Gas Law'sm1
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Transcript of Gas Law'sm1
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EH 1
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Air is the mostimportant gas toliving things on the
Earth. The atmosphere of
the Earth is amixture of
nitrogen, oxygen,water vapor, argon,and a few tracegases.
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Gases assume the volume and shape oftheir containers.
Gases are the most compressible state of
matter.
Gases will mix evenly and completely whenconfined to the same container.
Gases have much lower densities thanliquids and solids.
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Boyles Law is one of the laws in physics thatconcerns the behaviour of gases
When a gas is under pressure it takes up lessspace:
The higher the pressure, the smaller the volume Boyles Law tells us about the relationship between
the volume of a gas and its pressure at a constanttemperature
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The volume of a gas is inversely proportionalto the absolute pressure applied to it whenthe temperature is kept constant. Pis theabsolute pressure.
P
V1
contPV
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P1V1 = P2V2
Final pressure (N/m2
)
Final volume (m3
)Original volume (m3
)
Original pressure
(N/m2)
Robert Boyle (1627-1691). Son of Earl of
Cork, Ireland.
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A deep sea diver isworking at a depth wherethe pressure is 3.0atmospheres. He isbreathing out air bubbles.
The volume of each airbubble is 2.0 cm2. At thesurface the pressure is1.0 atmosphere. What isthe volume of each
bubble when it reachesthe surface?(assume thetemperature of the wateris constant)
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Real gases DO experience inter-molecular attractions
Real gases DO have volume
Real gases DO NOT have elastic collisions
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The volume of a gas is directly
proportional to absolutetemperature when the
pressure is kept constant
2
2
1
1
T
V
T
V
Temperature MUST be in KELVIN
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At constant volume theabsolute pressure of agas is directly
proportional to theabsolute Temperature. TP
kT
P
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You basicallytake BoylesCharles and
Gay-LussacsLaw andcombine themtogether.
Moles areconstant
T
PV
T
VP
T
PVk
T
VPk
kTVPTVP
o
oo
o
oo
oooooo
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Pure helium gas isadmitted into a leakproof cylindercontaining a movablepiston. The initialvolume, pressure,
and temperature ofthe gas are 15 L, 2.0atm, and 300 K. Ifthe volume isdecreased to 12 Land the pressureincreased to 3.5 atm,find the finaltemperature of thegas.
)2)(15(
)300)(5.3)(12(
T
VP
PVT
TT
PV
T
VP
oo
o
o
oo
T= 420 K
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Combining Boyles and Charles laws allowsfor developing a single equation:
P*V = n*R*T
P = pressure
V = volume
n = number of molesR = universal gas constant
T = temperature
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nRTPV
KmolatmLR
R
nTPV
/0821.0
KmolJ8.31ConstantGasUniversalR
alityproportionofconstant
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PV = nRT
The conditions 0 0C and 1 atm are calledstandard temperature and pressure (STP).
Experiments show that at STP, 1mole of an ideal gas occupies
22.414 L.
R =
PV
nT =
(1 atm)(22.414L)
(1 mol)(273.15 K)
R= 0.082057 L atm / (mol K)
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R can also be expressed asR = 8.314 J/mol K
The value of the constantdepends on the units used todefine the other variables.
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P*V = n*R*T Learn it!
Use it!
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PV = nRT
V =nRT
P
T= 0 0C = 273.15 K
P = 1atm
n= 49.8 g x1 mol HCl
36.45 g HCl= 1.37 mol
V= 1 atm
1.37 x 0.0821 x
273.15 K
V= 30.7 L
[R= 0.082057 L atm /(mol K)]
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PV = nRT
V =nRT
P
V= 1 atm x1.01 x105
1.37 x 8.314 x 273.15 K
V= 0 .0307 m3
T= 0 0C =273.15 K
P = 1 atm =1.01 x105 Pa
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How many particles are in 49.8 g of HCl atSTP?
How many particles are in 1 mole of HCl?
n= 49.8 g x1 mol HCl
36.45 g HCl= 1.37 mol
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When measured at STP, a quantity of gas hasa volume of 500 dm3. What volume will itoccupy at
0 oC and 93.3 kPa?
1 1 2 2
1 2
PV PV T T
P1 = 101.3 kPa
T1 = 273 KV1 = 500 dm
3
P2 = 93.3 kPaT2 = 0
oC + 273 = 273 K
V2 = X dm3
(101.3 kPa) x (500 dm
3
) = (93.3kPa) x (V2)273 K 273 K
V2 = 542.9 dm3
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When a scuba diver is severalhundred feet under water, thehigh pressures cause N2 from thetank air to dissolve in the blood.If the diver rises too fast, thedissolved N2 will form bubbles inthe blood, a dangerous andpainful condition called "thebends". Helium, which is inert,less dense, and does not dissolvein the blood, is mixed with O2 inscuba tanks used for deepdescents.