Unformatted text preview: 4.3 of the text, the time t that elapses between successive
collisions of one atom with one wall of the container is Chapter 14 Problems t= 2L
vrms 751 (1) This is because we assume that, between successive collisions with a given wall, the atom
travels to the opposite wall and back, which is a total distance of 2L. We will use
1 mv 2 = 3 kT (Equation 14.6) to determine the rms speed of the neon atoms for use in
rms
2
2
Equation (1).
There are three identical pairs of walls, so on average 1/3 of the NA atoms in the container
collide with a given wall in the time t. Therefore, the rate at which atoms collide with a
given wall is
N
(2)
Collision rate = A
3t
SOLUTION Substituting Equation (1) into Equation (2) yields
Collision rate = Solving 1 mv 2
rms
2 1
2 NA
3t = NA 2L
3
v rms = N A vrms (3) 6L = 3 kT (Equation 14.6) for the rms speed of the atoms, we obtain
2
2
mvrms = 3
2 kT or 2
vrms = 3kT
m or
vrms = 3kT
m (4) Substituting Equation (4) into Equation (3) yields
Collision rate = N A vrms
6L = NA
6L 3kT
m Therefore, the rate at which neon atoms collide with each wall of the container is ( 6.022 ×1023 atoms ) 3 (1.38 ×10−23 J/K ) ( 293 K ) = 2.01×1026 atoms/s
Collision rate =
6 ( 0.300 m ) 3.35 ×10−26 kg (5) 752 THE IDEAL GAS LAW AND KINETIC THEORY 45. REASONING The mass m of oxygen that diffuses in a time t through a trachea is given by
Equation 14.8 as m = ( DA ∆C ) t / L , where ∆C is the concentration difference between the
two ends of the trachea, A and L are, respectively, its crosssectional area and length, and D
is the diffusion constant. The concentration difference ∆C is the higher concentration C2
minus the lower concentration C1, or ∆C = C2 − C1 . SOLUTION Substituting the relation ∆C = C2 − C1 into Equation 14.8 and solving for C1
gives L m C1 = C2 − DA t We recognize that m/t is the mass per second of oxygen diffusing through the trachea. Thus,
the oxygen concentration at the interior end of the trachea is 1.9 × 10−3 m
3
−12
C1 = 0.28 kg/m3 − (1.7 × 10 kg/s ) = 0.14 kg/m
2)
−5 2
−9 (1.1 × 10 m /s ) ( 2.1 × 10 m ______________________________________________________________________________
46. REASONING The mass m of methane that diffuses out of the tank in a time t is given by
( DA ∆C ) t (Equation 14.8), where D is the diffusion constant for methane, A is the
m=
L
crosssectional area of the pipe, L is the length of the pipe, and ∆C = C1 − C2 is the
difference in the concentration of methane at the two ends of the pipe. The higher
concentration C1 at the tankend of the pipe is given, and the concentration C2 at the end of
the pipe that is open to the atmosphere is zero.
SOLUTION Solving m = ( DA ∆ C ) t (Equation 14.8) for the crosssectional area A, we L obtain
A= mL
( D ∆C ) t The elapsed time t must be converted from hours to seconds: 3600 s 4
t = 12 hr = 4.32 ×10 s 1 hr ( ) Therefore, the crosssectional area A of the pipe is (1) Chapter 14 Problems A= 47. 753 (9.00 ×10−4 kg ) (1.50 m )
= 2.29 × 10−3 m 2
2
3
3
4
−5
( 2.10 ×10 m /s )( 0.650 kg/m − 0 kg/m )( 4.32 ×10 s ) SSM REASONING AND SOLUTION
According to Fick's law of diffusion
(Equation 14.8), the mass of ethanol that diffuses through the cylinder in one hour (3600 s)
is ( DA ∆ C )t (12.4 × 10 –10 m 2 /s)(4.00 × 10 –4 m 2 )(1.50 kg/m 3 )(3600 s)
=
= 1.34 × 10 –7 kg
(0.0200 m)
L
______________________________________________________________________________
m= 48. REASONING According to the kinetic theory of gases, the average kinetic energy KE of
an atom in an ideal gas is related to the Kelvin temperature of the gas by KE = 3 k T
2
(Equation 14.6). Since the temperature is the same for both gases, the A and B atoms have
the same average kinetic energy. The average kinetic energy is related to the rmsspeed by
2
KE = 1 mvrms . Since both gases have the same average kinetic energy and gas A has the
2
smaller mass, it has the greater rmsspeed. Therefore, gas A has the greater diffusion rate. SOLUTION For a fixed temperature, the ratio RA/RB of the diffusion rates for the two
types of atoms is equal to the ratio vrms, A/vrms, B of the rmsspeeds. According to
Equation 14.6, it follows that vrms = 2KE/m , so that v
RA
= rms, A =
RB
vrms, B 2KE
mA
2KE
mB = mB
=
mA 2.0 u
= 1.4
1.0 u ______________________________________________________________________________
49. REASONING AND SOLUTION
a. As stated, the time required for the first solute molecule to traverse a channel of length L
2
is t = L /(2 D ) . Therefore, for water vapor in air at 293 K, where the diffusion constant is D = 2.4 × 10
is –5 2 m /s , the time t required for the first water molecule to travel L = 0.010 m ( 0.010 m )
L2
=
= 2.1 s
2 D 2 2.4 × 10−5 m 2 /s
2 t= ( ) 754 THE IDEAL GAS LAW AND KINETIC THEORY b. If a water molecule were traveling at the translational rms speed v rms for water, the time t
it would take to travel the distance L = 0.010 m would be given by t = L / vrms , where, according to Equatio...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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