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Why gases are studied separately

- Many common compounds exist as gases
- Gases transport matter and energy across the globe (i.e., weather)
- Compared with those of liquids and solids, the behavior of gases is easiest to model.

Physical Characteristics of Gases

- Gases assume the volume and shape of their containers.
- Gases are the most compressible state of matter.
- Gases will mix evenly and completely when confined to the same container. (No solubility rules!)
- Gases have much lower densities than liquids and solids.
- Density of a gas given in g/L (vice g/mL for liquids)

NO2

Area

mass

Fig. 10.2

760 mm Hg

Pressure =

(force = mass x acceleration)

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 101,325 Pa = 101.325 kPa

1 atm = 760 mm Hg = 760 torr

Height of column ∝1/density

Pressure-Volume Relationship: Boyle’s Law

P1V1 = P2V2

- Temperature-Volume Relationship: Charles’s Law
V ∝ T

Volume-Amount Relationship: Avogadro’s Law

V ∝ n

The Ideal Gas Law: PV = nRT

P1·V1

=

355 mm Hg

P2

A sample of argon gas occupies a volume of 500. mL at a pressure of 626 mm Hg. What is the volume of the gas (in mm Hg) if the pressure is reduced at constant temperature to 355 mm Hg?

P1 = 626 mm Hg

P2 = 356 mm Hg

V1 = 500. mL

V2 = ?

P1·V1 = P2·V2

V2 =

= 882 mL

As T increases

V increases

Variation of gas volume with temperature

at constant pressure: Charles’ Law

Fig 10.8

Temperature must be in Kelvin

T (K) = t (°C) + 273.15

V∝T

V = constant ·T

Constant pressure

Constant amount of gas

6.60 L · 296.0 K

V2·T1

=

13.20 L

V1

A sample of hydrogen gas occupies 13.20 L at 22.8 °C. At what temperature will the gas occupy half that volume if the pressure remains constant?

V1 = 13.20 L

V2 = 1.60 L

T1 = 296.0 K

T2 = ?

T1 = 22.8 (°C) + 273.15 (K) = 296.0 K

T2 =

= 148 K

148 K - 273.15 = -125 °C

Avogadro’s Law

Fig 10.10

V∝ number of moles (n)

V = constant ·n

Constant pressure

Constant temperature

C3H8 + 5O2 3CO2 + 4H2O

1 mole C3H8 3 mole CO2

1 volume C3H8 3 volumes CO2

Propane burns in oxygen to form carbon dioxide and water vapor. How many volumes of carbon dioxide are obtained from one volume of propane at the same temperature and pressure?

At constant T and P

The conditions 0 °C and 1 atm are called

standard temperature and pressure (STP).

Experiments show that at STP,

1 mole of an ideal gas

occupies 22.414 L:

Fig 10.13 Comparison of Molar Volumes at STP

- One mole of an ideal gas occupies 22.41 L STP
- One mole of various real gases at STP occupy:

V =

n = (49.8 g)

= 1.37 mol

(36.45 g HCl)

(1.37 mol)(0.08216 ) (273.15 K)

V =

1 atm

nRT

L•atm

P

mol•K

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

T = 0 °C = 273.15 K

P = 1 atm

PV = nRT

V = 30.6 L

PV = nRT useful when P, V, n, and T do not change

Modify equation when P, V, and/or T change:

- Initial state (1) of gas:

Combined Gas Law

- Final state (2) of gas:

Eqn [10.8]

V

P(MM)

=

RT

dRT

P

Density (d) Calculations

m is the mass of the gas in g

d =

MM is the molar mass of the gas

Molar Mass (MM) of a Gaseous Substance

d is the density of the gas in g/L

MM =

- What happens to the density of a gas if:
- it is heated at constant volume?
- It is compressed at constant temperature?
- Additional gas is added at constant volume?

What is the volume of CO2 produced at 37.0°C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

6 mol CO2

g C6H12O6 mol C6H12O6 mol CO2V CO2

x

1 mol C6H12O6

1 mol C6H12O6

x

180 g C6H12O6

L•atm

mol•K

nRT

(0.1867 mol) (0.0821 ) (310.15 K)

=

P

1.00 atm

Volumes of Gases in Chemical Reations

= 0.1867 mol CO2

5.60 g C6H12O6

V =

= 4.75 L

Gas Mixtures and Partial Pressures

V and T are constant

P1

P2

Ptotal= P1 + P2

Dalton’s Law of Partial Pressures

PA =

nART

nBRT

V

V

PB =

nA

nB

nA + nB

nA + nB

XB =

XA =

ni

mole fraction (Xi) =

nT

Consider a case in which two gases, A and B, are in a container of volume V.

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB

PA = XAPT

PB = XBPT

Pi = Xi PT

Dalton’s Law of Partial Pressures

2KClO3 (s) 2KCl (s) + 3O2 (g)

PT = PO + PH O

2

2

Fig. 10.16 Collecting a water-insoluble gas over water

Bottle full of oxygen gas and water vapor

2KClO3 (s) 2KCl (s) + 3O2 (g)

PT = PO + PH O

2

2

Sample Exercise 10.12 p 413

A 0.811 g sample of KClO3 is partially decomposed to produce

oxygen gas over water. The volume of gas collected is 0.250 L

at 26 °C and 765 torr total pressure. How many grams of O2

are collected?

PT

PO = - PH O

2

2

PV = nRT

Kinetic Molecular Theory of Gases

- A gas is composed of widely-separated molecules. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
- Gas molecules are in constant random motion.
- Collisions among molecules are perfectly elastic.
- The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins.
- KE ∝ T

Fig 10.18 The Effect of Temperature on Molecular Speeds

urms≡

root-mean-square

speed

Hot molecules are fast, cold molecules are slow.

The distribution of speeds

of three different gases

at the same temperature

3RT

urms =

(MM)

Fig 10.19 The Effect of Molecular Mass on Molecular Speeds

R = 8.314 J/(mol K)

Heavy molecules are slow, light molecules are fast.

An “Ideal Gas” Volume of molecules themselves is negligible compared to volume of container

- An ideal gas “obeys” PV = nRT

- i.e., calculated value ≈ experimental value

Assumptions:

- Gas molecules do not exert any force (attractive or repulsive) on each other
- i.e., collisions are perfectly elastic

- i.e., the molecules are considered to be points

Deviations from Ideal Behavior

- Assumptions made in the kinetic-molecular model:
- negligible volume of gas molecules themselves
- no attractive forces between gas molecules
These breakdown at high pressure and/or low temperature.

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