Unformatted text preview: resh air contains approximately 21% oxygen, the total number of oxygen molecules
in a normal breath is (0.21) nN A or
(0.21)(1.94 ×10−2 mol)(6.022 ×1023 mol −1 )= 2.5 ×1021 Chapter 14 Problems 757 ______________________________________________________________________________
55. SSM REASONING According to Equation 11.1, the mass density ρ of a substance is
defined as its mass m divided by its volume V: ρ = m / V . The mass of nitrogen is equal to
the number n of moles of nitrogen times its mass per mole: m = n (Mass per mole). The
number of moles can be obtained from the ideal gas law (see Equation 14.1) as
n = (PV)/(RT). The mass per mole (in g/mol) of nitrogen has the same numerical value as its
molecular mass (which we know).
SOLUTION Substituting m = n (Mass per mole) into ρ = m / V , we obtain ρ= m n (Mass per mole)
=
V
V (1) Substituting n = (PV)/(RT) from the ideal gas law into Equation 1 gives the following result: PV
n (Mass per mole) RT
ρ=
=
V (Mass per mole) P (Mass per mole) =
RT
V
5 The pressure is 2.0 atmospheres, or P = 2 (1.013 × 10 Pa). The molecular mass of nitrogen
is given as 28 u, which means that its mass per mole is 28 g/mol. Expressed in terms of
kilograms per mol, the mass per mole is g 1 kg Mass per mole = 28 mol 103 g The density of the nitrogen gas is g 1 kg 2 (1.013 ×105 Pa ) 28 mol 103 g P(Mass per mole) ρ=
=
= 2.2 kg/m3
[8.31 J/ ( mol ⋅ K )] ( 310 K )
RT
_____________________________________________________________________________________________ 56. REASONING The diffusion of the glycine is driven by the concentration difference
∆C = C2 – C1 between the ends of the tube, where C2 is the higher concentration and
C1 = C2 − ∆C is the lower concentration. The concentration difference is related to the mass
m DA ∆C
rate of diffusion by Fick’s law:
(Equation 14.8), where D is the diffusion
=
t
L
constant of glycine in water, and A and L are, respectively, the crosssectional area and
length of the tube. 758 THE IDEAL GAS LAW AND KINETIC THEORY SOLUTION Solving m DA ∆C
(Equation 14.8) for ∆C, we obtain
=
t
L m
L ∆C = t DA (1) Substituting Equation (1) into C1 = C2 − ∆C yields m
L C1 = C2 − ∆C = C2 − t DA (2) We note that the length of the tube is L = 2.0 cm = 0.020 m. Substituting this and the other
given values into Equation (2), we find that the lower concentration is
C1 = 8.3 × 10−3 kg/m3 − 57. ( 0.020 m ) ( 4.2 × 10−14 kg/s ) (1.06 ×10−9 m 2 /s )(1.5 × 10−4 m 2 ) = 3.0 × 10−3 kg/m3 SSM REASONING AND SOLUTION Using the expressions for v 2 and ( v ) given in
the statement of the problem, we obtain:
2 1
3 a. 2
2
2
v 2 = (v1 + v2 + v3 ) = b. (v) 2 1
(3.0
3 m/s)2 + (7.0 m/s) 2 + (9.0 m/s)2 = 46.3 m 2 /s 2 2 2 = 1 (v1 + v2 + v3 ) = 1 (3.0 m/s + 7.0 m/s + 9.0 m/s) = 40.1 m 2 /s 2
3 3 v 2 and ( v ) are not equal, because they are two different physical quantities.
______________________________________________________________________________
2 58. REASONING The translational rmsspeed vrms is related to the Kelvin temperature T by
1
2 2
3
mvrms = 2 kT (Equation 14.6), where m is the mass of the oxygen molecule and k is Boltzmann’s constant. Solving this equation for the rmsspeed gives vrms = 3kT / m . This
relation will be used to find the ratio of the speeds.
SOLUTION The rmsspeeds in the ionosphere and near the earth’s surface are ( vrms )ionosphere = 3k Tionosphere
m and ( vrms )earth's surface = 3kTearth's surface
m Chapter 14 Problems 759 Dividing the first equation by the second gives ( vrms )ionosphere ( vrms )earth's surface 3kTionosphere
= m
3kTearth's surface
m = Tionosphere
Tearth's surface = 3 = 1.73 ____________________________________________________________________________________________ 59. SSM REASONING The mass m of nitrogen that must be removed from the tank is equal
to the number of moles withdrawn times the mass per mole. The number of moles
withdrawn is the initial number ni minus the final number of moles nf, so we can write m= −n
( ni 24) ( Mass per mole )
14 f3 (1) Number of
moles withdrawn The final number of moles is related to the initial number by the ideal gas law.
SOLUTION From the ideal gas law (Equation 14.1), we have nf = Pf V
RT and ni = PV
i
RT Note that the volume V and temperature T do not change. Dividing the first by the second
equation gives
Pf V
nf
P
P RT
=
=f
or
nf = ni f PV
ni
Pi
i Pi RT
Substituting this expression for nf into Equation 1 gives P m = ( ni − nf ) ( Mass per mole ) = ni − ni f ( Mass per mole ) Pi The molecular mass of nitrogen (N2) is 2 (14.0067 u) = 28.0134 u. Therefore, the mass per
mole is 28.0134 g/mol. The mass of nitrogen that must be removed is 760 THE IDEAL GAS LAW AND KINETIC THEORY P
m = ni − ni f Pi ( Mass per mole ) g 25 atm = 0.85 mol − ( 0.85 mol ) 28.0134 = 8.1 g
mol 38 atm ______________________________________________________________________________
3 60. REASONING AND SOLUTION Since the density of aluminum is 2700 kg/m , the
number of atoms of aluminum per cubic meter is, using the given data, 2700 × 103 g/m3 23
−1
28
3
N /V = 6.022 × 10 mol = 6.0 × 10 atoms/m 26.981...
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 Spring '13
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 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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