Unformatted text preview: nner tube, we
can apply Boyle’s law:
PVi = Pf Vf
i (14.3) We will use Equations (1) and Equation 14.3 to determine the distance d = hi − hf through
which the pump handle moves before air begins to flow. We note that the final pressure Pf is
equal to the pressure of air in the inner tube. SOLUTION Since the crosssectional area A of the piston does not change, substituting
Equations (1) into Equation 14.3 yields
Pi A hi = Pf A hf Pi hi = Pf hf or (2) Solving Equation (2) for the unknown final height hf, we obtain
hf = Pi hi
Pf (3) With Equation (3), we can now calculate the distance d = hi − hf that the biker pushes down
on the handle before air begins to flow from the pump to the inner tube: 736 THE IDEAL GAS LAW AND KINETIC THEORY d = hi − hf = hi − P
= hi 1 − i
P
Pf f Phi
i 1.0 × 105 Pa = ( 0.55 m ) 1 − 2.4 × 105 Pa = 0.32 m 23. SSM WWW REASONING AND SOLUTION
a. Since the heat gained by the gas in one tank is equal to the heat lost by the gas in the
other tank, Q1 = Q2, or (letting the subscript 1 correspond to the neon in the left tank, and
letting 2 correspond to the neon in the right tank) cm1∆T1 = cm2 ∆T2 ,
cm1 (T − T1 ) = cm2 (T2 − T )
m1 (T − T1 ) = m2 (T2 − T )
Solving for T gives T= m2T2 + m1T1
m2 + m1 (1) The masses m1 and m2 can be found by first finding the number of moles n1 and n2. From
the ideal gas law, PV = nRT, so n1 = ( 5.0 ×105 Pa ) ( 2.0 m3 )
PV1
1
=
= 5.5 ×102 mol
RT1 [8.31 J/(mol ⋅ K) ] ( 220 K ) This corresponds to a mass m1 = (5.5×10 2 mol) ( 20.179 g/mol ) = 1.1×104 g = 1.1×101 kg .
Similarly, n2 = 2.4 × 102 mol and m2 = 4.9 × 103 g = 4.9 kg. Substituting these mass values
into Equation (1) yields
T= ( 4.9 kg ) ( 580 K ) + (1.1× 101 kg ) ( 220 K )
=
( 4.9 kg ) + (1.1×101 kg ) 3.3 × 102 K b. From the ideal gas law,
nRT ( 5.5 ×102 mol ) + ( 2.4 ×102 mol ) [8.31 J/ ( mol ⋅ K )] ( 3.3 ×102 K ) =
= 2.8 × 105Pa
3
V
2.0 m + 5.8 m3
______________________________________________________________________________
P= Chapter 14 Problems 737 24. REASONING Since the temperature of the confined air is constant, Boyle's law applies,
and PsurfaceVsurface = PhVh , where P
surface and Vsurface are the pressure and volume of the
air in the tank when the tank is at the surface of the water, and Ph and Vh are the pressure
and volume of the trapped air after the tank has been lowered a distance h below the surface
of the water. Since the tank is completely filled with air at the surface, Vsurface is equal to
the volume Vtank of the tank. Therefore, the fraction of the tank's volume that is filled with
air when the tank is a distance h below the water's surface is
Vh
Vtank = Vh
Vsurface = Psurface
Ph We can find the absolute pressure at a depth h using Equation 11.4. Once the absolute
pressure is known at a depth h, we can determine the ratio of the pressure at the surface to
the pressure at the depth h.
SOLUTION According to Equation 11.4, the trapped air pressure at a depth h = 40.0 m is
Ph = Psurface + ρ gh = (1.01×105 Pa) + (1.00 ×103 kg/m3 )(9.80 m/s 2 )(40.0 m) = 4.93 × 105 Pa
where we have used a value of ρ = 1.00 × 103 kg/m3 for the density of water. The desired
volume fraction is 1.01 × 10 5 Pa
= 0.205
Vtank
Ph
4.93 × 10 5 Pa
______________________________________________________________________________
Vh = Psurface = 25. REASONING The mass (in grams) of the air in the room is the mass of the nitrogen plus
the mass of the oxygen. The mass of the nitrogen is the number of moles of nitrogen times
the molecular mass (in grams/mol) of nitrogen. The mass of the oxygen can be obtained in a
similar way. The number of moles of each species can be found by using the given
percentages and the total number of moles. To obtain the total number of moles, we will
apply the ideal gas law. If we substitute the Kelvin temperature T, the pressure P, and the
volume V (length × width × height) of the room in the ideal gas law, we can obtain the total
PV
number nTotal of moles of gas as nTotal =
, because the ideal gas law does not distinguish
RT
between types of ideal gases.
SOLUTION Using m to denote the mass (in grams), n to denote the number of moles, and
M to denote the molecular mass (in grams/mol), we can write the mass of the air in the room
as follows: 738 THE IDEAL GAS LAW AND KINETIC THEORY m = mNitrogen + mOxygen = nNitrogen M Nitrogen + nOxygen M Oxygen We can now express this result using f to denote the fraction of a species that is present and
nTotal to denote the total number of moles:
m = nNitrogen M Nitrogen + nOxygen M Oxygen = f Nitrogen nTotal M Nitrogen + f Oxygen nTotal M Oxygen According to the ideal gas law, we have nTotal = PV
. With this substitution, the mass of
RT the air becomes ( ) ( m = f Nitrogen M Nitrogen + fOxygen M Oxygen nTotal = f Nitrogen M Nitrogen + f Oxygen M Oxygen ) PV
RT The mass per mole for nitrogen (N2) and for oxygen (O2) can be obtained from the periodic
table on the inside of the back cover of the text. They are, respectively, 28.0 and
32.0 g/mol. The temperature of 22 ºC must be expressed on the Kelvin scale as 295 K (see
Equation 12.1). The mass of the air is, then, ( m = f Nitrogen M Nitrogen + fOxygen M Ox...
View
Full Document
 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details