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Gases Chapter 10 Gases

Chapter 10 Gases

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Chapter 10 Gases. Characteristics of Gases. Unlike liquids and solids, gases expand to fill their containers; are highly compressible; have extremely low densities. F A. P =. Pressure. Pressure is the amount of force applied to an area. - PowerPoint PPT Presentation

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Page 1: Chapter 10 Gases

Gases

Chapter 10Gases

Page 2: Chapter 10 Gases

Gases

Characteristics of Gases

• Unlike liquids and solids, gases– expand to fill their containers;– are highly compressible;– have extremely low densities.

Page 3: Chapter 10 Gases

Gases

• Pressure is the amount of force applied to an area.

Pressure

• Atmospheric pressure is the weight of air per unit of area.

P =FA

Page 4: Chapter 10 Gases

Gases

Units of Pressure

• Pascals– 1 Pa = 1 N/m2

• Bar– 1 bar = 105 Pa = 100 kPa

Page 5: Chapter 10 Gases

Gases

Units of Pressure

• mm Hg or torr–These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury.

• Atmosphere–1.00 atm = 760 torr = 101.3 kPa

Page 6: Chapter 10 Gases

Gases

Manometer

This device is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

Page 7: Chapter 10 Gases

Gases

Standard Pressure

• Normal atmospheric pressure at sea level is referred to as standard pressure.

• It is equal to– 1.00 atm–760 torr (760 mm Hg)–101.325 kPa

Page 8: Chapter 10 Gases

Gases

Boyle’s Law

The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

Page 9: Chapter 10 Gases

Gases

As P and V areinversely proportional

A plot of V versus P results in a curve.

Since

V = k (1/P)This means a plot of V versus 1/P will be a straight line.

PV = k

Page 10: Chapter 10 Gases

Gases

Examples• What would the final volume be if 247 mL of gas at

22ºC is heated to 98ºC , if the pressure is held constant?

Page 11: Chapter 10 Gases

Gases

Charles’s Law

• The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

A plot of V versus T will be a straight line.

• i.e., VT

= k

Page 12: Chapter 10 Gases

Gases

Examples• At what temperature would 40.5 L of gas at 23.4ºC

have a volume of 81.0 L at constant pressure?

(.wav)

Page 13: Chapter 10 Gases

Gases

Avogadro’s Law

• The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.

• Mathematically, this means V = kn

Page 14: Chapter 10 Gases

Gases

Examples• A deodorant can has a volume of 175 mL and a

pressure of 3.8 atm at 22ºC. What would the pressure be if the can was heated to 100.ºC?

• What volume of gas could the can release at 22ºC and 743 torr?

Page 15: Chapter 10 Gases

Gases

Ideal-Gas Equation

V = 1/P (Boyle’s law)V = T (Charles’s law)V = n (Avogadro’s law)

• So far we’ve seen that

• Combining these, we get

V =nTP

Page 16: Chapter 10 Gases

Gases

Ideal-Gas Equation

The constant of proportionality is known as R, the gas constant.

Page 17: Chapter 10 Gases

Gases

Ideal-Gas Equation

The relationship

then becomes

nTP

V =

nTP

V = R

or

PV = nRT

Page 18: Chapter 10 Gases

Gases

Examples• Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at

24.8ºC What is the mass of Kr?

• A sample of gas has a volume of 4.18 L at 29ºC and 732 torr. What would its volume be at 24.8ºC and 756 torr?

Page 19: Chapter 10 Gases

Gases

Examples• Mercury can be produced by the following

decomposition: • HgO • What volume of oxygen gas can be produced from

4.10 g of mercury (II) oxide at 400.ºC and 740 torr?

Page 20: Chapter 10 Gases

Gases

Densities of Gases

If we divide both sides of the ideal-gas equation by V and by RT, we get

nV

PRT

=

Page 21: Chapter 10 Gases

Gases

• We know that–moles molecular mass = mass

Densities of Gases

• So multiplying both sides by the molecular mass ( ) gives

n = m

PRT

mV

=

Page 22: Chapter 10 Gases

Gases

Densities of Gases

• Mass volume = density

• So,

Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas.

PRT

mV

=d =

Page 23: Chapter 10 Gases

Gases

Molecular Mass

We can manipulate the density equation to enable us to find the molecular mass of a gas:

Becomes

PRT

d =

dRTP =

Page 24: Chapter 10 Gases

Gases

Examples• Complete and balance the following equation:• NaHCO3 + HCl

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

• If 27 L of gas are produced at 26ºC and 745 torr when 2.6-L of HCl are added what is the concentration of HCl?

Page 25: Chapter 10 Gases

Gases

Examples• Consider the following reaction

• NH3 + O2 NO2 + H2O (balance first!)

• What volume of NO2 at 1.0 atm and 1000.ºC can be produced from 10.0 L of NH3 and excess O2 at the same temperature and pressure?

