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Gases - SFASU

Transcript Gases - SFASU

Topic 5 Gases

Gases have several characteristics that distinguish them from liquids and solids. For one, gases can be compressed into smaller containers or volumes. Gas molecules have plenty of space between them thereby allowing for compression while solid and liquid molecules are touching thereby restricting their compressibility. Gases also can relate pressure (P), volume (V), temperature (T), and number of mols (n) with a fair degree of accuracy by empirical relationships.

Gas Laws

We will examine the quantitative relationships, or

empirical laws

, governing gases.

Basically, we will learn about equations that are available to calculate P, V, T, & n for gases. Note: these equations are for gases only!

What is pressure?

Pressure

Pressure - force exerted per unit area of surface by molecules in motion.

P = Force/unit area

1 atmosphere (atm) = 14.7 psi 1 atmosphere = 760 mm Hg = 760 Torr (memorize) 1 atmosphere = 101,325 Pascals 1 Pascal = 1 kg/m .

s 2 1 atm = 0.101325 MPa = 1.01325 bar (helpful in HW) 3

Chemist have units of pressure based on the mercury barometer (mm Hg).

Barometer – device for measuring the pressure of the atmosphere.

It consists of a glass tube about 1 meter in length filled with mercury, Hg, and inverted in a dish of the same liquid metal. At sea level, the Hg in the tube is at a height of about 760 mm Hg above the level in the dish. This height is a direct measure of atmospheric pressure. 760 mm Hg is referred to as standard pressure at 25 o C. We also refer to standard pressure as being equal to 1 atm (1 atm = 760 mm Hg).

A similar process for measuring pressure in a vessel is done by using a sealed u-type flask called a manometer.

A mercury barometer

Pressure Conversions The pressure of gas in a flask is 797.7 mmHg. What is the pressure in atm?

For this type of problem, you need to perform dimensional analysis to cancel units.

Note: 760 mm Hg is an exact number and does not affect significant digits in answer.

HW 37 code: pressure

The Empirical Gas Laws

All gases behave quite simply; This allows relationships to be determined by holding any two physical properties constant (P, V, T, or n) which leads us to the empirical gas laws. The gas property of compressibility lead to the discovery of Boyle’s Law:

The volume of a sample of gas at a given temperature varies inversely with the applied pressure. V a 1/P (at constant moles (n) and T)

meaning, PV = constant

P 1  V

and therefore

1  P 2  V 2 6

A Problem to Consider

V 1 P 1

A sample of chlorine gas has a volume of 1.8 L at 1.0 atm.

P 2

If the pressure increases to 4.0 atm (at constant temperature), what would be the new volume?

We assume constant number of mols of gas since there is no evidence of gas escaping. Since we are at constant temperature, it allows us to use Boyle’s Law; we know the pressure and volume at a given temperature and a new pressure later on so we can easily solve for the new volume, V 2 . P 1 V 1 = P 2 V 2 rearranging to solve for the new volume, V 2 Notes: 1.) as long as the pressures or volumes are the same unit, they will cancel out. 2.) Since the pressure increased and we know the volume will have the inverse effect, we anticipate the volume to decrease which it does (1.8 L to 0.45 L).

HW 38

code: boyle

The Empirical Gas Laws

Temperature also affects gas volume. A gas contracts when cooled & expands when heated which gets us to Charles’s Law:

of gas The volume occupied by any sample at constant pressure and mols is directly proportional to its absolute temperature, meaning Kelvin scale.

V a T abs (at constant n and P) V T

and therefore

V 1 T 1  V 2 T 2

where T must be in K.

A Problem to Consider

A sample of methane gas that has a volume of

V 1 T 1 T 2

3.8 L at 5.0

°C is heated to 86.0°C at constant pressure. Calculate its new volume.

We assume constant number of mols of gas since there is no evidence of gas escaping. Since we are at constant pressure, it allows us to use Charles’s Law; we know the temperature and volume at a given pressure and a new temperature later on so we can easily solve for the new volume, V 2 .

V 1 T 1  V 2 T 2

rearranging to solve for the new volume, V 2 and using Kelvin temperature we obtain Notes: 1.) you must use Kelvin temperature: K = o C + 273.15 K 2.) Since the temperature increased and we know the volume will also increase, we anticipate the volume to increase which it does (3.8 L to 4.9 L).

