# Physical Characteristics Of Gases

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Information about Physical Characteristics Of Gases

Published on June 5, 2008

Author: shawnschlueter

Source: slideshare.net

## Description

Presented by Allie Schoen

Physical Characteristics of Gases Contents Kinetic Molecular Theory Define pressure & units Gas Laws with relationship between pressure, temperature, volume, and quantity of gases.

Kinetic Molecular Theory The kinetic molecular theory is based on the idea that particles of matter are always in motion. The theory provides a model of an ideal gas. ideal gas = imaginary gas that perfectly fits all the assumptions of the kinetic molecular theory .

The kinetic molecular theory is based on the idea that particles of matter are always in motion. The theory provides a model of an ideal gas.

ideal gas = imaginary gas that perfectly fits all the assumptions of the kinetic molecular theory .

5 Assumptions of kinetic molecular theory 5 Assumptions of kinetic molecular theory 1. Gases made of tiny particles far apart relative to their size. 2. Collisions between gas particles and between particles and container walls are elastic collisions. elastic collisions = one in which there is no net loss of kinetic energy 3. Gas particles are in continuous random motion .

5 Assumptions of kinetic molecular theory

1. Gases made of tiny particles far apart relative to their size.

2. Collisions between gas particles and between particles and container walls are elastic collisions.

elastic collisions = one in which there is no net loss of kinetic energy

3. Gas particles are in continuous random motion .

4. There are no forces of attraction or repulsion between gas particles . 5. The average kinetic energy of gas particles depends on the temperature of the gas. K E = ½mv 2 m=mass v=velocity All gases at the same temperature have the same average kinetic energy . Thus, the smaller mass particles-have higher velocity . **Gases behave like ideal gases if the pressure is not very high or the temperature is not very low .**

4. There are no forces of attraction or repulsion between gas particles .

5. The average kinetic energy of gas particles depends on the temperature of the gas.

K E = ½mv 2

m=mass

v=velocity

All gases at the same temperature have the same average kinetic energy .

Thus, the smaller mass particles-have higher velocity .

**Gases behave like ideal gases if the pressure is not very high or the temperature is not very low .**

Properties of Real Gases real gas = a gas that does not behave according to the assumptions of the kinetic molecular theory 1. no definite shape or no definite volume 2. expands to fill the container 3. low density 4. compressible 5. exhibits diffusion and effusion diffusion = spreading out of particles from high concentration to low effusion = process that particles pass through a small opening

1. no definite shape or no definite volume

2. expands to fill the container

3. low density

4. compressible

5. exhibits diffusion and effusion

diffusion = spreading out of particles from high concentration to low

effusion = process that particles pass through a small opening

Kinetic molecular theory will hold true for real gases if the pressure is not very high and the temperature is not very low. Plus nonpolar gases will hold closer to the kinetic theory than polar gases . Which gases would deviate significantly from the kinetic molecular theory? He, O 2 , H 2 , H 2 0, N 2 , HCl, NH 3 ? H 2 0, HCl, NH 3 (Polar gases) To describe a gas, 4 measurements must be stated: volume, temperature, pressure, number of gas particles

Kinetic molecular theory will hold true for real gases if the pressure is not very high and the temperature is not very low. Plus nonpolar gases will hold closer to the kinetic theory than polar gases .

Which gases would deviate significantly from the kinetic molecular theory?

He, O 2 , H 2 , H 2 0, N 2 , HCl, NH 3 ?

H 2 0, HCl, NH 3 (Polar gases)

To describe a gas, 4 measurements must be stated:

volume, temperature, pressure, number of gas particles

Pressure & Units pressure = force per area pressure = force area newton = SI unit of force 1N = 1kg . m/s 2 Units of pressure Standard pressure = pressure at sea level and 0 o C 1 atmosphere = 101.3 kilopascals = 760millimeters of Hg = 760 torr 1 atm =101.3 kPa =760 mm of Hg = 760 torr Pressure is measured by a barometer or manometer.

pressure = force per area

pressure = force

area

newton = SI unit of force

1N = 1kg . m/s 2

Units of pressure

Standard pressure = pressure at sea level and 0 o C

1 atmosphere = 101.3 kilopascals = 760millimeters of Hg = 760 torr

1 atm =101.3 kPa =760 mm of Hg = 760 torr

Pressure is measured by a barometer or manometer.

