRTP P e

=

Assume that the temperature remains constant at T = 250 K and the molar mass of ammonia is M = 0.017 kg mol-1. These values can be substituted into the above equation.

( 0.017)(9.807)(8000)(8.3145)(250)(400 Torr)P e

=

0.642(400 Torr)P e=

210 TorrP =



1-52

1.25. Pilots are well aware that in the lower part of the atmosphere the temperature decreases linearly with altitude. This dependency may be written as T = T0 az, where a is a proportionality constant, z is the altitude, and T0 and T are the temperatures at ground level and at altitude z, respectively. Derive an expression for the barometric equation that takes this into account. Work to a form involving ln (P/P0).

Solution:

Given: linear dependency of temperature on altitude: T = T0 az

Required: Derive an expression for the barometric equation taking linearity of temperature increase into account.

Beginning with the Barometric Distribution Law equation (Eq. 1.74), and substituting for T from the linear dependency of temperature on altitude equation;

dP Mg dzP RT

=

( )0dP Mg dzP R T az

=

This is a differential equation. In order to solve this, let ( )0x T az= then we have:

dx adz= and rearrangement gives dxdza

=

Integration of the expression is then as follows (and with the proper substitutions):

0

1 lno

x

x

dx xax a x

=

0

0

1 lno

x

x

Tdxax a T az

=


1-53

Integration of the LHS between the values 0P and P (with the final substitution) gives:

0

0 0

ln ln T azP MgP Ra T

=



1-54

1.26. An ideal gas thermometer and a mercury thermometer are calibrated at 0 C and at 100 C. The thermal expansion coefficient for mercury is:

04 9 10 2

1 ( / )

1.817 10 5.90 10 3.45 10

PV TV

=

= + +

where is the value of the Celsius temperature and V0 = V at = 0. What temperature would appear on the mercury scale when the ideal gas scale reads 50 C?

Solution:

Given: Thermometers: 1 20 C, 100 CT T= =

Thermal expansion coefficient for mercury: 4 9 10 20

1 ( / ) 1.817 10 5.90 10 3.45 10PV TV = = + +

Required: Hg when ideal gas scale reads 50 C

In the case of a mercury column, we assign its length the value 100l when it is at thermal equilibrium with boiling water vapor at 1atm pressure. The achievement of equilibrium with melting ice exposed to 1atm pressure establishes the length, 0l . Assuming a linear relationship between the temperature and the thermometric property (length) we can write;

( )( )

0

100 0

(100 C)l l

l l

=

(Eq. 1.15)

This expression can be tailored to the situation given above by;

( )( )

50 0 HgHg

100 0 Hg

(100 C)V V

V V

=

Since 0

1 ( / )PV TV = then we can integrate the expression with respect to to get;


1-55

50

50 0 00V V V d =

which can be evaluated as follows (divide the second and third terms by 2 and 3 respectively)

50 50504 9 2 10 300 0 0

1.817 10 2.95 10 1.15 10 V + +

00.009 107 V=

The same can be done for the denominator in the equation ( )( )

50 0 HgHg

100 0 Hg

(100 C)V V

V V

=

100 1001004 9 2 10 300 0 0

1.817 10 2.95 10 1.15 10x V + +

00.018 31V=

Now these two values can be substituted into the above equation to get;

0Hg

0

0.009 107 (100 C)0.018 31

VV

=

Hg 49.7 C =



1-56

1.27. It takes gas A 2.3 times as long to effuse through an orifice as the same amount of nitrogen. What is the molar mass of gas A?

Solution: Given: Gas A: teffusion = 2.3 times longer than nitrogen Required: Gas AM This particular problem makes use of Grahams Law of Effusion which states that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:

rate(gas 1) (gas 2) (gas 2) (gas 2)rate(gas 2) (gas 1) (gas 1) (gas 1)

t Mt M

= = =

Since we can easily determine the molar mass of nitrogen (N2), we can make the appropriate substitutions and solve for the molar mass of Gas A.