HW 39 code: charles

The Empirical Gas Laws

Amontons’ Law (also know as Gay Lussac’s Law) :

The pressure exerted by a gas at constant volume and number of mols is directly proportional to its absolute temperature, K. P a T abs (at constant n and V) P T

and therefore

P 1 T 1  P 2 T 2

where T must be in K.

A Problem to Consider

P 1 T 1

An aerosol can has a pressure of 1.4 atm at 25 °C. What pressure would it attain at 1200 °C, assuming the volume remained constant?

T 2 We assume constant number of mols of gas since there is no evidence of gas escaping. Since we are at constant volume, it allows us to use Amontons’ Law; we know the temperature and pressure at a given volume and a new temperature later on so we can easily solve for the new pressure, P 2 .

P

T

1 

P

T

rearranging to solve for the new pressure, P 2 and using Kelvin temperature we obtain HW 40 code: temp Notes: 1.) you must use Kelvin temperature: K = o C + 273.15 K 2.) Since the temperature increased and we know the pressure will also increase, we anticipate the pressure to increase which it does (1.4 atm to 6.9 atm).

The Empirical Gas Laws

Combined Gas Law:

In the event that all three parameters, P, V, and T, are changing, their combined relationship is defined as follows (at constant n): PV a T abs ( at constant n) PV T

and therefore

P 1

1 T 1  P 2

2 T 2

where T must be in K.

Note: the combined gas law involves all three previous laws discussed and can be used to solve any problems involving constant n. If any of the variables are constant, they will simply cancel out in the calculation.

A Problem to Consider

V 1 T 1

A sample of carbon dioxide gas occupies 4.5 L at 30 °C and

P 1

650 mm Hg. What volume would it occupy at 800 mm Hg

and 200 °C?

P 2 We assume constant number of mols of gas since there is no evidence of gas escaping. Since this problem involves one set of conditions as compared to another set of conditions, we will use the combined gas law; we know the temperature, volume, and pressure of the gas and are given a new temperature and pressure so we can easily solve for the new volume, V 2 .

P 1

1 T 1  P 2

2 T 2

rearranging to solve for the new volume, V 2 and using Kelvin temperature we obtain HW 41 code: combine

The Empirical Gas Laws

Avogadro’s Law:

At a given temperature and pressure, the volume of a gas sample is directly proportional to the number of moles of gas particles in the gas sample.



n at constant T & P This implies that if we have twice as many gas particles, the gas would occupy twice as much volume at the same T & P. Another way of stating Avogadro’s law is to say: at the same T & P, equal volumes of any gas contain equal number of particles.

The volume of one mole of any gas is called the

molar gas volume, V m

and has the units of L/mol.

The Empirical Gas Laws

Volumes of gases are often compared at standard temperature and pressure ( STP ), chosen to be 0 o C (273 K) and 1 atm pressure.

At STP, the molar volume,

V m

, that is, the volume occupied by one mole of any gas, is

22.4 L/mol

Conversion factor for gases at STP: 1 mol of any gas = 22.4 L

1 mol H 2 at STP has volume of 22.4L

1 mol of O 2 at STP has a volume of 22.4 L 1 mol of any gas at STP has a volume of 22.4 L HW 42 code: molar

The Ideal Gas Law

From the empirical gas laws, we have shown that volume varies in proportion to pressure, absolute temperature, and moles.

V V V

 

1/P Boyle' s Law T

abs

Charles' Law



n Avogadro' s Law

combining gives

V  n

T abs

/P 16

The Ideal Gas Law

V  n

T abs

/P This implies that there must exist a proportionality constant governing these relationships meaning they must be equal based on some constant

V



" R"

(

nT

abs

P

)

where “R” is the proportionality constant referred to as the

ideal gas constant

(independent of gas species).

The Ideal Gas Law

We can calculate R at STP using the ideal gas law because the only unknown is “R”:

V

1 mol of any gas has a volume of 22.4 L at 273 K and 1 atm of pressure 

" R"

( nT abs P ) rearranging to

R



VP nT R



(22.4

L)(1.00

atm) (1.00

mol)(273 K)



0.0821

L



atm mol



K

Note: memorize this common value for R. There are other values of R that can also be used; however, caution should be taken in the differences in units.

The Ideal Gas Law

Thus, the

ideal gas equation

, is usually expressed in the following form:

PV



nRT

P is pressure (in atm) V is volume (in liters) n is number of atoms (in moles) R is universal gas constant - 0.0821 L

atm/mol

K T is temperature (in Kelvin)

Note: R value with correct units to cancel with units given for the other variables.

