# Ms. Lina Ai

Chapter 5

1

Elements that exist as gases at 250C and 1 atmosphere

2

3

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. Gases have much lower densities than liquids and solids.

NO2 gas

4

Force Pressure = Area

(force = mass x acceleration)

Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa Barometer For measuring atmospheric pressure 5

The force experienced by any area exposed to Earth’s atmosphere is equal to the weight of that column of air above it.

10 miles

0.2 atm

4 miles Sea level

0.5 atm 1 atm

6

Manometers Used to Measure Gas Pressures

closed-tube open-tube Patm

h = PHg

Pgas>Patm Pgas = Patm + PHg

Measures pressures below atm. pressure Measures pressures ≥ atm. pressure 7

What is the pressure in atmospheres in a cabin if the barometer reading is 672 mmHg? 672 mmHg x 1 atm/760 mmHg = 0.884 atm 8

closed-tube

The hight, h, of liquid level depends on the density of that liquid

For two liquids A and B h(A)/h(B) = d(A)/d(B) (hxd)B = (hxd)A

Measures pressures below atm. pressure

9

Referring to the following scheme: A liquid (d=0.845 g/ml) has a height of 17.5 mm in an open-tube manometer at 745.0 mmHg of atmospheric pressure. Calculate the gas pressure in torr if the density of mercury is 13.6 g/ml? Given: d(L) = 0.845 g/ml dHg = 13.6 g/ml h(L) = 17.5 mm Patm = 745 mmHg From the scheme Pgas>Patm Pgas = Patm + PHg (hxd)L = (hxd)Hg hHg = 17.5 x 0.854 / 13.6 = 1.10 mmHg Pgas = 745.0 + 1.10 = 746.1 torr

17.5 mm

10

Boyle’s law The pressure of a fixed amount of gas maintained at constant temperature is inversely proportional to the volume of the gas PV = 1 P = pressure (atm) n V = Volume (L or mL) 1 = constant For a sample of gas under two different sets and conditions at constant temperature, we have P1V1 = 1

= P2V2

1

= nRT

P1V1 = P2V2

11

Apparatus for Studying the Relationship Between Pressure and Volume of a Gas

P α 1/V P = K1 x 1/V

As P (h) increases

V decreases

12

Boyle’s Law

P ∝ 1/V P x V = constant P1 x V1 = P2 x V2 Constant temperature Constant amount of gas 13

Summary of Gas Laws Boyle’s Law

14

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg V1 = 946 mL P2 = P1 x V1 V2 P2 = ? V2 = 154 mL

726 mmHg x 946 mL = = 4460 mmHg 154 mL

15

Charles’s and Gay-Lussac’s law OR Charles’s law The volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas For a given sample of gas at constant pressure, we can compare two sets of volume-temperature conditions

V1/T1 = V1/T1 = PαT P = 3T

2

= V2/T2 = V2/T2

T = temperature (K) V = volume (L or mL) 2 = nR/P

2

Another form of Charles’s law shows that at constant amount of gas and volume the pressure is proportional to temperature

P/T =

3

3

= constant

Two sets of pressure-volume conditions P1/T1 = P2/T2

3

= nR/V

16

Charles Law

17

Variation in Gas Volume with Temperature at Constant Pressure

As T increases

V increases

18

Variation of Gas Volume with Temperature at Constant Pressure Notice that the temperature at which the gas has zero volume is zero Kelvin = -273.15 °C

Charles’ & Gay-Lussac’s Law

V∝T V = constant x T V1/T1 = V2 /T2

Temperature must be in Kelvin

T (K) = t (°C) + 273.15

19

A sample of carbon monoxide gas occupies 3.20 L at 125 °C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 /T1 = V2 /T2 V1 = 3.20 L T1 = 398.15 K V2 = 1.54 L T2 = ?

T1 = 125...

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