• What volume of O2 measured at STP will be consumed when 10.0 kg NH3 is reacted?

Page 26: Chapter 10 Gases

Gases

Dalton’s Law ofPartial Pressures

• The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone.

• In other words,

Ptotal = P1 + P2 + P3 + …

Page 27: Chapter 10 Gases

Gases

Partial Pressures

• When one collects a gas over water, there is water vapor mixed in with the gas.

• To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

Page 28: Chapter 10 Gases

Gases

Examples• The partial pressure of nitrogen in air is 592 torr. Air

pressure is 752 torr, what is the mole fraction of nitrogen?

• What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm?

Page 29: Chapter 10 Gases

Gases

Examples

3.50 L

O2

1.50 L

N2

2.70 atm• When these valves are opened, what is each partial

pressure and the total pressure?

4.00 L

CH4

4.58 atm 0.752 atm

Page 30: Chapter 10 Gases

Gases

Example• N2O can be produced by the following reaction

NH4NO3(s) N2O(g) + 2 H2O(l)

• What volume of N2O collected over water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g of NH4NO3? (the vapor pressure of water at 22ºC is 21 torr)

Page 31: Chapter 10 Gases

Gases

Kinetic-Molecular Theory

This is a model that aids in our understanding of what happens to gas particles as environmental conditions change.

Page 32: Chapter 10 Gases

Gases

Main Tenets of Kinetic-Molecular Theory

Gases consist of large numbers of molecules that are in continuous, random motion.

Page 33: Chapter 10 Gases

Gases

Main Tenets of Kinetic-Molecular Theory

The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained.

Page 34: Chapter 10 Gases

Gases

Main Tenets of Kinetic-Molecular Theory

Attractive and repulsive forces between gas molecules are negligible.

Page 35: Chapter 10 Gases

Gases

Main Tenets of Kinetic-Molecular Theory

Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.

Page 36: Chapter 10 Gases

Gases

Main Tenets of Kinetic-Molecular Theory

The average kinetic energy of the molecules is proportional to the absolute temperature.

Page 37: Chapter 10 Gases

Gases

Effusion

Effusion is the escape of gas molecules through a tiny hole into an evacuated space.

Page 38: Chapter 10 Gases

Gases

Effusion

The difference in the rates of effusion for helium and nitrogen, for example, explains a helium balloon would deflate faster.

Page 39: Chapter 10 Gases

Gases

Diffusion

Diffusion is the spread of one substance throughout a space or throughout a second substance.

Page 40: Chapter 10 Gases

Gases

• (KE)avg = NA(1/2 mu2)

• [m is mass of one gas particle in kg]• (KE)avg = 3/2 RT so…..

urms = (3RT/Mkg)1/2

• Where Mkg is the molar mass in kilograms, and R has the units 8.314 kg-m2 / s2-mol-K or J/K-mol.

• The velocity will be in m/s

Page 41: Chapter 10 Gases

Gases

Graham's Law

KE1 KE2=

1/2 m1v12 1/2 m2v2

2=

=m1

m2

v22

v12

m1m2

v22

v12

=v2

v1

=

Page 42: Chapter 10 Gases

Gases

Example • Calculate the root mean square velocity of carbon

dioxide at 25ºC.

Page 43: Chapter 10 Gases

Gases

Examples• A compound effuses through a porous cylinder 1.60

times faster than chlorine. What is it’s molar mass?

Page 44: Chapter 10 Gases

Gases

• If 0.00251 mol of NH3 effuses through a hole in 2.47 min, how much HCl would effuse in the same time?

Page 45: Chapter 10 Gases

Gases

• A sample of N2 effuses through a hole in 38 seconds. what must be the molecular weight of gas that effuses in 55 seconds under identical conditions?

Page 46: Chapter 10 Gases

Gases

Real Gases

In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

Page 47: Chapter 10 Gases

Gases

Real Gases

Even the same gas will show wildly different behavior under high pressure at different temperatures.

Page 48: Chapter 10 Gases

Gases

Deviations from Ideal Behavior

The assumptions made in the kinetic-molecular model (negligible volume of gas molecules themselves, no attractive forces between gas molecules, etc.) break down at high pressure and/or low temperature.

Page 49: Chapter 10 Gases

Gases

Corrections for Nonideal Behavior

• The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account.

• The corrected ideal-gas equation is known as the van der Waals equation.

Page 50: Chapter 10 Gases

Gases

The van der Waals Equation

) (V − nb) = nRTn2aV2(P +

Page 51: Chapter 10 Gases

Gases

Example• Calculate the pressure exerted by 0.5000 mol Cl2 in a

1.000 L container at 25.0ºC• Using the ideal gas law.

Page 52: Chapter 10 Gases

Gases

• Van der Waals equation given:a = 6.49 atm L2 /mol2

b = 0.0562 L/mol