Section 5 Gases

1Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General

Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

Section 5

Gases



Gas Laws

• In the first part of this chapter we will examine the quantitative relationships, or empirical laws, governing gases.

• First, however, we need to understand the concept of pressure.



Pressure

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

– 1 atmosphere (atm) = 14.7 psi– 1 atmosphere = 760 mm Hg = 760 Torr– 1 atmosphere = 101,325 Pascals– 1 Pascal = 1 kg/m.s2

–1 atm = 0.101325 MPa = 1.01325 bar

P = Force/unit area



A mercury barometer.



Pressure Conversions

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

HW 37



The Empirical Gas Laws• Boyle’s Law: The volume of a sample of gas

at a constant temperature varies inversely with the applied pressure.

V 1/P [constant moles (n) and T] or

2211 VPVP



A Problem to Consider• A sample of chlorine gas has a volume of 1.8 L

at 1.0 atm. If the pressure increases to 4.0 atm (at constant temperature), what would be the new volume?

HW 38



The Empirical Gas Laws

• Charles’s Law: The volume occupied by any sample of gas at constant pressure is directly proportional to its absolute temperature.

V Tabs (constant moles and P)

or

2

2

1

1TV

TV



A Problem to Consider

• A sample of methane gas that has a volume of 3.8 L at 5.0°C is heated to 86.0°C at constant pressure. Calculate its new volume.

HW 39



The Empirical Gas Laws• Gay-Lussac’s Law: The pressure exerted by

a gas at constant volume is directly proportional to its absolute temperature.

P Tabs (constant moles and V)

or

2

2

1

1TP

TP



A Problem to Consider• 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?

HW 40




• 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):

2

22

1

11T

PT

P VV



A Problem to Consider• A sample of carbon dioxide gas occupies 4.5

L at 30°C and 650 mm Hg. What volume would it occupy at 800 mm Hg and 200°C?

HW 41



• The volume of one mole of gas is called the molar gas volume, Vm.

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

The Empirical Gas Laws• Avogadro’s Law: Equal volumes of any two gases

at the same temperature and pressure contain the same number of molecules.

nV at constant T & P



– At STP, the molar volume, Vm, that is, the volume occupied by one mole of any gas, is 22.4 L/mol

at STP, 1 mol of any gas = 22.4 L


• Avogadro’s Law

HW 42



The Ideal Gas Law

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

Law sBoyle' 1/PV

Law sAvogadro' nV Law Charles' TV abs

/PnV absT



The Ideal Gas Law• This implies that there must exist a proportionality

constant governing these relationships.

)( PnTabs R""V

where “R” is the proportionality constant referred to as the ideal gas constant (independent of gas).

/PnV absT



The Ideal Gas Law

• At STP, only missing “R”: 1 mol, 22.4 L, 273 K, 1 atm

nTVP R

K) mol)(273 (1.00atm) L)(1.00 (22.4 R

KmolatmL 0.0821



The Ideal Gas Law

• Thus, the ideal gas equation, is usually expressed in the following form:

nRT PV 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/K.molT is temperature (in Kelvin)



A Problem to Consider• An experiment calls for 3.50 moles of chlorine

gas, Cl2. What volume would this be if the gas volume is measured at 34°C and 2.45 atm?

HW 43



Molecular Weight Determination

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

mass molecular massmoles

or

mMmn



Molecular Weight Determination

• If we substitute this in the ideal gas equation, we obtain

RT)(PVmM

mIf we solve this equation for the molecular mass, we obtain

PVmRT Mm



A Problem to Consider• A 9.25 gram sample of an unknown gas

occupied a volume of 5.75 L at 25°C and a pressure of 1.08 atm. Calculate its molecular mass. Which of the following gases is most likely to be the unknown gas - N2, O2, or HCl?



Density Determination

• If we look again at our derivation of the molecular mass equation,

RT)(PVmM

mwe can solve for m/V, which represents density.

RTPM

Vm mD




• Calculate the density of ozone gas, O3 (Mm = 48.0g/mol), at 50°C and 1.75 atm of pressure.

HW 44



Stoichiometry Problems Involving Gas Volumes

• Suppose you heat 0.0100 mol of potassium chlorate, KClO3, in a test tube. How many liters of oxygen can you produce at 298 K and 1.02 atm?

)g(O 3 KCl(s) 2 (s)KClO 2 23



Stoichiometry Problems Involving Gas Volumes

Many air bags are inflated with N2 gas by the following rxn:

6NaN3 (s) + Fe2O3 (s) 3 Na2O (s) + 2Fe (s) + 9N2 (g)

How many grams of NaN3 would be needed to provide 75.0 L of N2 gas at 25oC and 748 mmHg?

HW 45



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.

....PPPP cbatot




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

tot

A

tot

AA P

Pnn Aof fraction Mole

– The mole fraction of a component gas is the fraction of moles of that component in the total moles of gas mixture.




• The partial pressure of a component gas, “A”, is then defined as

totAA P P – Applying this concept to the ideal gas equation,

we find that each gas can be treated independently.

RTn VP AA



A Problem to ConsiderA 10.0 L flask contains 1.031 g O2 and 0.572 g CO2 gases

at 18oC. What are the partial pressures of O2 and CO2? What is the total pressure? What is the mole fraction of O2 in the mixture?

HW 46



Collecting Gases “Over Water”

• A useful application of partial pressures arises when you collect gases over water.

– As gas bubbles through the water, the gas becomes saturated with water vapor.

– The partial pressure of the water in this “mixture” depends only on the temperature (vapor pressure of water).



A Problem to Consider• Suppose a 156 mL sample of H2 gas was

collected over water at 19oC and 769 mm Hg. What is the mass of H2 collected?


Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. HW 47 & 48



Kinetic-Molecular Theoryof gases

A simple model based on the actions of individual atoms

• Volume of particles can be neglected but volume of container cannot.

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

• No inherent attractive or repulsive forces• When molecules collide, the collisions are elastic

(total KE remains constant).• The average kinetic energy of a collection of particles

is proportional to the temperature (K) – higher T, greater KE



Molecular Speeds; Diffusion and Effusion

• The root-mean-square (rms) molecular speed, u – 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:

mM3RT u




• Diffusion is the transfer of a gas through space or another gas over time.

• 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 effusion and molecular mass (inversely proportional). Graham’s Law:

mM1 effusion of Rate



The rate of 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).

3.) the average molecular speed (affected by temp and molar mass)

Therefore, temp, conc., molar mass, and size of hole affects rate of effusion.




• According to Graham’s law, the rate of effusion or diffusion is inversely proportional to the square root of its molecular mass. (for same container at constant T & P). Following relationship allows for comparison of gases:

A gas of MB gas of M

B"" gas ofeffusion of RateA"" gas ofeffusion of Rate

m

m




• How much faster would H2 gas effuse through an opening than methane, CH4?

)(HM)(CHM

CH of RateH of Rate

2m

4m

4

2

8.2g/mol 2.0g/mol 16.0

CH of RateH of Rate

4

2

So hydrogen effuses 2.8 times faster than CH4

HW 49

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Section 5 Gases