STP STP = standard temperature and pressure temperature = average kinetic energy of the particles Standard temperature = O o C or 273 K Convert the following: 35 o C = ___________ K 231 K _______ o C

STP = standard temperature and pressure

temperature = average kinetic energy of the particles

Standard temperature = O o C or 273 K

Convert the following: 35 o C = ___________ K

231 K _______ o C

Pressure conversion problems Convert the following 0.830 atm = _____________ mm of Hg .830 atm X 760 mm of Hg = 631 mm of Hg 1 1 atm 98.1 kPa = ______________ torr 98.1kPa X 760 torr = 736 torr 1 101.3 kPa 755 mm of Hg = ____________ kPa 755 mm of Hg X 101.3 kPa = 101 kPa 1 760 mm of Hg

Convert the following

0.830 atm = _____________ mm of Hg

.830 atm X 760 mm of Hg = 631 mm of Hg

1 1 atm

98.1 kPa = ______________ torr

98.1kPa X 760 torr = 736 torr

1 101.3 kPa

755 mm of Hg = ____________ kPa

755 mm of Hg X 101.3 kPa = 101 kPa

1 760 mm of Hg

Open Manometer An open manometer is filed with mercury and connected to a container of hydrogen. The mercury level is 62 mm higher in the arm of the tube connected to the gas. Atmospheric pressure is 97.7 kPa. What is the pressure of the hydrogen in kilopascals? 62 mm X 101.3 kPa = 8.3 kPa 1 760 mm of Hg 97.7 kPa -8.3 kPa 89.4 kPa

An open manometer is filed with mercury and connected to a container of hydrogen. The mercury level is 62 mm higher in the arm of the tube connected to the gas. Atmospheric pressure is 97.7 kPa. What is the pressure of the hydrogen in kilopascals?

62 mm X 101.3 kPa = 8.3 kPa

1 760 mm of Hg

97.7 kPa

-8.3 kPa

89.4 kPa

Closed Manometer A closed manometer is connected to a container of nitrogen. The difference in the height of mercury in the two arms is 691 mm. What is the pressure of the nitrogen in kilopascals? 691 mm of Hg X 101.3 kPa = 92.1 kPa 1 760 mm of Hg

A closed manometer is connected to a container of nitrogen. The difference in the height of mercury in the two arms is 691 mm. What is the pressure of the nitrogen in kilopascals?

691 mm of Hg X 101.3 kPa = 92.1 kPa

1 760 mm of Hg

Boyle’s Law Boyle’s Law (temperature constant) P 1 V 1 = P 2 V 2 A balloon filled with helium gas has a volume of 500 mL at a pressure of 1 atm. The balloon is released and reaches an altitude of 6.5 km, where the pressure is 0.5 atm. Assuming that the temperature has remained the same, what volume does the gas occupy at this height? A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume of 2.93 L, what will its pressure be, assuming constant temperature? Divers know that the pressure exerted by the water increases about 100 kPa with every 10.2 m of depth. This means that at 10.2 m below the surface, the pressure is 201 kPa; at 20.4m, the pressure is 301 kPa; and so forth. Given that the volume of a balloon is 3.5 L at STP and that the temperature of the water remains the same, what is the volume 51 m below the water’s surface?

Boyle’s Law (temperature constant)

P 1 V 1 = P 2 V 2

A balloon filled with helium gas has a volume of 500 mL at a pressure of 1 atm. The balloon is released and reaches an altitude of 6.5 km, where the pressure is 0.5 atm. Assuming that the temperature has remained the same, what volume does the gas occupy at this height?

A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume of 2.93 L, what will its pressure be, assuming constant temperature?