-1nitrogen 28 g molM =

nitrogenA

nitrogen A

(gas 2)(gas 1)

tv Mv t M

= =

1A

nitrogen A

1 28 g mol2.3

vv M

= = 22 1

A

1 28 g mol2.3 M

=

2 1

A

1 28 g mol2.3 M

=

1

A 228 g mol

12.3

M

=

2 1A 1.5 10 g molM

= Back to Problem 1.27 Back to Top


1-57

1.28. Exactly 1 dm3 of nitrogen, under a pressure of 1 bar, takes 5.80 minutes to effuse through an orifice. How long will it take for helium to effuse under the same conditions?

Solution: Given: 3nitrogen 1 dm , 1 bar, 5.8 minV P t= = = Required: Het Using Grahams Law of effusion (as in previous problem 1.27) recall that effusion time is inversely proportional to the rate of effusion.

2 2

He He2

N N

rate(N )rate(He)

t Mt M

= =

Rearrange the above equation to isolate for the wanted variable, Het ;

2

2

HeHe N

N

Mt tM

=

We can determine the molar masses of both helium and nitrogen to get; 1

He

4 g mol5.80mint

=128 g mol

He 2.19 mint = Back to Problem 1.28 Back to Top


1-58

1.29. What is the total kinetic energy of 0.50 mol of an ideal monatomic gas confined to 8.0 dm3 at 200 kPa?

Solution: Given: Ideal monatomic gas: 0.50 moln = , 38.0 dmV = , 200 kPaP = Required:

totkE

This particular problem refers to the section of Kinetic Theory of Gases. Here, we are trying to determine the relationship between 2u and T, the mechanical variable of u of Eq. 1.41:

2

3NmuP

V=

which is the fundamental equation derived from the simple kinetic theory of gases. For our purpose of determining this relationship (kinetic energy and temperature), Eq. 1.41 may be converted into another useful form by recognizing that the average kinetic energy per molecule is defined as;

212

k mu=

Substitution of this expression into Eq. 1.41 then gives;

1 223 3

k kPV N N= =

At constant pressure, the volume of a gas is proportional to the number of molecules and the average kinetic energy of the molecules. Since N nL= then we can write;

23

kPV nL= and since kL is the total kinetic energy per mole of gas, then

23 k

PV nE=

32k

PVEn

=


1-59

The data given above can be substituted into the above equation to yield;

33 (200 kPa)(8.0 dm )2 0.5 molk

E =

14800 J molkE=

So for half a mole, the kinetic energy will be:

2400 JknE = Back to Problem 1.29 Back to Top


1-60

1.30. Nitrogen gas is maintained at 152 kPa in a 2.00-dm3 vessel at 298.15 K. If its molar mass is 28.0134 g mol1 calculate:

a. The amount of N2 present. b. The number of molecules present. c. The root-mean-square speed of the molecules. d. The average translational kinetic energy of each molecule. e. The total translational kinetic energy in the system.

Solution: Given: Nitrogen: P = 152 kPa, V = 2.00 dm3, T = 298.15 K, M = 28.0134 g mol1 Required: see above a through e Using the ideal gas law, PV nRT= we can solve for the number of moles present.

PVnRT

=( ) 3152 000 Pa (2.00 dm

3 3 3)(10 m dm

1

)

(8.3145 J K 1mol )(298.15 K )

0.1226 moln = We can now use Avogadros number in order to find the number of molecules present;

( )( )number molecules number of moles L= 0.1226 molN nL= = 23 1(6.022 10 mol )

227.38 10N = We can take the square root of Eq. 1.43 in order to find the root mean square speed of the molecules;

2 3RTuM

= (Eq. 1.43)

2 3RTuM

=

2 3(8.3145 J Ku =1mol )(298.15 K

1

)0.028 013 4 kg mol

2 1515.2 m su =


1-61

The average translational energy (for each molecule) is given by Eq. 1.44;

212

k mu=10.0280134 kg mol1

2

( ) 1 2

23 1

(515.2 m s )

6.022 10 mol

216.175 10 Jk = It is possible to find the total translational kinetic energy in the system by using the equation;

tot

32k

E nRT= (Eq. 1.49)

tot

3 (0.1226 mol2k

E = 1)(8.3145 J K 1mol )(298.15 K)

tot456 JkE =



1-62

1.31. By what factor are the root-mean-square speeds changed if a gas is heated from 300 K to 400 K?

Solution: Given: T1 = 300 K, T2 = 400 K Required: change in root-mean-square speeds Recall Eq. 1.43

2 3RTuM

=

Remember that in problem 1.29 we outlined the relationship between 2u and T. Using this information, it is possible to see that the following ratios are equivalent;

22 2

211

u TTu

=

Now we can determine the magnitude of change in root-mean-square speed when moving from a lower to a higher temperature.

