The Gaseous State - Part II Applications and Kinetic...

The Gaseous State - Part II

Applications and

Kinetic Molecular Theory

Molar Volume

Solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L

for 6.022 x 1023 molecules of any gas

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.

Molar Volume

d = = = = 1.25

1. Calculate the density of N2(g) at STP.

MM d

g molMM

22.4L

mol22.4 L mol

g L

28.02 g mol g

Lm V

mol g g x = mol

Density is directly proportional to molar mass

( )

( )gL

gmol

L atmmol K

( )( ) atm

( )K=


2. Calculate the density of N2 at 125°C and 755 mmHg .

P = 755 mmHg, t = 125 °C, dN2 = ?

P, MM, T, R d (P)(MM)(R)(T) d =

1 atm760. mmHg P = 755 mmHg x = 0.993 atm

T(K) = 273 + (125) = 398

d = (P)(MM)(R)(T)

= 0.852 g/L= (.993 atm)(28.01 )(0.0821 )(398 K)atm L

mol K

gmol

( )

( )gL

gmol

L atmmol K

( )( ) atm

( )K=


3. Calculate the density of a gas at 775 torr and 27 °C if 0.250 moles weighs 9.988 g.

m = 9.988g, n = 0.250 mol, P = 775 torr, T = 27ºC, density = ?

K = 273+ºC=273+(27)=300

P, n, T, R V V,m d

V = (n)(R)(T)(P) d = m

V


V = (n)(R)(T)(P)

= (.250 mol)(.0821 )(300 K)(1.02 atm)

atm Lmol K

= 6.04 L

3. Calculate the density of a gas at 775 torr and 27 °C if 0.250 moles weighs 9.988 g.

m = 9.988g, n = 0.250 mol, P = 775 torr, T = 27ºC, density = ?

V= 6.04 L

P, n, T, R V

V = (n)(R)(T)(P)

V,m d

d = mV

d = = = 1.65 g/L mV

9.99 g6.04 L

mol g

g x = mol

Molar mass can be determined from an experiment in which all of these

variables are known.

Density can be determined when mass and volume

are available.

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.

Determining the Molar Mass of an Unknown Volatile Liquid.

JBA Dumas (1800-1884)

n = PVRT

molar mass = gramsmole (n)

4. Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg.

m=0.311g, V=0.225 L, P=886 mmHg, T = 55 ºC, molar mass = ?

K = 273+ºC=273+(55)=328

P, V, T, R n n = (P)(V)(R)(T)


n = (P)(V)(R)(T)

n = (P)(V)(R)(T) = (1.17 atm)(0.225 L)

(0.0821 )(328 K)atm Lmol K

= 9.75 x 10-3 mol

MM = 31.9 g/mol

4. Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg.

m=0.311g, V=0.225 L, P=886 mmHg, T = 55 ºC, molar mass = ?

P, V, T, R n

n = (P)(V)(R)(T)

= 9.75 x 10-3 mol

n, m MM

n = (P)(V)(R)(T) MM = m

n

(0.311 g)(9.75 x 10-3 mol)MM =

Mixtures of Gases

When gases are mixed together, their molecules behave independently, but

All the gases in the mixture have the same volume.

All gases in the mixture are at the same temperature.

In certain applications, the mixture can be thought of as one gas

We can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules.

Partial PressureThe pressure of a single gas in a mixture of gases is

called its partial pressure.

We can calculate the partial pressure of a gas if

we know what fraction of the mixture it composes and the total pressure,or

we know the number of moles of the gas in a container of known volume and temperature.

Dalton’s Law of Partial Pressures - The sum of the partial pressures of all the gases in the mixture equals the total pressure

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

Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures

+Maintain volume at 5.00 L

Ptot = 3.90 atm, V = 8.70 L, T = 598 K, nXe = 0.175 mol , nNe = ?

5. A mixture of neon and xenon has a volume of 8.70 L at temperature of 598 K and 3.90 atm pressure. If the mixture contains 0.175 mol of xenon, how many moles of neon are present ?

nXe,V, T, R PXe

PXe =(0.175 mol)(.0821 )(598 K)8.70 atm

atm Lmol K

PXe = 0.988 atm

PXe = (nXe)(R)(T)V

Ptotal, PXe PNe

PNe = Ptotal - PXe

PNe = 3.90 atm-0.988 atm

PNe = 2.91 atm

Ptot = 3.90 atm, V = 8.70 L, T = 598 K, nXe = 0.175 mol , nNe = ?

