Chapter 5: The Gaseous State Bushra Javed CHM 2045 1

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Chapter 5: The Gaseous StateBushra JavedCHM 2045

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Contents1. Gas pressure & it’s units2. Empirical Gas Laws• Boyle’sLaw• Charles’Law• Combined gas Law• Avogadro’s Law• Dalton’s Law• Ideal Gas Law

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Contents

3. Kinetic-Molecular Theory

4. Molecular Speeds 5. Diffusion and Effusion Graham’s Law 6. Real Gases

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The Gaseous State• Gases differ from liquids and solids:• Have low densities, are compressible & can

expand• Physical condition of any gas can be defined• by four variables:• Pressure, P• Volume, V• temperature, T• amount or number of moles n.

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Pressure , P of a Gas

Pressure is the force exerted per unit area.It can be given by two equations:

The SI unit for pressure is the pascal, Pa.

A

FP

(pascal)Pasm

kg

ms

mkg

22

2

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Pressure , P of a GasOther Unitsatmosphere, atm

mmHgtorrbar

Atmospheric Pressure

• Atmospheric pressure results from the exertion & pressure of air molecules in the environment.

• A barometer is a device for measuring the pressure of the atmosphere.

• A manometer is a device for measuring the pressure of a gas or liquid in a vessel.

Atmospheric Pressure

5 | 8

Empirical Gas LawsAll gases behave quite simply with respect to temperature, pressure, volume, and molar amount.

By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties.

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Boyle’s LawPressure –Volume Relationship

The volume of a sample of gas at constant temperature varies inversely with the applied pressure.

The mathematical relationship:

In equation form:

5 | 10

PV

1

ffii

constant

VPVP

PV

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Boyle’s LawPressure –Volume Relationship

• When the volume decreases, the gas molecules collide with the container more often and the pressure increases.

• When the volume increases, the gas molecules collide with the container less often and the pressure decreases.

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Boyle’s Law

plot of V versus P for 1.000 g O2 at 0°C. This plot is nonlinear.

Boyle’s Law

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At one atmosphere the volume of the gas is 100 mL. When pressure is doubled, the volume is halved to 50 mL. When pressure is tripled, the volume decreases to one-third, 33 mL.

Boyle’s LawExample 1A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant?

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Vi = 38.7 mLPi = 751 mmHgTi = 21°C

Vf = ?Pf = 359 mmHgTf = 21°C

f

iif P

VPV

Boyle’s Law

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Vi = 38.7 mLPi = 751 mmHgTi = 21°C

Vf = ?Pf = 359 mmHgTf = 21°C

mmHg)(359

mmHg)mL)(751(38.7f V

= 81.0 mL

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Boyle’s LawExample 2A sample of methane, CH4, occupies a volume of 244.0 mL at 25°C and exerts a pressure of 1135.0 mmHg. If the volume of the gas is allowed to expand to 720.0 mL at 298 K, what will be thepressure of the gas? P2 = ?

a. 3350 mmHgb. 385 mmHgc. 4580 mmHgd. 0.0149 mmHg

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

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Absolute ZeroThe temperature -273.15°C is called absolute zero. It is the temperature at which the volume of a gas is hypothetically zero.

This is the basis of the absolute temperature scale, the Kelvin scale (K).

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Charles’s Law

The volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K).The mathematical relationship:In equation form:

TV

f

f

i

i

constant

T

V

T

VT

V

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A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume.

As the air inside warms, the balloon expands to its orginial size.

Charles’s LawExample 3You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your calculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27°C?

Vi = 79.4 mL

Pi = 760 mmHgTi = 0°C = 273 K

Vf = ?Pf = 760 mmHgTf = 27°C = 300. K

i

iff T

VTV

Vi = 79.4 mLPi = 760 mmHgTi = 0°C = 273 K

Vf = ?Pf = 760 mmHgTf = 27°C = 300. K

K)(273

mL)K)(79.4(300.f V

= 87.3 mL

Charles’s Law

i

iff T

VTV

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Charles’s LawExample 4The volume of a sample of gas measured at 10.0°C and 1.00 atm pressure is 6.00 L. What must the final temperature be in order for the gas to have a final volume of 7.00 L at 1.00 atm pressure? T2 = ? Convert °C into Kelvin then back into °C.

a. –30.4°Cb. 8.6°Cc. 11.7°Cd. 57.2°C

Combined Gas Law

The volume of a sample of gas at constant pressure is inversely proportional to the pressure and directly proportional to the absolute temperature.