A Problem to Consider

An experiment calls for 3.50 moles of chlorine gas, Cl 2 . What volume would this be if the gas volume is measured at 34 °C and 2.45 atm?

Since this problem involves P, V, T, and n, we will use the ideal gas law; we know the temperature, mols, and pressure of the gas so we can easily solve for the volume. PV



nRT rearranging to solve for the volume and using Kelvin temperature we obtain HW 43 code: ideal

Molecular Weight Determination

In section 3 we showed the relationship between moles and mass.

molar mass, M m 

mass, m moles, n



m M m

Molecular Weight Determination

If we substitute



m M m

PV



(

m M m

) RT

If we solve this equation for the molar mass, we obtain M m  mRT PV The procedure for determining the molar mass of a gas from measurements of m, P, T, and V is known as the Dumas method.

A Problem to Consider

of 5.75 L at 25 °C and a pressure of 1.08 atm. Calculate its molar mass. Which of the following gases is most likely to be the unknown gas - N 2 , O 2 , or HCl?

M m  mRT PV Molar mass: N 2 – 28.02 g/mol O 2 – 32.00 g/mol

Therefore, most likely unknown gas is HCl.

HCl – 36.46 g/mol 23

Density Determination

We can use the Dumas method, also known as the vapor density method, to calculate molar mass based on the density of a gas. If we look again at our derivation of the molar mass equation,

PV



(

) RT

M m we can solve for molar mass and insert density,d = m/V: M m  mRT VP  dRT P We can use the above equation to solve for molar mass; however, if we are at STP, we can simplify things further because we also know that any gas at STP has a molar volume of 22.4 L/mol meaning V m  22.4

L mol  V n  RT P therefore, M m = d V m @ STP 24

A Problem to Consider

Calculate the density of ozone gas, O 3 50 °C and 1.75 atm of pressure.

323 K

(M m = 48.0 g/mol), at We must use the Dumas method equation and rearrange to solve for density. Note: we can’t use the simpler equation because we are not at STP. M m  dRT P rearranges

d  PM m RT

Note: typical density units for a gas are g/L HW 44 code: vary

Stoichiometry Problems Involving Gas Volumes

2 KClO

(s)



2 KCl(s)



3 O

0.0100 mol ?L

( g )

Suppose you heat 0.0100 mol of potassium chlorate, KClO 3 , in a test tube. How many liters of oxygen can you produce at 298 K and 1.02 atm?

This type of problem is similar to other stoichiometry problems except for the fact that we are looking for the volume of O 2 generated instead of mols or mass. The mols of gas is related to the volume through the ideal gas law.

Stoichiometry: Ideal gas law (PV = nRT) solving for V:

Stoichiometry Problems Involving Gas Volumes

Many air bags are inflated with N 2 gas by the following rxn: 6NaN 3 (s) + Fe 2 O 3 (s)  3 Na 2 O (s) + 2Fe (s) + 9N 2 (g)

? g n

How many grams of NaN 3 N 2 would be needed to provide gas at 25 o C and 748 mmHg ?

298 K

75.0 L of

This problem gives us the information needed to calculate the mols of N 2 gas to inflate the air bag by using the ideal gas law. Once we have the mols of N 2 gas, we can calculate the mass of sodium azide needed to produce that amount of gas through stoichiometry.

Ideal gas law (PV = nRT) solving for n: Stoichiometry: HW 45 code: stoich

Partial Pressures of Gas Mixtures

Dalton’s Law of Partial Pressures:

the sum of all the pressures of all the different gases in a mixture equals the total pressure of the mixture. 28

Partial Pressures of Gas Mixtures

The composition of a gas mixture is often described in terms of its mole fraction.

The

mole fraction

of a component gas is the fraction of moles of that component in the total moles of gas mixture  A 

mole fraction of A



n

tot

note: no units on mole fraction

Mole fraction can also be calculated through partial pressure. 29

Partial Pressures of Gas Mixtures

By rearranging the partial pressure of a component gas, “A”, is then defined as P A   A  P total 30

A Problem to Consider

A 10.0 L flask contains 1.031 g O 2 and 0.572 g CO 2

291 K

gases at 18 o C. What are the partial pressures of O 2 What is the mole fraction of O 2 and CO 2 ? What is the total pressure? in the mixture?

We know that each gas can be treated independently by using the ideal gas law P O2 V = n O2 RT and P CO2 V = n CO2 RT and rearranging to solve for pressure.