5. A mixture of neon and xenon has a volume of 8.70 L at temperature of 598 K and 3.90 atm pressure. If the mixture contains 0.175 mol of xenon, how many moles of neon are present ?

nXe,V, T, R PXe Ptotal, PXe PNe

nNe = (PNe)(V)(R)(T)

PXe = 0.988 atm PNe = 2.91 atm

PNe,V, T, R nNe

= (2.91 atm)(8.70 L)(0.0821 )(598 K)atm L

mol K = 0.516 mol Ne

Mole FractionThe fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes.

The ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction, χ.

The partial pressure of a gas is equal to the mole fraction of that gas times the total pressure.

6. Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K.

mHe = 24.2 g, mO2 = 4.32 g, V = 12.5 L, T = 298 K

𝒙He, 𝒙O2, PHe, PO2, Ptotal = ?????

ntot,V, T, R Ptot

𝒙gas, Ptotal Pgas

ngas 𝒙gasmgas

PA = (𝒙A)(Ptotal)

(𝒙A) =ngas Antotal Ptotal = ntotal)(R)(T)

V

= (6.185 mol)(0.0821 )(298 K)(12.5 L)

atm Lmol K

6. Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K .

ntotal = 6.185 mol

1 mol He4.00 g He24.2 g He x = 6.05 mol He

1 mol O232.00 g O2

4.32 g O2 x = 0.135 mol O2

Ptotal = ntotal)(R)(T)V

= 12.1 atm

6.05 mol He6.185 moltotal

𝒙He = = 0.978

0.135 mol O26.185 moltotal

𝒙O2 = = 0.0218

PHe = (𝒙He)(Ptotal) = (0.978)(12.1) atm= 11.8 atm

PO2 = (𝒙O2)(Ptotal) = (0.0218)(12.1) atm= 0.264 atm

Collecting GasesGases are often collected by having them

displace water from a container.

Because water evaporates, there is also water vapor in the collected gas.

The partial pressure of the water vapor, called the vapor pressure, depends only on the

temperature, so you can use a table to find out the partial pressure of the water vapor in the

gas you collect.

Collecting Gas by Water Displacement

A chemical process

generates a gas.

Collection of a Gas Over Water

Pgas + Pwater

A chemical process

generates a gas.

Collection of a Gas Over Water

It is necessary to adjust the position of the bottle so that the water levels inside and outside the bottle are the same.

This makes the total pressure of the gaseous mixture in the bottle equal to

the atmospheric pressure as measured with a barometer.

Pgas = Pbar - Pwater

Pgas + Pwater

Vapor Pressure of Water vs Temperature

If you collect a gas sample with a total pressure of 758.2 mmHg* at 25 °C, the partial pressure of the water vapor will be 23.78 mmHg – so the partial pressure of the dry gas will be 734.4 mmHg. This is the pressure you would use to calculate the

amount of gas present.

7. 1.02 L of O2 collected over water at 20ºC with a total pressure of 755.2 mmHg. Find the mass of O2 collected.

V=1.02 L, P=755.2 mmHg, T=293 K, mass O2 = ?

Ptotal = PO2 + PH2O @20ºC

PO2 = Ptotal - PH2O @20ºC

PO2 = 755.2 mmHg - 17.55 mmHg

PO2 = 737.7 mmHg1 atm

760.0 mmHg PO2 = 737.7 mmHg x

PO2 = 0.9706 atm

n = (P)(V)(R)(T)

n = (0.9706 atm)(1.02 L)(0.0821 )(293 K)atm L

mol K

n = 0.0412 mol

0.0412 mol x 32.00 g O21.00 mol O2

= 1.32 g O2

Ptotal , PH2O PO2 PO2,V, T nO2 gO2

Reactions Involving Gases

The principles stoichiometry can be combined with the gas laws for reactions involving gases.

In reactions of gases, the amount of a gas is often given as a volume instead of moles.

The ideal gas law allows us to convert from the volume of the gas to moles; then we can use the

coefficients in the equation as a mole ratio.

