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