© 2012 Pearson Education, Inc. Chapter 10 Gases Dr. Subhash C. Goel South Georgia College Douglas,...

© 2012 Pearson Education, Inc.

Chapter 10

Gases

Dr. Subhash C. GoelSouth Georgia College

Douglas, GA

Lecture Presentation

Gases


Characteristics of Gases

• Unlike liquids and solids, gases– Expand to fill their containers.– Are highly compressible.– Have extremely low densities.– Examples:

He, Ne, Ar, Kr, Xe

H2, N2, O2, Cl2CO2, NH3, SO2, CH4

Tro, Principles of Chemistry: A Molecular Approach 3

Gases Pushing• Gas molecules are constantly in

motion.

• As they move and strike a surface, they push on that surface.push = force

• If we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting.pressure = force per unit area


The Effect of Gas Pressure• The pressure exerted by a gas can cause

some amazing and startling effects.

• Whenever there is a pressure difference, a gas will flow from area of high pressure to low pressure.The bigger the difference in pressure, the

stronger the flow of the gas.

• If there is something in the gas’s path, the gas will try to push it along as the gas flows.

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

Gases


Units of Pressure

• Pascals– 1 Pa = 1 N/m2

[1 N = 1 kg-m/s2]

• Bar– 1 bar = 105 Pa = 100 kPa

Gases


Units of Pressure• mmHg or torr

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

Gases


Standard Pressure• Normal atmospheric pressure at sea level

is referred to as standard pressure.

• It is equal to– 1.00 atm– 14.7 psi– 760 torr (760 mmHg)– 101.325 kPa

© 2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Sample Exercise 10.1 Converting Pressure Units

SolutionAnalyze In each case we are given the pressure in one unit and asked to convert it to another unit. Our task, therefore, is to choose the appropriate conversion factors.

Plan We can use dimensional analysis to perform the desired conversions.

Solve(a) To convert atmospheres to torr, we use the relationship 760 torr = 1 atm:

Note that the units cancel in the required manner.

(b) We use the same relationship as in part (a). To get the appropriate units to cancel, we must use the conversion factor as follows:

(c) The relationship 760 torr = 101.325 kPa allows us to write an appropriate conversion factor for this problem: 760 torr = 101.325

(a) Convert 0.357 atm to torr. (b) Convert 6.6 102 torr to atmospheres. (c) Convert 147.2 kPa to torr.

Gases


Manometer

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


Sample Exercise 10.2 Using a Manometer to Measure Gas Pressure

On a certain day a laboratory barometer indicates that the atmospheric pressure is 764.7 torr. A sample of gas is placed in a flask attached to an open-end mercury manometer and a meter stick is used to measure the height of the mercury in the two arms of the U tube. The height of the mercury in the open-end arm is 136.4 mm, and the height in the arm in contact with the gas in the flask is 103.8 mm. What is the pressure of the gas in the flask (a) in atmospheres, (b) in kilopascals?


Continued

Solve (a) The pressure of the gas equals the atmospheric pressure plus h:

We convert the pressure of the gas to atmospheres:

(b) To calculate the pressure in kPa, we employ the conversion factor Between atmospheres and kPa:

Sample Exercise 10.2 Using a Manometer to Measure Gas Pressure


Boyle’s Law

• Pressure of a gas is inversely proportional to its volume.constant T and amount of gasgraph P vs. V is curvegraph P vs. 1/V is straight line

• As P increases, V decreases by the same factor.

• P × V = constant

• P1 × V1 = P2 × V2

Robert Boyle (1627–1691)

Gases


P and V are Inversely Proportional


Boyle’s Law: A Molecular View• Pressure is caused by the molecules striking the

sides of the container.

• When you decrease the volume of the container with the same number of molecules in the container, more molecules will hit the wall at the same instant.

• This results in increasing the pressure.


Charles’s Law

• Volume is directly proportional to temperature.constant P and amount of gasgraph of V vs. T is straight line

• As T increases, V also increases.• Kelvin T = Celsius T + 273• V = constant × T

if T measured in Kelvin

Jacques Charles (1746–1823)


If the lines are extrapolated back to a volume of “0,” they all show the same temperature, –273.15 °C, called absolute zero.

If you plot volume vs. temperature for any gas at constant pressure, the points will all fall on a straight line.


Example 3: A gas has a volume of 2.57 L at 0.00 °C. What was the temperature at 2.80 L?

Since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does.

