Chapter 5 GasesI. Pressure
A. Properties of Gases• Expand to completely fill their container• Take the Shape of their container• Low Density, much less than solid or liquid state• Compressible• Mixtures of gases are always homogeneous• Fluid
B. Pressure = total force applied to a certain area1. larger force = larger pressure2. smaller area = larger pressure3. Gas pressure caused by gas molecules colliding with container or surface4. More forceful or frequent collisions mean higher gas pressure
AFP
C. Measuring Pressure1. Atmospheric Pressure is constantly present
a. Decreases with altitude because of less airb. Varies with weather conditions
2. Measuring pressure using a barometera. Column of mercury supported by air pressureb. Longer mercury column = higher pressurec. Force of the air on the surface of the mercury is balanced by the pull
of gravity on the column of mercury
3. Units of Gas Pressurea. atmosphere (atm) = 29.92 in Hg = 760 mm Hgb. Torr = mm Hg 1 atm = 760 torrc. Pascal (Pa) = 1 N/m2 1 atm = 101,325 Pa
4. Example: Convert 49 torr to atm and Pa
PatorrPa
torratmtorratm
torr 6500760
325,10149 064.0
7601
49
5. Measuring the Pressure of a Trapped Gas Samplea. Use an manometer to compare to atmospheric pressureb. Open-end manometer
i. if gas end lower than open end, Pgas = Pair + h
ii. if gas end higher than open end, Pgas = Pair – h
II. Simple Gas LawsA. Boyle’s Law = Pressure is inversely proportional to Volume (constant T, n)
1. PV = k or P1V1 = P2V2 (k = Boyle’s Law Constant)
2. As Pressure on a gas increases, the Volume decreases
3. Example: What is new V of 1.53L of SO2 at 5600Pa when changed to 15,000Pa?
V = k(1/P)
L
PaLPa
PVP
VVPVP 57.0000,15
53.1600,5
2
1122211
B. Charles’ Law = V is directly proportional to T (constant P, n)1. V = bT2. Another way to write Charles Law
3. As the temperature decreases, the volume of a gas decreases as well4. Absolute Zero
a. Theoretical temperature at which a gas would have zero V and Pb. 0 K = -273.2 °C = -459 °F (K = oC + 273)c. All gas law problems use Kelvin temperature scale!
5. Example: V = ? If 2.58L of gas at 15 oC is heated to 38 oC at same P?
2
2
1
1
TV
bTV
L
KLK
KLK
TVT
VTV
TV
79.2288
58.231127315
58.227338
1
122
2
2
1
1
C. Avogadro’s Law = V directly proportional to moles of gas (Constant T, P)1. V = an 2. Another way to write Avagodro’s Law:3. More gas molecules = larger volume4. Count number of gas molecules by moles = n5. One mole of any gas occupies 22.414 L (at 1 atm, 0 oC) = molar volume6. Equal volumes of gases contain equal numbers of molecules
7. Example: V of O3 = ? If we convert 11.2L (0.50 mol) O2 to O3?
a. 3O2(g) -----> 2O3(g)b.
c.
2
2
1
1
nV
anV
32
32 O mol 0.33
O mol 3O mol 2
O mol 0.50
L
molLmol
nVnV
nV
nV 4.7
50.02.1133.0
1
122
2
2
1
1
III. Ideal Gas Law A. By combining the constants from the 3 gas laws we can write a general
equation1. Each simple gas law holds something constant2. To consider changes in P, V, T, n at the same time, we combine constants
3. PV = nRT is called the Ideal Gas Law4. R is called the gas constant
a. The value of R depends on the units of P and Vb. Generally use R = 0.08206Latm/Kmol when P in atm and V in L
B. Only “Ideal gases” obey this law exactly1. Most gases obey when P is low (< 1 atm) and T is high (> 0°C)2. An ideal gas is only a hypothetical substance3. Constant conditions drop out of the equation to give simpler laws
a. If you hold n constantb.
nRTPV
PTn
RP
TnkbaanbT
Pk
V PT,nP,nT,
2
22
1
11
TVP
TVP
TPV
nRTPV nR
C. Ideal Gas Law Problems1. Example: How many moles of H2 gas if: 8.56L, 0 oC, 1.5atm?
2. Example: P = ? if 7.0 ml gas at 1.68 atm is compressed to 2.7 ml?
mol57.0273K
Kmolm0.08206Lat8.56L1.5atm
RTPV
nnRTPV
Law sBoyle' toReducesconstantnRTPV
4.4atm2.7ml
7.0ml1.68atmVVP
PVPVP2
1122211
3. Example: V = ? If 345 torr, -15 oC, 3.48L changes to 36 oC, 468torr?
4. Example: V = ? If 0.35mol, 13oC, 568torr changes to 56oC, 897torr
LLL
torratmtorr
KKmol
Latmmol
torratmtorr
KKmol
Latmmol3118
7601568
28608206.035.0
7601897
32908206.035.0ΔV
2
22
1
11
TVP
TVP
toReducesconstantnRT
PVnRTPV
L
KtorrLtorrK
07.3258468
48.3345309TPVPT
VTVP
TVP
12
1122
2
22
1
11
1
111
2
22212 P
TRnP
TRnV-VΔV
PnRT
VnRTPV
D. Gas Stoichiometry1. Molar volume = volume one mole of any gas occupies2. Standard Temperature and Pressure (STP) are 0 oC and 1atm
3. Real gases differ only slightly from the ideal molar volume
4. Example: n = ? For 1.75L of N2 at STP?
5. Example: V(CO2) = ? If 152g CaCO3 decomposes to CaO+CO2?
a. CaCO3(s) -------> CaO(s) + CO2(g)
b. Assume ideal gas behavior for CO2
Latm
KKmolLatmmol42.22
000.1)2.273)(/08206.0)(1(
PnRT
VnRTPV
molL
molL 0781.0
42.221
75.1
22
23
23
3
33
CO L 1.34mol
22.42LCO mol 52.1
CO mol 52.1CaCO 1mol
CO mol 1CaCO mol 52.1
CaCO 100.09gCaCO 1mol
CaCO 152
g
6. Example: a. 2.80 L of CH4 at 25 oC and 1.65atm
b. 35.0L of O2 at 31 oC and 1.25 atm
c. What is the volume of CO2 formed at 2.50 atm and 125 oC?
d. CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g)
e. Find Limiting reagent
f. Calculate the amount of CO2 product based on CH4 as limiting reagent
g. Use the ideal gas law to find the volume of this amount of CO2
mol
KKmolLatmLatm
OnmolKKmolLatm
LatmRTPV
CHn 75.1304/08206.0
0.3525.1)( 189.0
298/08206.080.265.1
)( 24
24
24 CO mol 0.189
CH mol 1molCO 1
CH mol189.0
Latm
KKmolLatmmol47.2
50.2)398)(/08206.0)(189.0(
PnRT
VnRTPV
7. Calculating Molar Mass of a Gas from density Example:a.
b. Example: d = 1.95g/L at 1.50atm and 27 oC. Molar Mass = ?c. Assume we have 1.0L of this gas
d. Calculate Molar Mass from its definition
Vm
d molgrams
MassMolar
gLLg
dVmVm
d 95.1195.1
mol
KKmolLatmLatm
0609.0300)/08206.0(
1)50.1(RTPV
nnRTPV
molgmolg
molgrams
MassMolar /0.320609.0
95.1