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, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details