V1 = 2.57 L, V2 = 2.80 L, t1 = 0.00 °C

t2, K and °C

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, V2, T2 T1

T(K) = t(°C) + 273.15,


Avogadro’s Law

• volume directly proportional to the number of gas moleculesV = constant × nwhile P and T remain constantmore gas molecules = larger

volume

• count number of gas molecules by moles

Amedeo Avogadro (1776–1856)

Gases


Avogadro’s Law• Equal volume of gases at constant P and T

have same number of moles of gas


Exercise 4: A 0.225-mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?

Since n and V are directly proportional, when the volume increases, the moles should increase, and it does.

V1 = 4.65 L, V2 = 6.48 L, n1 = 0.225 mol

n2, and added moles

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, V2, n1 n2

mol added = n2 – n1,

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

Gases


Ideal-Gas Equation

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

Gases


Ideal-Gas Equation

The relationship

then becomes

nTP

V

nTP

V = R

or

PV = nRT


Sample Exercise 5 Using the Ideal-Gas Equation

SolutionAnalyze We are given the volume (250 mL), pressure (1.3 atm), and temperature (31 C) of a sample of CO2 gas and asked to calculate the number of moles of CO2 in the sample.

SolveIn analyzing and solving gas law problems, it is helpful to tabulate the information given in the problems and then to convert the values to units that are consistent with those for R (0.08206 L-atm/mol-K). In this case the given values are

Remember: Absolute temperature must always be used when the ideal-gas equation is solved.

We now rearrange the ideal-gas equation (Equation 10.5) to solve for n

Calcium carbonate, CaCO3(s), the principal compound in limestone, decomposes upon heating to CaO(s) and CO2(g). A sample of CaCO3 is decomposed, and the carbon dioxide is collected in a 250-mL flask.

After decomposition is complete, the gas has a pressure of 1.3 atm at a temperature of 31 C. How many moles of CO2 gas were generated?

Plan Because we are given V, P, and T, we can solvethe ideal-gas equation for the unknown quantity, n.


Sample Exercise 6 Calculating the Effect of Temperature Changes on Pressure

SolutionAnalyze We are given the initial pressure (1.5 atm) and temperature (25 C) of the gas and asked for thepressure at a higher temperature (450 C).

Plan The volume and number of moles of gas do not change, so we must usea relationship connecting pressure and temperature. Converting temperature to the Kelvin scale and tabulating the given information, we have

Solve To determine how P and T are related, we start with the ideal-gas equation and isolate the quantities that do not change (n, V, and R) on one side and the Variables (P and T) on the other side.

Because the quotient P/T is a constant, we can write

(where the subscripts 1 and 2 represent the initial and final states, respectively).Rearranging to solve for P2 and substituting the given data give

The gas pressure in an aerosol can is 1.5 atm at 25 C. Assuming that the gas obeys the ideal-gas equation, what is the pressure when the can is heated to 450 C?


Sample Exercise 7 Calculating the Effect of Changing P and T on Gas Volume

SolutionAnalyze We need to determine a new volume for a gas sample when both pressure and temperature change.

Plan Let’s again proceed by converting temperatures to kelvins and tabulating our information.

Because n is constant, we can use Equation 10.8.

Solve Rearranging Equation 10.8 to solve for V2 gives

An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend until the pressure is 0.45 atm. During ascent, the temperature of the gas falls from 22 C to 21 C. Calculate the volume of the balloon at its final altitude.


Density at Standard Conditions

• Density is the ratio of mass to volume.

• Density of a gas is generally given in g/L.

• mass of 1 mole = molar mass

• volume of 1 mole at STP = 22.4 L


Practice 8: Calculate the density of N2(g) at STP.

The units and magnitude are reasonable.

N2

dN2, g/L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

MM d

Gases


Densities of Gases

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

nV

PRT

=

Gases


• We know that– Moles molecular mass = mass

Densities of Gases

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

n = m

PRT

mV

=

Gases


Densities of Gases

• Mass volume = density

• So,

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

PRT

mV

=d =


Sample Exercise 9 Calculating Gas Density

SolutionAnalyze We are asked to calculate the density of a gas given its name, its pressure, and its temperature. From the name we can write the chemical formula of the substance and determine its molar mass.

Plan We can use Equation 10.10 to calculate the density. Before we can do that, however, we must convert the given quantities to the appropriate units, degrees Celsius to kelvins and pressure to atmospheres. We must also calculate the molar mass of CCl4.

