Unformatted text preview: ygen ) PV
RT (1.01×105 Pa ) ( 2.5 m )( 4.0 m )(5.0 m ) = 0.79 ( 28.0 g/mol ) + 0.21( 32.0 g/mol ) 8.31 J/ ( mol ⋅ K ) ( 295 K ) = 5.9 ×104 g 26. REASONING The ideal gas law is PV = nRT. Since the number of moles is constant, this
PV
PV
equation can be written as
= nR = constant . Thus, the value of
is the same
T
T
initially and finally, and we can write P0V0
T0 = Pf Vf
Tf This expression can be solved for the final temperature Tf. (1) Chapter 14 Problems 739 We have no direct data for the initial and final pressures. However, we can deal with this by
realizing that pressure is the magnitude of the force applied perpendicularly to a surface
divided by the area of the surface. Thus, the magnitudes of the forces that the initial and
final pressures apply to the piston (and, therefore, to the spring) are given by Equation 11.3
as
(2)
F0 = P0 A
and
Ff = Pf A
14 244
4
3
14 244
4
3
Force applied by
final pressure Force applied by
initial pressure The forces in Equations (2) are applied to the spring, and the force FxApplied that must be
applied to stretch an ideal spring by an amount x with respect to its unstrained length is
given by Equation 10.1 as
FxApplied = kx (3) where k is the spring constant.
Lastly, we note that the final volume is the initial volume plus the amount by which the
volume increases as the spring stretches. The increased volume due to the additional
stretching is A ( xf − x0 ) . Therefore, we have
Vf = V0 + A ( xf − x0 ) (4) SOLUTION The final temperature can be obtained by rearranging Equation (1) to show
that PV (5)
Tf = f f T0 PV 0 0
Into this result we can now substitute expressions for P0 and Pf. These expressions can be
obtained by using Equations (2) in Equation (3) as follows (and recognizing that, for the
initial and final forces, P0A = FxApplied and Pf A = FxApplied): P0 A
= kx0
14243 and Pf A
= kxf
14243 (6) Force applied to
spring by final
pressure Force applied to
spring by initial
pressure Thus, we find that
P0 = kx0
A and Pf = kxf
A (7) Finally, we substitute Equations (7) for the initial and final pressure and Equation (4) for the
final volume into Equation (5). With these substitutions Equation (5) becomes 740 THE IDEAL GAS LAW AND KINETIC THEORY kxf
A PV Tf = f f T0 = PV 0 0 = V0 + A ( xf − x0 ) T0 x V + A ( x − x ) T f
0
f
00 =
kx0 x0V0 A V0 ( 0.1000 m ) 6.00 ×10−4 m3 + ( 2.50 ×10−3 m 2 ) ( 0.1000 m − 0.0800 m ) ( 273 K ) ( 0.0800 m ) ( 6.00 ×10−4 m3 ) = 3.70 × 102 K 27. SSM REASONING The graph that accompanies Problem 75 in Chapter 12 can be used
to determine the equilibrium vapor pressure of water in the air when the temperature is
30.0 °C (303 K). Equation 12.6 can then be used to find the partial pressure of water in the
air at this temperature. Using this pressure, the ideal gas law can then be used to find the
number of moles of water vapor per cubic meter.
SOLUTION According to the graph that accompanies Problem 75 in Chapter 12, the
equilibrium vapor pressure of water vapor at 30.0 °C is approximately 4250 Pa. According
to Equation 12.6, ( )( ) Partial
1
pressure of = Percent relative Equilibrium vapor pressure of ×
water at the existing temperature 100
humidity
water vapor = ( 55 )( 4250 Pa ) = 2.34 × 103 Pa
100 The ideal gas law then gives the number of moles of water vapor per cubic meter of air as
n
P
(2.34 × 10 3 Pa)
3
(14.1)
=
=
= 0.93 mol/m
V RT [8.31 J/(mol ⋅ K)](303 K)
______________________________________________________________________________
28. REASONING AND SOLUTION If the pressure at the surface is P1 and the pressure at a
depth h is P2, we have that P2 = P1 + ρ gh. We also know that P1V1 = P2V2. Then, V1 P2 P + ρ gh
ρ gh
=
=1
= 1+
V2 P
P
P
1
1
1 Chapter 14 Problems 741 Therefore, V1
(1.000 × 103 kg/m3 )(9.80 m/s 2 )(0.200 m)
= 1+
= 1.02
V2
1.01× 105 Pa
______________________________________________________________________________
29. REASONING The desired percentage is the volume the atoms themselves occupy divided
by the total volume that the gas occupies, multiplied by the usual factor of 100. The volume
VAtoms that the atoms themselves occupy is the volume of an atomic sphere ( 4 π r 3 , where r
3
is the atomic radius), times the number of atoms present, which is the number n of moles
times Avogadro’s number NA. The total volume VGas that the gas occupies can be taken to
be that calculated from the ideal gas law, because the atoms themselves occupy such a small
volume.
SOLUTION The total volume VGas that the gas occupies is given by the ideal gas law as
VGas = nRT/P, where the temperature and pressure at STP conditions are 273 K and 1.01 × 105 Pa. Thus, we can write the desired percentage as
Percentage = = VAtoms
VGas 4π
3 ( 4 π r 3 ) nN A ×100 = ( 4 π r 3 ) NA P ×100
3
3
×100 =
nRT
P RT ( 2.0 ×10−10 m ) ( 6.022 ×1023 mol−1 )(1.01×105 Pa ) ×100 = 0.090 %
3 8.31 J/ ( mol ⋅ K ) ( 273 K ) 30. REASONING AND SOLUTION We need to determine the amount of He inside the
balloon. We begin by using Archimedes’ principle; the balloon is being buoyed up by a
force equal to the weight of the air displa...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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