Divers know that the pressure exerted by the water increases about 100 kPa with every 10.2 m of depth. This means that at 10.2 m below the surface, the pressure is 201 kPa; at 20.4m, the pressure is 301 kPa; and so forth. Given that the volume of a balloon is 3.5 L at STP and that the temperature of the water remains the same, what is the volume 51 m below the water’s surface?

Solution to Boyle’s Law 1. 500 mL X 1 atm = V 2 X 0.5 atm 1000 mL = V 2 2. 7.40 L X 1.26 atm = 2.93 L X P 2 3.18 atm = P 2 100 kPa = P2 10.2 m 51 5 x 10 2 kPa 101.3 kPa (atmospheric pressure) +500 kPa 600 kPa 3.5 L X 101.3 kPa = V 2 X 600 kPa 0.6 L = V 2

1. 500 mL X 1 atm = V 2 X 0.5 atm

1000 mL = V 2

2. 7.40 L X 1.26 atm = 2.93 L X P 2

3.18 atm = P 2

100 kPa = P2

10.2 m 51

5 x 10 2 kPa

101.3 kPa (atmospheric pressure)

+500 kPa

600 kPa

3.5 L X 101.3 kPa = V 2 X 600 kPa

0.6 L = V 2

Charles’s Law Charles’s Law (pressure is constant) V 1 = V 2 T 1 T 2 Temperature must be in Kelvin !! A helium-filled balloon has a volume of 2.75 L at 20.0 o C. The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature in K? In o C? A gas at 65 o C occupies 4.22 L. At what Celsius temperature will the volume be 3.87 L, assuming the same pressure? Answer to Problems 2.75L = 2.46L 293.0 K T 2 T 2 = 262 K = -11 o C 2. 4.22L = 3.87 L 338K T 2 T 2 = 3.10 X 10 2 K = 37.0 o C

Charles’s Law (pressure is constant)

V 1 = V 2

T 1 T 2

Temperature must be in Kelvin !!

A helium-filled balloon has a volume of 2.75 L at 20.0 o C. The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature in K? In o C?

A gas at 65 o C occupies 4.22 L. At what Celsius temperature will the volume be 3.87 L, assuming the same pressure?

2.75L = 2.46L

293.0 K T 2

T 2 = 262 K = -11 o C

2. 4.22L = 3.87 L

338K T 2

T 2 = 3.10 X 10 2 K = 37.0 o C

Gay Lussac’s Law Gay Lussac’s Law (volume is constant) P 1 = P 2 T 1 T 2 Before a trip from New York to Boston, the pressure in an automobile tire is 1.8 atm at 20 o C. At the end of the trip, the pressure gauge reads 1.9 atm. What is the Celsius temperature of the air inside the tire? (Assume tires with constant volume.) 1.8 atm = 1.9 atm T 2 = 309 K = 36 o C 293 K T 2 At 120.0 o C, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at 205 o C, assuming constant volume? 1.07 atm = P 2 P 2 = 1.30 atm 393k 478 K A sample of helium gas has a pressure of 1.20 atm at 22 o C. At what Celsius temperature will the helium reach a pressure of 2.00 atm ? 1.20 atm = 2.00 atm T 2 = 492 K = 219 o C 295K T 2

Gay Lussac’s Law (volume is constant)

P 1 = P 2

T 1 T 2

Before a trip from New York to Boston, the pressure in an automobile tire is 1.8 atm at 20 o C. At the end of the trip, the pressure gauge reads 1.9 atm. What is the Celsius temperature of the air inside the tire? (Assume tires with constant volume.)

1.8 atm = 1.9 atm T 2 = 309 K = 36 o C

293 K T 2

At 120.0 o C, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at 205 o C, assuming constant volume?

1.07 atm = P 2 P 2 = 1.30 atm

393k 478 K

A sample of helium gas has a pressure of 1.20 atm at 22 o C. At what Celsius temperature will the helium reach a pressure of 2.00 atm ?