P,V,T of Gas A mol of Gas A mol of Gas B P,V,T of Gas B

The Big Picture of Stoichiometry

Moles of A

Moles of B

Mole to

Mole Ratio*

Grams of A

Grams of B

Molar Mass

Particles of A

Particles of B

Avogadro’s Number

Liters of a Solution of A

Liters of a Solution of B

MolarityMolarity

grams or L of gas A

grams or L of gas B

Gas Laws Gas Laws

8. What volume of H2 is needed to make 37.5 g of CH3OH at 738 mmHg and 355 K? CO(g) + 2 H2(g) → CH3OH(g)

mCH3OH = 37.5g, P = 738 mmHg, T=355 K , VH2 = ?

g CH3OH mol CH3OH mol H2 P,n,T,R V

1 mol CH3OH32.04 g CH3OH

2 mol H21 mol CH3OH V = (n)(R)(T)

P

37.5 g CH3OH X

X

= 2.23 mol H2

1 mol CH3OH32.04 g CH3OH

2 mol H21 mol CH3OH V = (2.23 mol)(.0821 )(355 K)

0.971 atm

atm Lmol K

= 66.9 L H2

738 mm Hg X 1 atm760 mm Hg

= 0.971 atm

L H2 mol H2 mol H2O g H2O

9. How many grams of H2O form when 1.24 L H2 reacts completely with O2 at STP? O2(g) + 2 H2(g) → 2 H2O(g)

VH2 = 1.24 L, P = 1.00 atm, T = 273 K massH2O = ?

1 mol H222.4 L H2

2 mol H22 mol H2O

18.02 g H2O1 mol H2O

10.0 g HgO X X

10. What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s) → 2 Hg(l) + O2(g)

mHgO = 10.0g; P = 0.750 atm; T = 313 K VO2 = ?

g HgO mol HgO mol O2

1 mol HgO216.59 g

1 mol O22 mol HgO

1 mol HgO216.59 g HgO

1 mol O22 mol HgO

= 0.0231 mol O2

V = (0.0231 mol)(.0821 )(313 K)0.750 atm

atm Lmol K

= 0.791 L O2

V = (n)(R)(T)P

P,n,T,R V

4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)

11. Consider the reaction:

How many liters of NO are produced from 10 L of NH3 at STP?

10 LHow many liters of NO are produced from 10 L of O2 at STP?

8 L

How many liters of H2O are produced from 10 L of NH3 at STP?15 L

How many liters of H2O are produced from 10 L of O2 at STP?12 L

Does the “STP” matter? Only the moles !!!!

Kinetic Molecular TheoryParticle volume - The volume of an individual gas particle is small compared to that of its container. Therefore, gas particles are considered to have mass, but no volume.

Particle motion - Gas particles are in constant straight-line motion, except for when they collied with each other or the sides of the container.

Particle collisions - Collisions between gas particles are perfectly elastic. The total kinetic energy of the particles is constant.

There is a lot of empty space between the gas particles compared to the size of the particles. Gases are highly compressible.

The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature.

“Pressure” exerted on the sides of the container is the result of the collisions of all the gas particles present.

Gas Laws Explained by

Kinetic Molecular Theory

Boyle's law: A Kinetic Theory View

“The volume of a gas is inversely proportional to

the pressure.”

Decreasing the volume forces the molecules into a smaller space.

Since the velocity of the molecules does not change, more

molecules will collide with the container at any one instant,

increasing the pressure.

Charles' law: A Kinetic Theory View

“The volume of a gas is directly proportional to the

absolute temperature.”

Increasing the temperature increases the average speed of particles, causing them to hit the wall harder and more

frequently on average.

Since the external pressure remains constant, to keep the internal

pressure constant, the volume must increase.

Amonton's law: A Kinetic Theory View

“The pressure exerted by a gas in a closed container is directly proportional to the

absolute temperature.”

Increasing the temperature increases the average speed of particles, causing them to hit the wall harder and more

frequently on average.

If the volume remains constant, the pressure exerted by the particles

increases with increasing temperature.

1.00 atm

1.37 atm

Avogadro’s law: A Kinetic Theory View

“The volume of a gas is directly proportional to the number of gas molecules.”

Velocity of the molecules does not change. Increasing the number of

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

To keep the pressure constant, the volume must then increase .

Dalton’s law: A Kinetic Theory View

“The sum of the partial pressures of the gases in a

mixture equals the total pressure of the mixture.”

Gas molecules are negligibly small and don’t interact. The molecules behave independently of each other, each gas contributing its own collisions to the

container with the same average kinetic energy.

Because the average kinetic energy is the same, the total pressure is the sum

of the pressures of the separate collisions.

