This preview shows page 1. Sign up to view the full content.
Unformatted text preview: __________________________________________
40. REASONING The average kinetic energy KE of each molecule in the gas is directly
proportional to the Kelvin temperature T, according to KE = 3 kT (Equation 14.6), where k
2
is the Boltzmann constant. Solving for the temperature, we obtain
T= () 2 KE
3k (1) The average kinetic energy KE of one molecule is the total average kinetic energy
of all the molecules divided by the total number N of molecules:
KE ( )total KE = ( KE )total
N (2) We will determine the number N of molecules in the gas by multiplying the number n of
moles by Avogadro’s number NA:
N = nN A (3) ( )total of all the molecules is equal to the kinetic energy The total average kinetic energy KE KEbullet of the bullet, which has a mass m and a speed v. Thus, according to Equation 6.2, 748 THE IDEAL GAS LAW AND KINETIC THEORY ( KE )total = KE bullet = 1 mv 2
2 (6.2) SOLUTION Substituting Equation (2) into Equation (1) yields () KE total 2 2 KE
N
2 KE =
total
T=
=
3k
3k
3kN () () (4) Substituting Equation (3) and Equation 6.2 into Equation (4), we obtain T= ( )total = 2 ( 1 mv2 ) =
2 2 KE 3kN 3k ( nN A ) mv 2
3knN A (5) The Boltzmann constant k is equal to the ratio of the universal gas constant R to Avogadro’s
number: k = R N A . Making this substitution into Equation (5) yields the Kelvin
temperature of the gas:
T= 41. ( 3 SSM WWW ) 8.0 ×10−3 kg ( 770 m/s )
mv 2
mv 2
=
=
= 95 K
3Rn 3 8.31 J/ ( mol ⋅ K ) ( 2.0 mol ) R n NA
NA 2 REASONING The rmsspeed vrms of the sulfur dioxide molecules is 2
3
related to the Kelvin temperature T by 1 mvrms = 2 kT (Equation 14.6), where m is the mass
2
of a SO2 molecule and k is Boltzmann’s constant. Solving this equation for the rmsspeed
gives vrms = 3kT
m (1) The temperature can be found from the ideal gas law, Equation 14.1, as T = PV / ( nR ) ,
where P is the pressure, V is the volume, n is the number of moles, and R is the universal gas
constant. All the variables in this relation are known. Substituting this expression for T into
Equation (1) yields Chapter 14 Problems vrms 749 PV 3k nR = 3 k PV
=
m
nmR The mass m of a single SO2 molecule will be calculated in the Solutions section.
SOLUTION Using the periodic table on the inside of the text’s back cover, we find the
molecular mass of a sulfur dioxide molecule (SO2) to be
32.07 u 3
u)
14 244 + 2 (15.9994 3 = 64.07 u
4
14 244
4
Mass of a single
sulfur atom Mass of two
oxygen atoms Since 1 u = 1.6605 × 10−27 kg (see Section 14.1), the mass of a sulfur dioxide molecule is 1.6605 ×10−27 kg −25
m = ( 64.07 u ) kg = 1.064 × 10
1u The translational rmsspeed of the sulfur dioxide molecules is
vrms = = 3 k PV
nm R
3 (1.38 × 10−23 J/K )( 2.12 ×104 Pa )( 50.0 m3 )
= 343 m/s
( 421 mol ) (1.064 × 10−25 kg ) 8.31 J/ ( mol ⋅ K ) ______________________________________________________________________________
42. REASONING
Power is the rate at which energy is produced, according to
Energy
(Equation 6.10b), so the time required for the engine to produce a certain
Power =
t
amount of energy is given by
t= Energy
Power (1) Assuming that helium (a monatomic gas) behaves as an ideal gas, its internal energy U can
be found from U = 3 nRT (Equation 14.7), where n is the number of moles of helium in the
2 container, R = 8.31 J/ ( mol ⋅ K ) is the universal gas constant, and T is the Kelvin
temperature. We will determine the quantity nRT in Equation 14.7 from the pressure P and
volume V of the helium via the ideal gas law P V = nRT (Equation 14.1). 750 THE IDEAL GAS LAW AND KINETIC THEORY SOLUTION The engine must produce an amount of energy equal to the internal energy
U = 3 nRT (Equation 14.7) of the helium, so from Equation (1) we have that
2
3 nRT
Energy
U
t=
=
=2
Power Power Power (2) Substituting P V = nRT (Equation 14.1) into Equation (2) yields t= 3 nRT
2 Power = 3PV
( Power )
2 (3) Using the equivalence 1 hp = 746 W in Equation (3), we obtain t= 43. 3PV
3 ( 6.2 ×105 Pa ) ( 0.010 m3 )
=
= 5.0 × 101 s
( Power )
746 W 2 2 0.25 hp 1 hp ( ) SSM WWW REASONING AND SOLUTION
a. Assuming that the direction of travel of the bullets is positive, the average change in
momentum per second is
∆p/∆t = m∆v/∆t = (200)(0.0050 kg)[(0 m/s) – 1200 m/s)]/(10.0 s) = –120 N b. The average force exerted on the bullets is F = ∆p/∆t. According to Newton’s third law,
the average force exerted on the wall is − F = 120 N .
c. The pressure P is the magnitude of the force on the wall per unit area, so
P = (120 N)/(3.0 × 10–4 m2) = 4.0 × 10 5 Pa ______________________________________________________________________________
44. REASONING Because the container holds 1.000 mol of neon, we know that the number of
neon atoms inside is equal to Avogadro’s number: N = NA. These NA atoms are continually
bouncing back and forth between the walls of the cubical container, with an rms speed vrms
2
3
found from 1 mvrms = 2 kT (Equation 14.6), where m is the mass of a single neon atom, k is
2
the Boltzmann constant, and T is the Kelvin temperature. Following the development of
kinetic theory given in Section 1...
View
Full
Document
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details