1.20 atm = 2.00 atm T 2 = 492 K = 219 o C

295K T 2

Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 The volume of a gas is 27.5 mL at 22.0 o C and 0.974 atm. What will the volume be at 15.0 o C and 0.993 atm? 2. A 700.0 mL gas sample of at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.0 o C. What is the new pressure of the gas in kPa? Solution .974 atm . 27.5mL = . 993atm . V 2 295K 288K V 2 = 26.3 mL 101.3kPa . 700.0mL = P 2 . 200.0mL 273 K 303 K P 2 = 394 kPa

P 1 V 1 = P 2 V 2

T 1 T 2

The volume of a gas is 27.5 mL at 22.0 o C and 0.974 atm. What will the volume be at 15.0 o C and 0.993 atm?

2. A 700.0 mL gas sample of at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.0 o C. What is the new pressure of the gas in kPa?

Solution

.974 atm . 27.5mL = . 993atm . V 2

295K 288K

V 2 = 26.3 mL

101.3kPa . 700.0mL = P 2 . 200.0mL

273 K 303 K

P 2 = 394 kPa

Dalton’s Law of Partial Pressure Dalton’s Law of Partial Pressure = total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases Gases collected by water displacement contain water vapor thus exerts a pressure which is affected by the temperature. Atmospheric pressure = pressure of dry gas + water vapor pressure (refer to chart) Some hydrogen gas is collected over water at 20.0 o C. The levels of water inside and outside the gas-collection bottle are the same. The partial pressure of hydrogen is 742.5 torr. What is the barometric pressure at the time the gas is collected? 742.5 torr +17.5 torr 760.0 torr Helium gas is collected over water at 25 o C. What is the partial pressure of the helium, given that the barometric pressure is 750.0 mm Hg? 750.0 mm Hg -23.8 mm Hg 726.2 mm Hg

Dalton’s Law of Partial Pressure = total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases

Gases collected by water displacement contain water vapor thus exerts a pressure which is affected by the temperature.

Atmospheric pressure = pressure of dry gas + water vapor pressure

(refer to chart)

Some hydrogen gas is collected over water at 20.0 o C. The levels of water inside and outside the gas-collection bottle are the same. The partial pressure of hydrogen is 742.5 torr. What is the barometric pressure at the time the gas is collected?

742.5 torr

+17.5 torr

760.0 torr

Helium gas is collected over water at 25 o C. What is the partial pressure of the helium, given that the barometric pressure is 750.0 mm Hg?

750.0 mm Hg

-23.8 mm Hg

726.2 mm Hg

Review Materials Internet Gas Law Tutorial tutorials with diagrams Online quizes Boyle's Law Demonstator Charles' and Lussacs Demonstrator

Internet Gas Law Tutorial

tutorials with diagrams

Online quizes

Boyle's Law Demonstator

Charles' and Lussacs Demonstrator

TEKS 2cde,3ae,4bcd,7ab 2c express and manipulate chemical quantities using scientific conventions and mathematical procedures 2d organize, analyze, evaluate, make inferences, predict trends from data 2e communicate valid conclusions 3a analyze, review, and critique scientific explanations 3e research and describe the history of chemistry and contributions of scientists 4b analyze samples of solids, liquids, and gases 4c investigate and identify properties of mixtures and pure substances 4d describe the physical and chemical characteristics of an element using periodic table 7a describe relationships among temperature, particle number, pressure and volume of gases contained within a closed system 7b illustrate data obtained from investigations with gases in a closed system and determine if the data are consistent with the Universal Gas Law

2c express and manipulate chemical quantities using scientific conventions and mathematical procedures

2d organize, analyze, evaluate, make inferences, predict trends from data

2e communicate valid conclusions

3a analyze, review, and critique scientific explanations

3e research and describe the history of chemistry and contributions of scientists

4b analyze samples of solids, liquids, and gases

4c investigate and identify properties of mixtures and pure substances

4d describe the physical and chemical characteristics of an element using periodic table

7a describe relationships among temperature, particle number, pressure and volume of gases contained within a closed system

7b illustrate data obtained from investigations with gases in a closed system and determine if the data are consistent with the Universal Gas Law

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