Solve The absolute temperature is 125 + 273 = 398 K. The pressure is (714 torr) (1 atm/760 torr) = 0.939. The molar mass of CCl4 is 12.01 + (4) (35.45) = 153.8 g/mol. Therefore,

Check If we divide molar mass (g/mol) by density (g/L), we end up with L/mol. The numerical value is roughly 154/4.4 = 35 . That is in the right ballpark for the molar volume of a gas heated to 125 C at near atmospheric pressure, so our answer is reasonable.

What is the density of carbon tetrachloride vapor at 714 torr and 125 C?

Gases


Molecular Mass

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

becomes

PRT

d =

dRTP =


Sample Exercise 10 Calculating the Molar Mass of a Gas

SolutionAnalyze We are given the temperature (31 C) and pressure (735 torr) for a gas, together with information to determine its volume and mass, and we are asked to calculate its molar mass.

Solve The gas mass is the difference between the mass of the flask filled with gas and the mass of the evacuated flask:

The gas volume equals the volume of water the flask can hold, calculated from the mass and density of the water. The mass of the water is the difference between the masses of the full and evacuated flask:

Rearranging the equation for density (d = m/V),we have

A large evacuated flask initially has a mass of 134.567 g. When the flask is filled with a gas of unknown molar mass to a pressure of 735 torr at 31 C, its mass is 137.456 g. When the flask is evacuated again

137.456 g 134.567 g = 2.889 g

1067.9 g 134.567 g = 933.3 g

and then filled with water at 31 C, its mass is 1067.9 g. (The density of water at this temperature is 0.997 g/mL.) Assuming the ideal-gas equation applies, calculate the molar mass of the gas.

Plan We need to use the mass information given to calculate the volume of the container and the mass of the gas in it. From this we calculate the gas density and then apply Equation 10.11 to calculate the molar mass of the gas.


Sample Exercise 10 Calculating the Molar Mass of a Gas

Continued

Knowing the mass of the gas (2.889 g) and its volume (0.936 L), we can calculate the density of the gas:

After converting pressure to atmospheres and temperature to kelvins, we can use Equation 10.11 to calculate the molar mass:

2.889 g/0.936 L = 3.09 g/L

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 + …


The partial pressure of each gas in a mixture can be calculated using the ideal

gas law.


Sample Exercise 11: Applying Dalton’s Law of Partial Pressures

SolutionAnalyze We need to calculate the pressure for two gases in the same volume and at the same temperature.

Solve We first convert the mass of each gas to moles:

We use the ideal-gas equation to calculate the partial Pressure of each gas:

According to Dalton’s law of partial pressures (Equation 10.12), the total pressure in the vessel is the sum of the partial pressures:

A mixture of 6.00 g O2(g) and 9.00 g CH4(g) is placed in a 15.0-L vessel at 0 C. What is the partial pressureof each gas, and what is the total pressure in the vessel?

Pt = PO2 + PCH4

= 0.281 atm + 0.841 atm = 1.122 atm

Plan Because each gas behaves independently, we can use the ideal gas equation to calculate the pressure each would exert if the other were not present. The total pressure is the sum of these two partial pressures.

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.

• Ptotal = Pgas + Pwater


Sample Exercise 12 Calculating the Amount of Gas Collected over Water

Solution(a) Analyze We need to calculate the number of moles of O2 gas in a container that also contains water vapor.

Plan We are given values for V and T. To use the ideal-gas equation to calculate the unknown, nO2, we must know

the partial pressure of O2 in the system. We can calculate this partial pressure from the total pressure (765 torr) and the vapor pressure of water.

Solve The partial pressure of the O2 gas is the difference between the total pressure and the pressure of the water vapor at 26 C, 25 torr (Appendix B):

PO2 = 765 torr 25 torr = 740 torr

We use the ideal-gas equation to calculate the number of moles of O2:

When a sample of KClO3 is partially decomposed in the setup shown in Figure 10.15, the volume of gas collected is 0.250 L at 26 C and 765 torr total pressure. (a) How many moles of O2 are collected? (b) How many grams of KClO3 were decomposed?


Continued

(b) Analyze We need to calculate the number of moles of reactant KClO3 decomposed.

Plan We can use the number of moles of O2 formed and the balanced chemical equation to determine the number of moles of KClO3 decomposed, which we can then convert to grams of KClO3.

Solve From Equation 10.17, we have 2 mol .