2

1

400 1.33300

TT

= =

2

1

1.15TT

=



1-63

1.32. The collision diameter of N2 is 3.74 1010 m at 298.15 K and 101.325 kPa. Its average speed is 474.6 m s1. Calculate the mean free path, the average number of collisions ZA experienced by one molecule in unit time, and the average number of collisions ZAA per unit volume per unit time for N2.

Solution:

Given: 1A-1

A0 298.15 K, 101.33.74 1 25 kP0 m a, 474.6 m s, T Pd u = = ==

Required: A AA, ,Z Z

The mean free path is given by Eq. 1.68;

22 A A

Vd N

Using the ideal gas law, PV nRT , and solving for V, a useful expression for the mean free path can be obtained;

nRTVP

Giving the mean free path as,

22 A A

nRTPd N

ANLn

, where AN is the number of particles

22 A

RTd LP


1-64

18.3145 J K

=1mol( ) 298.15 K( )

( )1 23 1202 6.022 13.74 10 0 mm ol ( )( )8 2 1

2 2 1 2

10 J m Pawhere 1 J kg m s and 1 Pa 1 kg m s

1

101325 Pa

6.53

kg1 Ja

7

=1 P

= =

=

2 2m s

1kg 1 2m s 3

8 2 36.537

1 m

10 m m

=

=

86.54 10 m =

The average number of collisions ZA experienced by one molecule in unit time, also known as the collision frequency for one molecule is given by Eq. 1.66;

2A 1A A

A2 (SI unit :s )d u NZ

V =

Using the ideal gas law PV nRT and solving for V, a useful expression for ZA can be obtained.

nRTVP

2AA A

A2 d u N PZ

nRT

=

ANLn



1-65

( ) ( )

2A

210

AA

1 23 1

A

3.74 10

2

2 474.6 m s 6.022 10 m lm o

d u LPZRT

Z

=

=

( )( )1

101325 Pa

8.3145 J K 1mol( ) 298.15 K( )

3 1

2 2 1 2

1A

where 1 J 1 kg m s and 1

7 259 759 289 m

Pa 1 kg

s P

a

m s

1kg1 PaJ

J

1

Z

=

= =

=1 2m s

1kg 2 2m s3

9 3A 7.26 0

1 m

1 m

Z

=

=

1 3s m

9 1A 7.26 10 sZ

=

The average number of collisions ZAA per unit volume per unit time for N2, also known as the collision density is given by Eq. 1.65;

2 2A 3 1A A

AA 2

2 (SI unit :m s )2d u NZV

=

Using the ideal gas law PV nRT and solving for V, a useful expression for ZAA can be obtained;

nRTVP

( )

2 2 2AA A

AA 22

2d u N PZ

nRT

=

ANLn


( )

2 2 2AA

AA 22d u L PZ

RT

=

Solving for ZAA,


1-66

( ) ( )121 23 1A

0

A

474.6 m s 6.022 10 mo3.74 10 m lZ

=

( ) ( )2 21

101325 Pa

2 8.3145 J K 1mol( )2 298.15 K( )34

2

3 2 2

2 2 1 2

22

1AA

2

10 m Pa J

where 1 J kg m s and 1 Pa 1 kg m

8.934 67 s

s

1kgPa1 J

1

Z

= =

=

=

2 4m s

21kg 4 -4m s1

AA

6

34 3 68.934 6

1m

10 m m7 sZ

=

=

-1AA

34 -3108.9 m3 s Z =



1-67

1.33. Express the mean free path of a gas in terms of the variables pressure and temperature, which are more easily measured than the volume.

