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11 dilutions . 11. SSM REASONING Both gases fill the balloon to the same pressure P, volume V, and
temperature T. Assuming that both gases are ideal, we can apply the ideal gas law
PV = nRT to each and conclude that the same number of moles n of each gas is needed to
fill the balloon. Furthermore, the number of moles can be calculated from the mass m (in
m
grams) and the mass per mole M (in grams per mole), according to n =
. Using this
M
expression in the equation nHelium = nNitrogen will allow us to obtain the desired mass of
nitrogen.
SOLUTION Since the number of moles of helium equals the number of moles of nitrogen,
we have 730 THE IDEAL GAS LAW AND KINETIC THEORY mHelium
M
14 Helium 3
4
244 = Number of moles
of helium mNitrogen
M Nitrogen
14244
4
3
Number of moles
of nitrogen Solving for mNitrogen and taking the values of mass per mole for helium (He) and nitrogen
(N2) from the periodic table on the inside of the back cover of the text, we find mNitrogen = M Nitrogen mHelium
M Helium = ( 28.0 g/mol ) ( 0.16 g ) = 1.1 g
4.00 g/mol 12. REASONING The number n of moles of an ideal gas that occupy a volume V is determined
by the Kelvin temperature T and pressure P of the gas, as we see from the ideal gas law
P V = nRT (Equation 14.1), where R = 8.31 J/ ( mol ⋅ K ) is the universal gas constant. We
will use Equation 14.1 to determine both the number n1 of moles of air present in the oven at
the initial temperature T1 and pressure P1, and also the final number n2 of moles of air at the
final temperature T2 and pressure P2. Because air is free to move in or out of the oven, the
air pressure inside the oven is the same as the air pressure of the environment outside. The
number ∆n of moles that leave the oven is then the difference between n1 and n2: ∆n = n1 − n2 (1) We note that the volume V of the oven is the same for both the initial and the final
conditions.
SOLUTION Solving P V = nRT (Equation 14.1) for the number n of moles of air in the
oven at a given temperature T and pressure P, we obtain n= PV
RT (2) Substituting Equation (2) into Equation (1), we obtain an expression for the number ∆n of
moles that leave the oven as it heats up: ∆n = n1 − n2 = PV P2V V P P2 1
−
= 1− RT1 RT2 R T1 T2 Using the given values of temperature and pressure in Equation (3), we find that (3) Chapter 14 Problems 731 0.150 m3 1.00 × 105 Pa 9.50 ×104 Pa ∆n = − = 2.31 mol 296 K
453 K 8.31 J/ ( mol ⋅ K ) 13. REASONING Since the absolute pressure, volume, and temperature are known, we may
use the ideal gas law in the form of Equation 14.1 to find the number of moles of gas.
When the volume and temperature are raised, the new pressure can also be determined by
using the ideal gas law.
SOLUTION
a. The number of moles of gas is ( )( ) 1.72 × 10 Pa 2.81 m
PV
n=
=
= 201 mol
RT 8.31 J/ ( mol ⋅ K ) ( 273.15 + 15.5 ) K 5 3 (14.1) b. When the volume is raised to 4.16 m3 and the temperature raised to 28.2 °C, the pressure
of the gas is
P= nRT ( 201 mol ) 8.31 J/ ( mol ⋅ K ) ( 273.15 + 28.2 ) K = 1.21 × 105 Pa
=
3
V
4.16 m ( ) (14.1) ______________________________________________________________________________
14. REASONING According to the ideal gas law, PV = nRT , the temperature T is directly
proportional to the product PV, for a fixed number n of moles. Therefore, tanks with equal
values of PV have the same temperature. Using the data in the table given with the problem
statement, we see that the values of PV for each tank are (starting with tank A):
100 Pa ⋅ m3 , 150 Pa ⋅ m3 , 100 Pa ⋅ m3 , and 150 Pa ⋅ m3 . Tanks A and C have the same
temperature, while B and D have the same temperature.
SOLUTION The temperature of each gas can be found from the ideal gas law,
Equation 14.1:
( 25.0 Pa ) 4.0 m3
PAVA
TA =
=
= 120 K
nR
( 0.10 mol ) 8.31 J /( mol ⋅ K ) ( ) ( ) ( 30.0 Pa ) 5.0 m3
PBVB
TB =
=
= 180 K
nR
( 0.10 mol ) 8.31 J /( mol ⋅ K ) 732 THE IDEAL GAS LAW AND KINETIC THEORY TC = PCVC TD = PDVD nR nR ( 20.0 Pa ) ( 5.0 m3 )
=
=
( 0.10 mol ) 8.31 J /( mol ⋅ K ) 120 K ( 2.0 Pa ) ( 75 m3 )
=
=
( 0.10 mol ) 8.31 J /( mol ⋅ K ) 180 K ______________________________________________________________________________
15. REASONING AND SOLUTION The ideal gas law gives P T 65.0 atm 297 K (
3
3
V2 = 1 2 V1 = 1.00 m ) = 67.0 m
P2 T1 1.00 atm 288 K ______________________________________________________________________________
16. REASONING We can use the ideal gas law, Equation 14.1 (PV = nRT) to find the number
of moles of helium in the Goodyear blimp, since the pressure, volume, and temperature are
known. Once the number of moles is known, we can find the mass of helium in the blimp.
SOLUTION The number n of moles of helium in the blimp is, according to Equation 14.1, n= PV (1.1×105 Pa)(5400 m3 )
=
= 2.55 × 105 mol
RT [8.31 J/(mol ⋅ K)](280 K) According to the periodic table on the inside of the text’s back cover, the atomic mass of
helium is 4.002 60 u. Therefore, the mass per mole is 4.002 60 g/mol. The mass m of
helium in the blimp is, then,...
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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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