The mathematical relationship:

In equation form:

P

TV

f

ff

i

ii

constant

T

VP

T

VPT

PV

Combined Gas LawExample 5Divers working from a North Sea drilling platform experience pressure of 5.0 × 101 atm at a depth of 5.0 × 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C? Vf or V2 = ?

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Vi = 5.0 LPi = 5.0 × 101 atmTi = 4°C = 277 K

Vf = ?Pf = 1.0 atmTf = 11°C = 284 K

fi

iiff PT

VPTV

1

f(284 K)(5.0 10 atm)(5.0 L)

(277 K)(1.0 atm)V

= 2.6 102 L

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Combined Gas LawExample 6If 7.75 L of radon gas is at 1.55 atm and –19 °C,what is the volume at STP? Standard Temperature and Pressure (STP)

The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure(a)4.65 L(b)5.37 L(c)8.33 L(d)12.9 L

Avogadro’s LawAmedeo Avogadro (1776–1856)

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• Volume directly proportional to the number of gas moleculesV = constant x nconstant P and Tmore gas molecules = larger volume

• Count number of gas molecules by moles

• Equal volumes of gases contain equal numbers of moleculesthe gas doesn’t matter

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Avogadro’s Law

Example 6If 1.00 mole of a gas occupies 22.4 L at STP, what volume would 0.750 moles occupy?

Ans: 16.8 L

Standard Conditions

• Because the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions– STP

• Standard pressure = 1 atm• Standard temperature = 273 K

– 0 °C30

Molar Volume• Solving the ideal gas equation for the volume of

1 mol of gas at STP gives 22.4 L– 6.022 x 1023 molecules of gas– notice: the gas is immaterial

• We call the volume of 1 mole of gas at STP the molar volume– it is important to recognize that one mole measures

of different gases have different masses, even though they have the same volume

31Tro: Chemistry: A Molecular Approach, 2/e

Molar Volume


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Ideal Gas Law/Ideal gas equation

Combining the three LawsV α 1/ P (Boyle’s Law)V α T (Charles’ Law)V α n (Avogadro’s Law)

V α T . n / Por V = R . T . n / PWhere R = molar gas constant

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Ideal Gas Equation

• Rearranging the above equation gives:• PV = nRT R = molar gas constant = 0.0821L .atm/ (mol.K)

R = 0.0821 atm.L/(K.mol)R = 8.3145 J/(K.mol)

R = 8.3145kg. m2 /(.K.mol)

Ideal Gas Law Equation Example 7A 50.0-L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder? Hint: Find mol first.

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V = 50.0 LP = 17.1 atmT = 23 C = 296 K

RT

PVn

K)(296Kmol

atmL0.08206

L)atm)(50.0(17.1

n

mol

g28.02mol35.20mass

Ansmass = 986 g

Density at Standard Conditions

• Density is the ratio of mass to volume• Density of a gas is generally given in g/L• The mass of 1 mole = molar mass• The volume of 1 mole at STP = 22.4 L


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Density at Standard ConditionsExample 8:Calculate the density of N2(g) at STP

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Gas Density and Molar MassGas Density Using the ideal gas law, it is possible to calculate the moles in 1 L at a given temperature and pressure. The number of moles can then be converted to grams (per liter).Molar MassTo find molar mass, find the moles of gas, and then find the ratio of mass to moles.

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Gas Density and Molar MassExample 8What is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm?Note: Since density of gases is measured in g/L, you can assume the volume is 1L. Hint: Find mols first, then convert to grams ;Mm = 16.04 g/mol

a) 1.71g/Lb) 1.72g/Lc) 4.5g/Ld) 0.04g/L

Molar Mass of a Gas


• One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law

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Gas Density and Molar MassExample 9An unknown gas occupies a volume of 4.75 L at 1227 °C and 5.00atm. If the mass is 5.45 g, what is the molar mass of gas?(R = 0.0821 atm•L/mol•K) Hint: Find mols first n = 0.1928

(a)21.5 g/mol(b) 23.8 g/mol(c )28.3 g/mol(d)141 g/mol

Gas Mixtures

Dalton found that in a mixture of unreactive gases, each gas acts as if it were the only gas in the mixture as far as pressure is concerned.

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Originally (left), flask A contains He at 152 mmHg and flask B contains O2 at 608 mmHg. Flask A is then filled with oil forcing the He into flask B (right). The new pressure in flask B is 760 mmHg.

Dalton’s Law of Partial Pressures

Partial PressureThe pressure exerted by a particular gas in a mixture.

Dalton’s Law of Partial PressuresThe sum of the partial pressures of all the different gases in a mixture is equal to the total pressure of the mixture:

P = PA + PB + PC + . . .