Solution:

Given: 22 A A

Vd N

Required: mean free path in terms of P and T

The mean free path is given by;

22 A A

Vd N

(Eq. 1.68)

Using the ideal gas law PV nRT and solving for V, a useful expression for the mean free path can be obtained;

nRTVP

22 A A

nRTPd N

ANLn


22 A

RTd LP



1-68

1.34. Calculate ZA and ZAA for argon at 25 C and a pressure of 1.00 bar using

d = 3.84 1010 m obtained from X-ray crystallographic measurements.

Solution:

Given: 1 50 298.3.84 15 K10 m, , 10 Pa A T Pd = ==

Required: A AA,Z Z

ZA is given by Eq. 1.66;

2A 1A A

A2 (SI unit :s )d u NZ

V =

Using the ideal gas law PV nRT and solving for V, a useful expression for ZA can be obtained.

nRTVP

2AA A

A2 d u N PZ

nRT

=

ANLn


2AA

A2 d u LPZ

RT

=

To solve for ZA the speed must first be found. The average speed is given in the Key Equations section of the chapter;


1-69

-1

8

8 8.3145 J K

RTuM

u

=

=1mol( ) 298.15 K( )

39.948 g 1mol 3 110 kg g ( )1

2 2since 1 J 1 kg

1

m s

58 021.4434 J kg

158 021.443 kg4

u

u

=

=

=

( )2 12 m s kg 1397.519 m su =

Solving for ZA gives;

( ) ( )

2

A

1 2321 1

A

03.84 10 m

2

2 397.519 m s 6.022 10 mol

AAd u LPZRT

Z

=

=

( )( )1

10 000 Pa

8.3145 J K 1mol( ) 298.15 K( )1

A2 2 1 2

3 1

where 1 J 1 kg m s and 1 Pa

6 326 37614

1 kg

9 m s Pa

m s

1kg1 Pa1 J

JZ

=

= =

=1 2m s

1kg 2 2m s3

9 3A 6.33 0

1 m

1 m

Z

=

=

1 3s m

9 1A 6.33 10 sZ

=

ZAA is given by Eq. 1.65;


1-70

2 2A 3 1A A

AA 2

2 (SI unit :m s )2d u NZV

=

Using the ideal gas law PV nRT and solving for V, an expression for ZAA is as follows;

nRTVP

( )

2 2 2AA A

AA 22

2d u N PZ

nRT

=

ANLn


( )

2 2 2AA

AA 22d u L PZ

RT

=

Solving for ZAA to get;

( ) ( )12 23 -1A

0

A

1 397.519 m s 6.0223.8410 10 molmZ

=

( ) ( )2 21

10 000 Pa

2 8.3145 J K 1mol( )2 298.15 K( )2

34 3 2 2

2 2 1 2

22

2

1AA 10 m Pa J

since 1 J 1 kg m s and 1 Pa 1 kg m

7.68

s

413 s

1kg1 Pa1 J

Z

= =

=

=

2 4m s

21kg 4 4m s

1AA

6

34 3 67.68413

1 m

10 m msZ

=

=

3 1AA

4 310 m 7.68 sZ =



1-71

1.35. Calculate the mean free path of Ar at 20 C and 1.00 bar. The collision diameter

d = 3.84 1010 m.

Solution:

Given: 10520 C 293.15 K, 1.00 bar 10 Pa, 3.84 10 mAP dT= = = = =

Required:


22 A A

Vd N

Using the ideal gas law PV nRT and solving for V, a useful expression for the mean free path can be obtained.

nRTVP

22 A A

nRTPd N

ANLn


22 A

RTd LP


1-72

-18.3145 J K =

1mol( ) 293.15 K( )( )1 23 -1202 6.022 103.84 10 m mol ( )( )

8 2 1

2 2 1 2

10 J m Pasince 1 J 1 kg m s and 1 P

10 000

a 1 kg m s

1kg1 J1 Pa

Pa

6.1781

= =

=

=

2 2m s

1kg 1 2m s 3

8 2 36.1781

1 m

10 m m =

=

86.18 10 m =



1-73

1.36. Hydrogen gas has a molecular collision diameter of 0.258 nm. Calculate the mean free path of hydrogen at 298.15 K and (a) 133.32 Pa, (b) 101.325 k Pa, and (c) 1.0 108 Pa.

