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# How Do Gases Behave - PowerPoint PPT Presentation

How Do Gases Behave?. What is a solid, liquid or gas?. Help Marvin the Martian understand what a solid, liquid and gas are! Draw what solids, liquids, gases look like Describe physical/chemical properties What would happen if we changed pressure? What would happen if we changed temperature?.

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• Help Marvin the Martian understand what a solid, liquid and gas are!

• Draw what solids, liquids, gases look like

• Describe physical/chemical properties

• What would happen if we changed pressure?

• What would happen if we changed temperature?

• Pressure = Force/Area

• 1 atmosphere (atm)

• 760 Torr

• 760 mmHg

• 1.01 Bar

• 101,327 Pascal

• 101.3 Kpa

• 14.7 lbs/in2

• Measured with

a barometer

• Column of mercury to measure pressure.

• h is how much lower or higher the pressure is than outside.

• Pgas = Patm - h

• Pgas = Patm + h

h

h

• Average Kinetic Energy (1/2 mv2) of an atom or molecule

• Measured in Fahrenheit, Celsius or Kelvin (SI)

• F = (C x 1.8) + 32

• K = C + 273

• 0 Kelvin: absolute zero (atom stops moving completely)

• Is there a maximum temperature in the universe?

• Theory explains why ideal gases behave the way they do.

• Assumptions that simplify the theory, but don’t work in real gases.

• The particles are so small we can ignore their volume.

• The particles are in constant motion and their collisions cause pressure.

• The particles do not affect each other, neither attracting or repelling.

• The average kinetic energy is proportional to the Kelvin temperature.

• The molecules move in straight path and all collisions are elastic

• An ideal gas or perfect gas is a hypothetical gas consisting of identical particles of:

• Negligible volume

• With no intermolecular forces

• Atoms or molecules undergo perfectly elastic collisions with the walls of the container

• Ideal gas law calculations are favored at low pressures and high temperatures.

• Real gases existing in reality do not exhibit these exact properties, although the approximation is often good enough to describe real gases.

What is Boyle’s Law?

• In the mid 1600's, Robert Boyle studied the relationship between the pressure P and the volume V of a confined gas held at a constant temperature.

• Boyle observed that the product of the pressure and volume are observed to be nearly constant.

• The product of pressure and volume is exactly a constant for an ideal gas.

• P * V = constant

• This relationship between pressure and volume is called Boyle's Lawin his honor.

V

P (at constant T)

V

1/P (at constant T)

O2

PV

CO2

P (at constant T)

20.5 L of nitrogen at 25ºC and 742 torr are compressed to 9.8 atm at constant T. What is the new volume?

P1V1=P2V2

(0.98 atm)(20.5 L) = (9.8 atm)(V2)

20.09 atm*L/9.8 atm= V2

2.1 L = V2

30.6 mL of carbon dioxide at 740 torr is expanded at constant temperature to 750 mL. What is the final pressure in kPa?

(30.6mL)(740Torr)=(750 mL)(P2)

22,644 mL*Torr/750 mL= P2

30.2 Torr =P2

(30.2 Torr) (1 atm/760 Torr) (101.3 kPa/1 atm)=

3059.3 kPa/760= 4.03 kPa

What is Charles’ Law?

• The relationship between temperature and volume, at a constant number of moles and pressure, is called Charles and Gay-Lussac's Law in honor of the two French scientists who first investigated this relationship.

• Charles did the original work, which was verified by Gay-Lussac. They observed that if the pressure is held constant, the volume V is equal to a constant times the temperature T, or:

• V / T= constant

He

CH4

H2O

V (L)

H2

-273.15ºC

T (ºC)

What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant?

247 ml/295 K = X ml/371 K

91,637 mL * K = 295 X * K

91,637 mL * K/295 K= X

310 mL = X

If the volume of oxygen at 21 °C is 785 L, at what temperature would oxygen occupy 804 L?

785 L/294 K = 804 L/X K

785 X = 236,376

X = 236,376/785

X= 301 K = 28 °C

• Combining Charles’s Law and Boyle’s Law in a single statement:

• P1V1/T1 = P2V2/T2

• 39.8 mg of caffeine gives 10.1 mL of nitrogen gas at 23°C and 746 mmHg. What is the volume of nitrogen at 0°C and 760 mmHg?

• First change temperature to Kelvin

• V1 = 10.1mL P1 = 746 mmHg K1 = 296 K

• V2 = ? P2 = 760 mmHg K2 = 273 K

10.1 * 746/296 = V2 * 760/273

V2 = 9.14 mL

• Gay-Lussac Law

• At constant volume, pressure and absolute temperature are directly related.

• P/T = k (constant)

• At constant temperature and pressure, the volume of gas is directly related to the number of moles.

• V /n= k (n is the number of moles)

What equation would we get if we combined them all?

• Combining Boyle’s Law, Charles’ law & Avogadro’s Law we derive the Ideal Gas Law:

• P V = n R T

• P = Pressure (atm)

• V = Volume (L)

• n = # moles (0.0821 L atm /mol K)

• R = Gas Constant

• T = Temperature (K)

• Ideal gas law calculations are favored at low pressures and high temperatures

• Tells you about a gas NOW.

• The other laws tell you about a gas when it changes.

• Example:

• If we had 1.0 mol of gas at 1.0 atm of pressure at 0°C (STP), what would be the volume?