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Example 10A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample? Hint: Convert grams to molsThen use ideal gas equation, PV= nRT to find partial pressure of each gas.

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mL10

L1mL100.0

K308Kmol

atmL0.08206

Ng28.01

Nmol1Ng0.0830

3

2

22

N2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

Og32.00

Omol1Og0.0194

3

2

22

O2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

COg44.01

COmol1COg0.00640

3

2

22

CO2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

OHg18.01

OHmol1OHg0.00441

3

2

22

OH2P

atm0.749

atm0.153

atm0.0368

atm0.0619


OHCOON 2222PPPPP

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atm0.7492N P

atm0.1532O P

atm0.03682CO P

atm0.0619OH2P

P = 1.00 atm

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Dalton’s Law of Partial PressuresCollecting Gas Over WaterGases are often collected over water. The result is a mixture of the gas and water vapor.

The total pressure is equal to the sum of the gas pressure and the vapor pressure of water.

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The partial pressure of water depends only on temperature .(See Table 5.6).

The pressure of the gas can then be found using Dalton’s law of partial pressures.

Collecting Gas by Water Displacement


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Dalton’s Law of Partial PressuresExample 10If carbon dioxide gas is collected over water at 25 °C and775 torr, what is the partial pressure of the CO2?The vapor pressure of water at 25 °C is 23.8 torr.a) 23.8 torrb) 750 torrC) 751 torrd) 775 torre) 799 torr

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mole fraction(X )

The concentration of any individual gas in a gas mixture can be expressed as a mole fraction(X )

Mole fraction(X) = Moles of component Total Moles in the mixture

Mole fraction of component 1,for example ,isX = n1 / n1 + n2 + n3 + ... = n1 / n total

mole fraction(X )

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Example 11The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570.0 mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air?

mmHg103.0 2OP

mmHg40.02CO P

mmHg570.02N P

mmHg47.0OH2P

mole fraction(X )

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OHCOON 2222PPPPP

570.0 mmHg103.0 mmHg

40.0 mmHg47.0 mmHg

P = 760.0 mmHg

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Mole fraction of N2

Mole fraction of H2OMole fraction ofCO2

Mole fraction of O2

mmHg760.0

mmHg47.0

mmHg760.0

mmHg40.0

mmHg760.0

mmHg103.0

mmHg760.0

mmHg570.0

mole fraction(X )

X N2 = 0.7500

X O2 = 0.1355

X CO2 = 0.0526

X H2O = 0.0618

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Kinetic-Molecular Theory

2K 2

1mvE

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A theory, developed by physicists, that is based on the assumption that a gas consists of molecules in constant random motion.Kinetic energy is related to the mass and velocity:

m = massv = velocity

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Kinetic-Molecular Theory Postulates of the Kinetic Theory1. Gases are composed of molecules whose sizes are

negligible.

2. Molecules move randomly in straight lines in all directions and at various speeds.

3. The forces of attraction or repulsion between two molecules (intermolecular forces) in a gas are very weak or negligible, except when the molecules collide.

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4. When molecules collide with each other, the collisions are elastic.

5. The average kinetic energy of a molecule is proportional to the absolute temperature

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Kinetic-Molecular Theory • An elastic collision occurs when the two

objects "bounce" apart when they collide.

• In an elastic collision, both momentum and kinetic energy are conserved. Almost no energy is lost to sound, heat, or deformation.

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Kinetic-Molecular Theory The Kinetic-Molecular Theory explains Boyle’s Law• Compressing a gas makes the V smaller but

does not alter the KE avg of the molecules since T is constant.

• Though the speed of the particles remains constant, the frequency of collisions increases because the container is smaller.

• Therefore, P increases as V decreases.

Molecular Speeds

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

5 | 63mM

RTu

3

Kinetic Energy and Molecular Velocities

• Average kinetic energy of the gas molecules depends on the average mass and velocity– KE = ½mv2

• Gases in the same container have the same temperature, therefore they have the same average kinetic energy

• If they have different masses, the only way for them to have the same kinetic energy is to have different average velocities– lighter particles will have a faster average velocity than more

massive particles

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Molecular Speed vs. Molar Mass• To have the same average kinetic energy,

heavier molecules must have a slower average speed

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Temperature and Molecular Velocities _• KEavg = ½NAmu2

– NA is Avogadro’s number• KEavg = 1.5RT

– R is the gas constant in energy units, 8.314 J/mol K∙• 1 J = 1 kg m∙ 2/s2

• Equating and solving we get– NA mass = molar mass in kg/mol∙

• As temperature increases, the average velocity increases66

Molecular Velocities• All the gas molecules in a sample can travel at different

speeds• However, the distribution of speeds follows a statistical

pattern called a Boltzman distribution• Ee talk about the “average velocity” of the molecules,

but there are different ways to take this kind of average• The method of choice for our average velocity is called

the root-mean-square method, where the rms average velocity, urms, is the square root of the average of the sum of the squares of all the molecule velocities