Total pressure is the sum of all gas pressures: Mole fraction may be calculated in several ways but the easiest way in this problem will be to use the partial pressure: HW 46

code: partial

Collecting Gases “Over Water”

A useful application of partial pressures arises when you collect gases over water. Basically, a gas displaces water in an amount equal in volume to the gas (V gas = V H2O collected).

As gas bubbles through the water, the gas becomes saturated with water vapor meaning that you must account for the water vapor pressure in the measured pressure.

The partial pressure of the water in this “mixture” depends only on the temperature (vapor pressure of water which can be looked up in reference books). 32

A Problem to Consider

Suppose a 156 mL sample of H 2 gas was collected over water at 19 o C and 769 mm Hg. What is the mass of H 2 collected?

First, we must determine the partial pressure of the hydrogen gas since the total pressure involves water vapor. Note that the water vapor pressure can be obtained from water vapor pressure tables in reference books at the correct temperature which in this case is 19 o C (16.5 mm Hg).

P H2 = P total - P H2O = 769 mm Hg – 16.5 mm Hg = 752 mm Hg

Suppose a 156 mL sample of H 2 mm Hg. What is the mass of H 2 gas was collected over water at 19 o C and 769 collected? 292 K Next, we can determine the mols of hydrogen gas through the ideal gas law, P H 2 V = n H 2 RT, by using the partial pressure of hydrogen gas and the volume of water displaced since it equals the volume of hydrogen gas generated. We also will need to make some conversions for units can cancel.

Lastly, we need to convert from mols to mass of hydrogen: HW 47 & 48 code: gas

Kinetic-Molecular Theory of gases A simple model based on the actions of individual atoms used to explain the behavior of ideal gases. There are five postulates to this theory:

• Volume of particles can be neglected but volume of container cannot (smaller container, higher probability of gas molecules having collisions).

• Particles are in constant random motion; move in straight lines in all directions and at various speeds (smaller mass, faster it moves, higher probability of collisions).

• • No inherent attractive or repulsive forces When molecules collide, the collisions are elastic (total KE remains constant; there may be a transfer of energy but none lost).

• The average kinetic energy of a collection of particles is proportional to the temperature (K) collisions) – higher T, greater KE (higher the temp, faster molecules move, higher probability of

Gas pressure is due to collisions of gas particles with a surface (container). If we add up the forces due to all the collisions of gas particles with the surface

(container), and divide the result by the area of the surface, we get the pressure.

Molecular Speeds; Diffusion and Effusion

The root-mean-square (rms) molecular speed,



– m/s,

is a type of average molecular speed, equal to the speed of a molecule having the average molecular kinetic energy. It is given by the following formula:



rms

3RT M m 36

Molecular Speeds; Diffusion and Effusion

Diffusion

is the transfer of a gas through space or another gas over time from a region of high concentration (more crowded) to a region of low concentration (less crowded)

Effusion

is the transfer of a gas through a membrane or orifice. The equation for the rms velocity of gases shows the following relationship between rate of diffusion and molecular mass (inversely proportional). Graham’s Law:

Rate of diffusion



1 M

expect larger molecules (larger M m ) to move slower and hence, slower diffusion rates.

The rate of diffusion/effusion of molecules from a container depends on three factors: 1.) cross-sectional area of the hole (the larger it is; the more likely molecules are to escape) 2.) the number of molecules per unit volume (conc of gas); the more crowded the molecules, the more likely for the molecules to diffuse.

3.) the average molecular speed (affected by temp and molar mass); faster the molecules move (smaller the M m ) and higher the temperature, the higher probability they will be able to diffuse.

The temp, conc., molar mass, and size of the actual hole involved in gas molecules escaping affects the rate of diffusion/effusion. Therefore, if you compare the diffusion/effusion of different gases that are the same concentration, in the same container, have the same T & P, then the one factor that will dictate the rate would

be the molar mass of species.

Molecular Speeds; Diffusion and Effusion

According to

Graham’s law,

the rate of effusion or diffusion is inversely proportional to the square root of its molecular mass . (for gases in same container at constant T & P). The following relationship allows for comparison of gases:

Rate of effusion of gas " A"



Rate of effusion of gas " B" M

of gas B M

of gas A

A Problem to Consider

How much faster would H 2 gas effuse through an opening than methane, CH 4 ?

We expect hydrogen gas to effuse faster than methane gas because it is lighter gas so we will put it in the numerator of our comparison equation: