Chapter 5 Gases I.Pressure A.Properties of Gases Expand to completely fill their container Take the Shape of their container Low Density, much less than.

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 area

1. larger force = larger pressure

2. smaller area = larger pressure

3. Gas pressure caused by gas molecules colliding with container or surface

4. More forceful or frequent collisions mean higher gas pressure

A

FP

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

Patorr

Patorratm

torr

atmtorr 6500

760

325,10149 064.0

760

149

5. Measuring the Pressure of a Trapped Gas Sample

a. Use an manometer to compare to atmospheric pressure

b. 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

Pa

LPa

P

VPVVPVP 57.0

000,15

53.1600,5

2

1122211

B. Charles’ Law = V is directly proportional to T (constant P, n)1. V = bT

2. Another way to write Charles Law

3. As the temperature decreases, the volume of a gas decreases as well

4. Absolute Zero

a. Theoretical temperature at which a gas would have zero V and P

b. 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

T

Vb

T

V

L

K

LK

K

LK

T

VTV

T

V

T

V79.2

288

58.2311

27315

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 volume

4. Count number of gas molecules by moles = n

5. One mole of any gas occupies 22.414 L (at 1 atm, 0 oC) = molar volume

6. 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

n

Va

n

V

32

32 O mol 0.33

O mol 3

O mol 2O mol 0.50

L

mol

Lmol

n

VnV

n

V

n

V4.7

50.0

2.1133.0

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

P

TnR

P

TnkbaanbT

P

kV PT,nP,

nT,

2

22

1

11

T

VP

T

VP

T

PV nRTPV nR

C. Ideal Gas Law Problems

1. 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?

mol57.0

273KKmol

m0.08206Lat

8.56L1.5atm

RT

PVnnRTPV

Law sBoyle' toReducesconstantnRTPV

4.4atm2.7ml

7.0ml1.68atm

V

VPPVPVP

2

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

torr

atmtorr

KKmol

Latmmol

torr

atmtorr

KKmol

Latmmol

3118

760

1568

28608206.0

35.0

760

1897

32908206.0

35.0ΔV

2

22

1

11

T

VP

T

VP toReducesconstantnR

T

PVnRTPV

L

Ktorr

LtorrK07.3

258468

48.3345309

TP

VPTV

T

VP

T

VP

12

1122

2

22

1

11

1

111

2

22212

P

TRn

P

TRnV-VΔV

P

nRTVnRTPV

D. Gas Stoichiometry

1. Molar volume = volume one mole of any gas occupies

2. 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(

P

nRT VnRTPV

molL

molL 0781.0

42.22

175.1

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 100.09g

CaCO 1molCaCO 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

KKmolLatm

LatmOnmol

KKmolLatm

Latm

RT

PVCHn 75.1

304/08206.0

0.3525.1)( 189.0

298/08206.0

80.265.1)( 24

24

24 CO mol 0.189

CH mol 1

molCO 1CH mol189.0

Latm

KKmolLatmmol47.2

50.2

)398)(/08206.0)(189.0(

P

nRT 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

V

md

mol

gramsMassMolar

gLL

gdVm

V

md 95.11

95.1

mol

KKmolLatm

Latm0609.0

300)/08206.0(

1)50.1(

RT

PV nnRTPV

molgmol

g

mol

gramsMassMolar /0.32

0609.0

95.1