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Temperature vs. Molecular Speed

• As the absolute temperature increases, the average velocity increases– the distribution

function “spreads out,” resulting in more molecules with faster speeds

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Molecular Speeds

When using the equation

R = 8.3145 J/(mol · K).T must be in KelvinsMm must be in kg/mol

mM

RTu

3

mM

RTu

3rms

T = 23°C = 296 KCO2 molar mass =

0.04401 kg/mol

Molecular SpeedsExample 12What is the rms speed of carbon dioxide molecules in a container at 23°C?

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mol

kg0.04401

K296Kmol

s

mkg

8.314532

2

rmsu

25

rms 2

m1.68 10

su

2rms

m4.10 10

su

2

2

s

mkgJ

Recall

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root-mean-square (rms) Speed

Example 13Calculate the root-mean-square velocity for the O2 molecules in a sample of O2 gas at 22.5°C.

(R = 8.3145 J/Kmol)

a) 132.4 m/sb) 15.18 m/sc) 479.9 m/sd) 277.1 m/s

Diffusion and Effusion• The process of a collection of molecules spreading out

from high concentration to low concentration is called diffusion

• The process by which a collection of molecules escapes through a small hole into a vacuum is called effusion

• The rates of diffusion and effusion of a gas are both related to its rms average velocity

• For gases at the same temperature, this means that the rate of gas movement is inversely proportional to the square root of its molar mass


Effusion


Graham’s Law of Effusion

• For two different gases at the same temperature, the ratio of their rates of effusion is given by the following equation:

Thomas Graham (1805–1869)


Graham’s Law of Effusion

Example 14Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor?

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Hydrogen will diffuse more quickly by a factor of 1.4.

2.016

4.002

4.002

12.016

1

HeRate

HRate 2

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Graham’s Law of EffusionExample 15Which of the following gases will have the slowest rate of effusion at constant temperature?

a)CF4

b)F2

c)H2

d)Ne

Ideal vs. Real Gases


• Real gases often do not behave like ideal gases at high pressure or low temperature

• Ideal gas laws assume1. no attractions between gas molecules2. gas molecules do not take up space

based on the kinetic-molecular theory• At low temperatures and high pressures these

assumptions are not valid

Real Gases

At high pressure, some of the assumptions of the kinetic theory no longer hold true:

1. At high pressure, the volume of the gas molecule (Postulate 1) is not negligible.

2. At high pressure, the intermolecular forces (Postulate 3) are not negligible.

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The Effect of Molecular Volume

• At high pressure, the amount of space occupied by the molecules is a significant amount of the total volume

• The molecular volume makes the real volume larger than the ideal gas law would predict

• van der Waals modified the ideal gas equation to account for the molecular volume– b is called a van der Waals constant and is

different for every gas because their molecules are different sizes

Johannes van der Waals (1837–1923)


Real Gas Behavior• Because real molecules attract each other, the

molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures

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The Effect of Intermolecular Attractions• At low temperature, the attractions between

the molecules is significant• The intermolecular attractions makes the real

pressure less than the ideal gas law would predict

• van der Waals modified the ideal gas equation to account for the intermolecular attractions– a is another van der Waals constant and is different

for every gas because their molecules have different strengths of attraction

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Van der Waals EquationExample 16a)Use the van der Waals equation to calculate the pressure exerted by 2.00 mol CO2 that has a volume of 10.0 L at 25°C. b)Compare this with value with the pressure obtained from the ideal gas law.

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n = 2.00 molV = 10.0 LT = 25°C = 298 K

For CO2:a = 3.658 L2 atm/mol2

b = 0.04286 L/mol

Van der Waals Equation

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n = 2.00 molV = 10.0 LT = 25°C = 298 K

L10.0

K)(298Kmol

atmL0.08206mol2.00

P

= 4.89 atm

V

nRTP

Ideal gas law:

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n = 2.00 molV = 10.0 LT = 25°C = 298 K

For CO2:a = 3.658 L2 atm/mol2

b = 0.04286 L/mol

Pactual = 4.79 atm

2

2

V

an

nbV

nRTP

2

2

22

L10.0

mol

atmL3.658mol2.00

mol

L0.04286mol2.00L10.0

K298Kmol

atmL0.08206mol2.00

P

atm0.146atm4.933 P

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Chapter 5: The Gaseous State Bushra Javed CHM 2045 1