Solution:

Given: 100.258 nm 2.58 10 m, 298.15 KAd T= = =

Required:


2A2 A

Vd N

Using the ideal gas law PV nRT and solving for V, a useful expression for the mean free path can be obtained.

nRTVP

2A2 A

nRTPd N

ANLn


22 A

RTd LP

Now using the fact that P = 133.32 Pa we can make the appropriate substitutions to get;


1-74

18.3145 J K

=1mol( ) 298.15 K( )

( )1 23 1202 6.022 12.58 10 0 mm ol ( )( )4 2 1

2 2 1 2

10 J m Pawhere 1 J 1 kg m s and 1 P

133.32

a 1 kg m s

1kg1 J1 P

Pa

1. 4

a

04

= =

=

=

2 2m s

1kg 1 2m s 3

4 2 31.044

1 m

10 m m

=

=

41.044 10 m =

With the next pressure (P = 101.325 kPa) we can use the same method as outlined above; 18.3145 J K

=1mol( ) 298.15 K( )

( )1 23 1202 6.022 12.58 10 0 mm ol ( )( )7 2 1

2 2 1 2

10 J m Pawhere 1 J 1 kg m s and 1 Pa

101325 Pa

11 kg m s

1kg1 J1

.

Pa

37

= =

=

=

2 2m s

1kg 1 2m s 3

7 2 31.37

1 m

10 m m

=

=

71.37 10 m =

For the final pressure 81.0 10 PaP = ;


1-75

-18.3145 J K =

1mol( ) 298.15 K( )( )1 23 1202 6.022 12.58 10 0 mm ol ( )( )8

10 2 1

2 2 1 2

10 J m Pawhere 1 J 1 kg m s and 1 Pa 1 k

1.0 10 Pa

g m s

1kg1 J1

.39

P

1

a

=

=

=

=

2 2m s

1kg 1 2m s 3

10 2 3

1 m

11 0 m m.39

=

=

1010 9 m1.3 =



1-76

1.37. In interstellar space it is estimated that atomic hydrogen exists at a concentration of one particle per cubic meter. If the collision diameter is 2.5 1010 m, calculate the mean free path . The temperature of interstellar space is 2.7 K.

Solution:

Given: 1A30 2.7 K,2.50 10 m, 1 particle mT Cd = ==

Required:


2A A2

Vd N

Concentration is given by;

ANCV

= , where AN is the number of particles

Now it is possible to solve for

( ) ( )

2

3210

18

12

1

2 1 particle m2.50 10 m

10 . m3 60

Ad C

=

=

=

183.60 10 m =

This is about a hundred times greater than the distance between the earth and the nearest star (Proxima Centauri)!



1-77

1.38. Calculate the value of Avogadros constant from a study made by Perrin [Ann. Chem. Phys., 18, 1(1909)] in which he measured as a function of height the distribution of bright yellow colloidal gamboge (a gum resin) particles suspended in water. Some data at 15 C are:

height, z/106 5 35 N, relative number of gamboge particles at height z 100 47 gamboge = 1.206 g cm3 water = 0.999 g cm-3 radius of gamboge particles, r = 0.212 106 m (Hint: Consider the particles to be gas molecules in a column of air and that the number of particles is proportional to the pressure.)

Solution:

Given: see above

Required: Avogadros number, L

Since we consider the gamboges particles to be proportional to the pressure, we can write;

dP Mg dzP RT

= (Eq. 1.74)

Here, g is the acceleration due to gravity. Taking the integral of both sides and simplifying gives;

0

ln N Mg zN RT

= where M mL=

This can then be substituted into the above equation to get;

0

ln N mLg zN RT

=

Solving for L;

0

lnRT NLmg z N

=


1-78

Density is given by mV

=

Rearranging for the mass of the gamboges particle then gives;

m V=

Where V is the volume of the gamboges particle. Since we know that volume is given by; 34

3rV =

We can then define mass as follows;

343rm =

Now Avogadros number can be expressed as;

