review of the gas laws pv = nrt. boyle’s law (isothermal & fixed amount) charles’s law...
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Review of the Gas Laws
PV = nRT
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PV = nRT
• Boyle’s Law (isothermal & fixed amount)
• Charles’s Law (isobaric & fixed amount)
• Avogadro’s Law (isothermal & isobaric)
• ????? Law (isochoric & fixed amount)
• ????? Law (isothermal & isochoric)
• ????? Law (isobaric & isochoric)
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Boyle’s Law
atmLVP
atmLVP
KmolK
atmLmolePV
nRTPV
4.22
4.22
)273(0821.0)1(
• Pressure and volume are inversely proportional.• As pressure increases, volume decreases.• If pressure increases by 2x, volume cuts in half.
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Charles’s Law
atmmolKatmL
mole
T
V
P
nR
T
V
1
0821.0)1(
• Temperature and volume are directly proportional.
• As temperature increases, volume also increases.
• If temperature increases by 2x, volume also doubles.
• Temperature must be measured in Kelvin.
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Avogadro’s Law
mol
L
n
V
atm
KmolKatmL
n
V
P
RT
n
V
4.22
)1(
)273(0821.0
• Moles of gas and volume are directly proportional.• As the number of moles increases, the volume also
increases.• If the number of moles increases by 2x, the volume also
doubles.
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????? Lawisothermal & isochoric
mol
atm
n
P
L
KmolKatmL
n
P
V
RT
n
P
1
4.22
)273(0821.0
• Moles of gas and pressure are directly proportional.
• As the moles of gas increase, the pressure also increases.
• If the number of moles of gas increases by 2x, the pressure also doubles.
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????? Lawisobaric & isochoric
• Moles of gas and temperature are inversely proportional.
• As the number of moles of gas increase, the temperature decreases.
molKTn
molKatmLLatm
nT
R
PVnT
273
0821.0
)4.22)(1(
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Units of Pressure
• 1 atm = 760 torr = 760 mmHg
• 1 atm = 101.325 kPa
• 1 bar = 105 Pa = 100 kPa
• 1 Pa = mN
2
1
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How does 1 atm = 101.325 kPa?Let the area of the base of a cylinder = 1 m2
Volume = area x height = 1 m2 x 0.76 m = 0.76 m3
Convert volume to cubic centimeters.
363
33 1076.0
1
10076.0 cmx
m
cmxm
Use the density of mercury and the acceleration due to gravity to calculate the weight of mercury in the column.
Nxs
mx
g
kgx
cm
gxcmx 3
2336 103.101
8.9
1000
16.131076.0
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ContinueHow does 1 atm = 101.325 kPa?
Pressure is force (or weight) per unit area. Divide the weight of mercury by the area it is resting on.
kPam
Nx
m
Nx
area
Nx101103.101
1
103.101103.1012
32
33
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Let the area of the cylinder = 1cm2
kPa
m
Nx
m
cmx
cm
N
A
N
Nors
mkg
s
mx
g
kgx
cm
gxcm
cmAxhV
cmA
cmh
101103.1011
100
1
13.1013.10
13.1013.108.9
1000
16.1376
76
1
76
23
2
2
2
2233
3
2
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Barometric Formula
As elevation increases, the height of the atmosphere decreases and its pressure decreases.
hhgP
Check units.
22
2
223 m
N
msm
kg
sm
kgmx
s
mx
m
kg
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Continue Derivation of Barometric Formula
Write in differential form. gdhdP
density V
Mmoles
volume
mass W
Rewrite PV = nRT asRT
P
V
n
Therefore,RT
PMW
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Continue Derivation of Barometric Formula
Substitute the expression for density into the differential eqn.
dhRT
gPMdP W
Divide both sides of the above equation by P and integrate.
dhRT
gM
P
dP W
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Continue Derivation of Barometric Formula
Integration of the left side and moving the constants outside the integral on the right side of the differential equation gives,
hRT
gMdh
RT
gMP WWln
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Continue Derivation of Barometric Formula
Evaluating the integral between the limits of Pi at zero height and Pf at height h, gives
RT
ghM
P
PW
i
f
ln
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Sample Problem Using the Barometric Formula
torrP
P
P
P
P
KmolKJ
msm
molkg
x
P
P
RT
ghM
P
P
f
i
f
i
f
i
f
W
i
f
464
6099.0
4944.0ln
)293(314.8
42678.9108.28ln
ln
23
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Dalton’s Law of Partial Pressures
Pressure is additive.baT PPP
Write each pressure as, V
nRTP
V
RTn
V
RTn
V
RTn baT
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ContinueDalton’s Law of Partial Pressures
V
RTn
V
RTn
V
RTn baT
Multiply through by V (the combined volume of the gases) and divide by R T.
baT nnn
Moles are indeed additive.
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Mole Fraction & Partial Pressure
T
bb
T
aa n
nXand
n
nX
T
T
T
ba
T
b
T
aba n
nor
n
nn
n
n
n
nXX
Therefore, 1 ba XX
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ContinueMole Fraction & Partial Pressure
Show that TbbTaa PXPPXP &
T
bb
T
aa P
PX
P
PX &
1
T
T
T
ba
T
b
T
aba P
P
P
PP
P
P
P
PXX