The molar mass of KClO3 is 122.6 g/mol. Thus, we can convert the number of moles of O2 from part (a) to moles of KClO3 and then to grams of KClO3:

Sample Exercise 10.12 Calculating the Amount of Gas Collected over Water

Gases


Kinetic-Molecular Theory

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

Gases


Main Tenets of Kinetic-Molecular Theory

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

Gases



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

Gases



Attractive and repulsive forces between gas molecules are negligible.

Gases



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.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energyKE = ½ mu2

Gases

© 2012 Pearson Education, Inc.48

Gas Laws Explained—Boyle’s Law

• Boyle’s law states that the volume of a gas is inversely proportional to the pressure.

• Decreasing the volume forces the molecules into a smaller space.

• More molecules will collide with the container at any one instant, increasing the pressure.

Gases


Gas Laws Explained—Charles’s Law

• Charles’s law states that the volume of a gas is directly proportional to the absolute temperature.

• Increasing the temperature increases their average speed, causing them to hit the wall harder and more frequently.– on average

• In order to keep the pressure constant, the volume must then increase.

Gases


Gas Laws Explained—Avogadro’s Law

• Avogadro’s law states that the volume of a gas is directly proportional to the number of gas molecules.

• Increasing the number of gas molecules causes more of them to hit the wall at the same time.

• In order to keep the pressure constant, the volume must then increase.

Gases


Gas Laws Explained—Dalton’s Law of Partial Pressures

• Dalton’s law states that the total pressure of a mixture of gases is the sum of the partial pressures.

• Kinetic-molecular theory says that the gas molecules are negligibly small and don’t interact.

• Therefore, the molecules behave independently of each other, each gas contributing its own collisions to the container with the same average kinetic energy.

• Since the average kinetic energy is the same, the total pressure of the collisions is the same.

Gases


Dalton’s Law and Pressure

• Since the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side.

Gases


Effusion

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

Gases


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

Copyright © Cengage Learning. All rights reserved. 5 | 55

Molecular SpeedsAccording to kinetic theory, molecular speeds vary over a wide range of values. The distribution depends on temperature, so it increases as the temperature increases.

Root-Mean Square (rms) Molecular Speed, u

A type of average molecular speed, equal to the speed of a molecule that has the average molecular kinetic energy

mM

RTu

3


When using the equation

R = 8.3145 J/(mol · K).

T must be in Kelvins

Mm must be in kg/mol


Graham’s Law of EffusionAt constant temperature and pressure, the rate of effusion of gas molecules through a particular hole is inversely proportional to the square root of the molecular mass of the gas.

mM

1moleculesofeffusionofrate

Gases


Graham's Law

M2

M1

=r1

r2

M1 and M2 are the molar masses of gases.This indicate that lighter gas has higher effusion rate.

Gases


Sample Exercise 13 Calculating a Root-Mean-Square Speed

SolutionSolve We must convert each quantity in our equation to SI units. We will also use R in units of J/mol-K (Table 10.2) to make the units cancel correctly.

Calculate the rms speed of the molecules in a sample of N2 gas at 25 C.


Sample Exercise 14 Applying Graham’s LawAn unknown gas composed of homonuclear diatomic molecules effuses at a rate that is 0.355 times the rate at which O 2 gas effuses at the same temperature. Calculate the molar mass of the unknown and identify it.

Because we are told that the unknown gas is composed of homonuclear diatomic molecules, it must be an element. The molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I2.


61

Practice 15—Calculate the ratio of the rate of effusion for oxygen to hydrogen.

O2, 32.00 g/mol; H2 2.016 g/mol

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

MMO2, MMH2 rateA/rateB

This means that, on average, the O2 molecules

are traveling at ¼ the speed of H2 molecules.

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.

Gases


Real Gases

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

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.

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.

Gases


The van der Waals Equation

) (V − nb) = nRTn2aV2(P +


Sample Exercise 16 Using the van der Waals Equation

SolutionSolve Substituting n = 1.000 mol, R = 0.08206 L-atm/mol-K, T = 273.2 K, V = 22.41 L, a = 6.49 L2-atm/mol2, and b = 0.0562 L/mol:

If 1.000 mol of an ideal gas were confined to 22.41 L at 0.0 C, it would exert a pressure of 1.000 atm. Use the van der Waals equation and Table 10.3 to estimate the pressure exerted by 1.000 mol of Cl2(g) in 22.41 L at 0.0 C.

© 2012 Pearson Education, Inc. Chapter 10 Gases Dr. Subhash C. Goel South Georgia College Douglas,...

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