• PV = nRT

• V = nRT/P

• V = (1.0 mol)(0.0821 L atm/mol K)(273 K)/(1.0 atm)

• V = 22.41 L

• 1 mole of ANY gas at STP will occupy 22.4 Liters of volume

• D = m/V

• Let M stand for molar mass

• M = m/n

• n = m/M

• PV = nRT

• PV = (m/M) RT

• P = mRT/VM = (m/V)(RT/M)

• P = d RT/M

• PM/RT = d (density)

• What is the density of ammonia at 23ºC and 735 torr?

• Units must be: atm, K

• 735 torr(1 atm/760 torr) = 0.967 atm

• 23 + 273 = 296 K

• Molar mass of NH3 = 17.0 g

d = 0.967 * 17.0 g

(0.0821 L* atm/mol * K)(296 K)

d = 0.676 g / L

• Reactions happen in moles

• At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occupies 22.42 L.

• If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.

• Consider the following reaction:

• Suppose you heat 0.0100 mol of potassium chlorate, KClO3, in a test tube. How many liters of oxygen can you produce at 298 K and 1.02 atm?

• Break it into 2 problems, one involving stoichiometry and the other using the ideal gas law

0.0100 mol KClO3 X 3 mol O2/2 mol KClO3

= 0.0150 mol O2

Now that you have the moles of oxygen use the ideal gas law to calculate the volume:

V = nRT/P

0.0150 mol x 0.0821 L * atm (K * mol) x 298 K

1.02 atm

V = 0.360 L

• Using the following reaction

• Calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25ºC and 2.00 atm.

n = PV/RT = (2.00 atm)(2.87 L)

(0.0821 L*atm/K*mol)(298 K)

n= 0.235 mol CO2

0.235 mol CO2 (1 mol NaHCO3) ( 84.0 g)

(1 mol CO2 ) (1 mol NaHCO3)

19.7 g NaHCO3

• The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container.

• The total pressure is the sum of the partial pressures.

• PTotal = P1 + P2 + P3 + P4 + P5 ...

• For each P = nRT/V

• PTotal = n1RT + n2RT + n3RT +... V VV

• In the same container R, T and V are the same.

• PTotal = (n1+ n2 + n3+...)RT V

• PTotal = (nTotal)RT V

• Ratio of moles of the substance to the total moles.

• symbol is Greek letter chi c

• Because pressure of a gas is proportional to moles, for fixed volume and temperature then,

• c1 = n1= P1nTotalPTotal

• A 1.00 L sample of dry air at 25°C and 786 mmHg contains 0.925 g N2, plus other gases including oxygen, argon and carbon dioxide.

• What is the partial pressure (in mmHg) of N2 in the air sample?

• What is the mole fraction and mole percent of N2 in the mixture?

• Convert grams into moles Mixture

0.925 g N2 x (1 mol N2/28.0g N2)

=0.0330 mol N2

• Substitute into ideal gas law

PN2 = nN2RT/V

=0.0330mol x 0.0821 L*atm/K*mol x 298

1.00 L

=0.807 atm = 613 mmHg

• The mole fraction of N Mixture2 in the air is

= PN2/P = 613 mmHg/786 mmHg

=0.780

• Mole percent equals mole fraction x 100

=0.780 x 100 = 78%

• Air contains 78.0 mole percent of N2

Vapor Pressure Mixture

• Water evaporates!

• When that water evaporates, the vapor has a pressure.

• Gases are often collected over water so the vapor pressure of water must be subtracted from the total pressure.

• Vapor pressure varies by temperature and must be given in the problem or in a table.

• First find the Partial Pressure. The vapor pressure of water at 19°C is 16.5 mmHg

P = PH2 + PH2O

PH2 = P - PH2O

PH2 =769 – 16.5 = 752 mmHg

• Use the ideal gas law to find the moles of hydrogen collected.

P: 752 mmHg x (1 atm/760 mmHg) = 0.989 atm

V: 156 mL x (1 L/1000 mL) = 0.156 L

T: 19 + 273 = 292 K

R: 0.0821 L*atm/K*mol

n: ?

• Solve for moles: at 19°C is 16.5 mmHg

n = PV/RT

= 0.989 x 0.156/0.0821 x 292

=0.00644 mol H2

• Convert moles to grams:

0.00644 mol H2 x (2.02g/1 mol H2)

=0.0130 g H2

What’s Diffusion and Effusion? at 19°C is 16.5 mmHg

• Only a few physical properties of gases depends on the identity of the gas.

• Diffusion - The rate at which two gases mix.

• Effusion - The rate at which a gas escapes through a pinhole into a vacuum.

What is at 19°C is 16.5 mmHgGraham’s Law?

• We know that Kinetic energy = 1/2 mv2

• If two bodies of unequal mass have the same kinetic energy, which moves faster?

• The lighter one!

• Thus, for two gases at the same temperature, the one with lower molecular mass will diffuse/effuse faster.

• The rate of effusion/diffusion of a gas is inversely proportional to the square root of its mass.

• Calculate the ratio of effusion rates of molecules of carbon dioxide and sulfur dioxide from the same container and at the same temperature and pressure:

Rate of effusion of CO2 = √Mm SO2

Rate of effusion of SO2 √Mm CO2

= √64.1/44.0 = 1.21

• In other words, carbon dioxide effuses 1.21 times faster than sulfur dioxide.