Kinetic Energy and Molecular Velocities

Average kinetic energy of the gas molecules depends on the average mass and velocity.

KEave = 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.

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·m2/s2)

Equating and solving we get 1.5RT = ½NAmu2

NA·mass = molar mass in kg/mol

As temperature increases, the average velocity increases

Temperature vs. Molecular Speed

As the absolute temperature increases,

the average velocity increases and the

distribution function “spreads out,” resulting in

more molecules with faster speeds.

KEavg = 1.5RT

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.

The method of choice for 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.

Molecular Speed vs. Molar MassTo have the same average kinetic energy, heavier

molecules must have a slower average speed.

molecular velocity

rela

tive

# o

f m

olec

ule

s

Calculate the rms velocity of O2 at 25 °C

MM, T urms

√3RTMMurms=

Practice – Calculate the rms velocity of CH4 (MM 16.04) at 25 °C

MM, T urms

Mean Free Path

Molecules in a gas travel in straight lines until they collide with another molecule or the

container.

The average distance a molecule travels between collisions is called the mean free path.

Mean free path decreases as the pressure increases.

Diffusion and EffusionThe 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.

Diffusion and Effusion

Diffusion is the mixing of gas molecules by random motion under conditions where molecular collisions occur.

Effusion is the escape of a gas through a pinhole without molecular collisions.

Graham’s Law of EffusionThomas Graham (1805–1869)

For two different gases at the same temperature, the ratio of their rates of

effusion is given by the following equation:

“The rate of gas movement is inversely proportional to the square root of its molar mass.”

Calculate the ratio of rate of effusion for oxygen to hydrogen.

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

This means that, on average, the O2 molecules are traveling at ¼ the speed of H2 molecules.

0.251

Calculate the molar mass of a gas that effuses at a rate 0.462 times N2.

MMgas A = ?

rateA/rateB, MMN2 MMunknown

MMN2 = 28.02 g/mol

(rate gas A) 2

(rate gas B)2(Molar Mass gas B)

(Molar Mass gas A)=

(0.482)2 = (28.02 g/mol)

(Molar Mass gas A)

(Molar Mass gas A) = (28.02 g/mol)

(0.482)2

(Molar Mass gas A) = 131 g/mol

Ideal vs. Real Gases

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

Ideal gas laws assume

1. no attractions between gas molecules

2. gas molecules do not take up space based on the kinetic-molecular theory

At low temperatures and high pressures these assumptions are not valid.

PV = nRT n = PV/RT = 1


This graph shows how real gas's behavior deviates from ideal behavior as pressure increases. If a gas were to behave perfectly ideally, then the ratio PV/RT would

equal exactly 1 for one mole of gas (dashed line).

This graph shows how a real gas's behavior deviates from ideal behavior as pressure increases. Each curve represents the behavior of the gas at a different temperature. If a gas were to behave perfectly ideally, then the ratio PV/RT would equal exactly 1 for one mole of gas (dashed line).


Real Gas Behavior

Because real molecules take up space, the molar volume of a real gas is larger than

predicted by the ideal gas law at high pressures.

The Effect of Molecular VolumeJohannes van der Waals (1837–1923)

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.

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.

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.

van der Waals’ Equation

Combining the equations to account for molecular volume and intermolecular attractions we get the

following equation used for real gases:

Van der Waals Constants for Some Common Gases

0.0340.2111.352.324.190.2441.391.366.493.592.254.175.46

HeNeArKrXeH2N2O2Cl2CO2CH4NH3H2O

0.02370.01710.03220.03980.05110.02660.03910.03180.05620.04270.04280.03710.0305

Gasa

atm*L2

mol2b

L

mol

(P +n2aV 2 )(V − nb) = nRTVan der Waals

equation for nmoles of a real gas adjusts P up adjusts V down

The volume taken up by the gas particles themselves is less important at lower pressure (a) than at higher pressure (b). As a result, the volume of a real gas at high pressure is somewhat larger than the ideal value.

Real Gas Behavior vs Ideal Gas Behavior

Ideal Gas Behavior vs Real Gas Behavior

· Particle size is small compared to space between particles.

· Interactions between particles are insignificant.

Ideal gas conditions · High temperature · Low pressure Non-ideal gas conditions

· Low temperature · High pressure

· Particle size is significant compared to space between particles.

· Interactions between particles are significant.

· Intermolecular interactions

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