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Unformatted text preview: ING Assuming that neon (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
2
neon, R = 8.31 J/ ( mol ⋅ K ) is the universal gas constant, and T is the Kelvin temperature.
We seek the increase ∆U = Uf − Ui in the internal energy of the neon as its temperature
increases from Ti to Tf. Because the gas is in a closed, rigid container, neither the number n
of moles nor the volume V of the neon change when its temperature is raised. Consequently,
we will employ the ideal gas law, P V = nRT (Equation 14.1), to determine the value of the
quantity nR from the volume V, the initial pressure Pi, and the initial Kelvin temperature Ti
of the neon. Chapter 14 Problems 745 SOLUTION From U = 3 nRT (Equation 14.7), the increase in the internal energy of the
2
gas is given by
∆ U = U f − U i = 3 nRTf − 3 nRTi =
2
2 3
2 nR (Tf − Ti ) Solving PV = nRTi (Equation 14.1) for the quantity nR yields nR =
i (1) PV
i
Ti . Substituting this result into Equation (1), we find that PV ∆U = 3 i (Tf − Ti ) =
2T i 3
2 (1.01×105 Pa ) ( 680 m3 )
293.2 K ( 294.3 K − 293.2 K ) = 3.9 ×105 J 36. REASONING AND SOLUTION To find the rmsspeed of the CO2 we need to first find the
temperature. We can do this since we know the rmsspeed of the H2O molecules. Using
1 m v2
rms
2 = 3 kT,
2 2
we can solve for the temperature to get T = m vrms /(3k), where the mass of an H2O molecule is (18.015 g/mol)/(6.022 × 1023 mol–1) = 2.99 × 10–23 g. Thus, ( 2.99 ×10−26 kg ) ( 648 m/s )2 = 303 K
T=
3 (1.38 ×10−23 J/K ) The molecular mass of CO2 is 44.01 u, hence the mass of a CO2 molecule is 7.31 × 10–26 kg.
The rmsspeed for CO2 is
3kT
3(1.38 ×10−23 J/K)(303 K)
=
= 414 m/s
m
7.31 × 10−26 kg
______________________________________________________________________________
vrms = 37. SSM
PV = REASONING
2 N ( 1 mv 2 )
rms .
3
2 The behavior of the molecules is described by Equation 14.5: Since the pressure and volume of the gas are kept constant, while the number of molecules is doubled, we can write PV2 = P2V2 , where the subscript 1 refers to
1
the initial condition, and the subscript 2 refers to the conditions after the number of
molecules is doubled. Thus,
2
3 1
2
1
2
2
N1 m (vrms )1 = N 2 m (vrms )2 2
3
2 or 2
2
N1 (vrms )1 = N 2 (vrms )2 The last expression can be solved for (vrms ) 2 , the final translational rms speed. 746 THE IDEAL GAS LAW AND KINETIC THEORY SOLUTION Since the number of molecules is doubled, N 2 = 2 N1 . Solving the last expression above for (vrms )2 , we find
N1 N1 463 m/s
= 327 m/s
N2
2 N1
2
______________________________________________________________________________
(vrms ) 2 = (vrms )1 = (463 m/s) = 38. REASONING According to the kinetic theory of gases, the average kinetic energy of an
atom is KE = 3 kT (Equation 14.6), where k is Boltzmann’s constant and T is the Kelvin
2
temperature. Therefore, the ratio of the average kinetic energies is equal to the ratio of the
Kelvin temperatures of the gases. We are given no direct information about the
temperatures. However, we do know that the temperature of an ideal gas is related to the
pressure P, the volume V, and the number n of moles of the gas via the ideal gas law,
PV = nRT. Thus, the ideal gas law can be solved for the temperature, and the ratio of the
temperatures can be related to the other properties of the gas. In this way, we will obtain the
desired kineticenergy ratio. SOLUTION Using Equation 14.6 and the ideal gas law in the form T = PV
, we find that
nR the desired ratio is 3
KE Krypton 2 kTKrypton
=3
=
kTArgon
KE Argon
2 PV
3k 2 n
R Krypton 3 k PV 2 n
R Argon = nArgon
nKrypton Here, we have taken advantage of the fact that the pressure and volume of each gas are the
same. While we are not given direct information about the number of moles of each gas, we
do know that their masses are the same. Furthermore, the number of moles can be
calculated from the mass m (in grams) and the mass per mole M (in grams per mole),
m
according to n =
. Substituting this expression into our result for the kineticenergy ratio
M
gives
m
nArgon
M Argon
M Krypton
KE Krypton
=
=
=
m
nKrypton
M Argon
KE Argon
M Krypton
Taking the masses per mole from the periodic table on the inside of the back cover of the
text, we find Chapter 14 Problems 747 KE Krypton M Krypton
83.80 g/mol
=
=
= 2.098
M Argon
39.948 g/mol
KE Argon ______________________________________________________________________________
39. REASONING The smoke particles have the same average translational kinetic energy as
2
the air molecules, namely, 1 mvrms = 3 kT , according to Equation 14.6. In this expression m
2
2
is the mass of a smoke particle, vrms is the rms speed of a particle, k is Boltzmann’s
constant, and T is the Kelvin temperature. We can obtain the mass directly from this
equation.
SOLUTION Solving Equation 14.6 for the mass m, we find ( ) −23
J/K ( 301 K )
3kT 3 1.38 ×10
m= 2 =
= 1.6 × 10−15 kg
2
vrms
2.8 × 10−3 m/s ( ) ____________________________________...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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