3 0

ln43

RT NLNr g z

=

Solving for Avogadros number,

18.3145 J KL

= ( )1mol 288.15 K ( )

34 (1.206 g cm3 30.999 g cm

33 kg m) 10

3g cm60.212 10 m

( ) ( )

( )

3 2

6 6

9.81 m s

1 100ln4735 10 m 5 10 m

23 17.439 74 10 molL =

23 17.44 10 molL =



1-79

1.39. Refer to Table 1.3 (p. 32) and write expressions and values for (a) the ratio 2 /u u , and (b) the ratio /ump. Note that these ratios are independent of the mass and the temperature. How do the differences between them depend on these quantities?

Solution:

Given: Table 1.3

Required: 2 /u u and / mp u

From Table 1.3 the root mean speed is 2 B3k Tum

= , and the average speed is B8k Tum

=

2B B

2B

3 8

3

k T k Tum mu

k Tuu

=

=m

m

B8 k T

2 38

uu

=

2

1.085uu

=

From Table 1.3 the average speed is B8k Tum

= and the most probable speed is Bmp2k Tu

m=


1-80

B B

mp

B

mp

8 2

8

k T k Tu m m

k Tu

=

=m

m

B2 k T

mp

mp

4

2

u

u

=

=

mp

1.128u

=

The differences between 2 and u u and between mp and u increase with T and decrease with m.



1-81

1.40. The speed that a body of any mass must have to escape from the earth is 1.07 104 m s1. At what temperature would the average speed of (a) a H2 molecule, and (b) an O2 molecule be equal to this escape speed?

Solution:

Given: 4 -11.07 10 m su =

Required: 2H

T , 2OT

Average speed, as listed in Table 1.3, is given by B8k Tum

=

By rearranging this equation, temperature can be described as;

2

B8muTk

=

The mass is given by MmL

= and by using this expression, the temperature can be simplified to;

2

B8u MTk L

= where, B

RkL

=

2

8u MT

R

=

a. Solving for2H

T ;


1-82

( )2

24 1

H

1.07 10 m s 2 1.00794 gT

=

1mol( ) 3 110 kg g 1 18 8.3145 J K mol ( )

2

2

2

2H

2H

2

2 1

2

where 1 J 1 kg m

10 898 m s kg J K

10 898 m s kg

s

T

T

=

=

=1 2 2kg m s ( )

2H

K

10 898 KT =

2

4H 1.09 10 KT =

b. Solving for 2O

T ;

( )2

24 -1

O

1.07 10 m s 2 15.9994 gT

=

-1mol( ) 3 -110 kg g-1 -18 8.3145 J K mol( )

2

2

2

2 -2 -1O

2 -2O

2where 1 J 1 kg m s

172 992 m s kg J K

172 992 m s kg

T

T

=

=

=1 2 2kg m s ( )

2O

K

172 992 KT =

2

5O 1.73 10 KT =



1-83

1.41. a. For H2 gas at 25 C, calculate the ratio of the fraction of molecules that have a speed 2u to the fraction that have the average speed . How does this ratio depend on the mass of the molecules and the temperature?

b. Calculate the ratio of the fraction of the molecules that have the average speed 100 C at 100 C to the fraction that have the average speed 25 C at 25 C. How does this ratio depend on the mass?

Solution:

Given: 25 C 298.15KT = =

Required: a. 1 21 2

u u

u u

dN dNN N

, where 1 22 ,u u u u= = , B8k Tu

m=

b. 1 21 2

u u

u u

dN dNN N

where 1 100 C 2 25 C,u u u u = = , B8k Tu

m=

a) The key words in this problem are ratio of the fractions, therefore we use the Boltzmann distribution. The Boltzmann distribution is

given by Eq. 1.91; 2

B

3/2/2 2

B

4 e2

mu k TdN m u duN k T

=

Solving for the ratio;

1 2

1 2

3/2

B

42

u u

u u

mk TdN dN

N N

=

21 B

2/21e mu k T u du

3/2

B

42

mk T

22 B

2/22e mu k T u du

( )2 21 2 B1 21 2

2/2 1

22

em u u k Tu u

u u

dN dN uN N u

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