The Gas State Gases are everywhere – atmosphere, environmental

processes, industrial processes, bodily functions Gases have unique properties from liquids and solids Gases are compressible (very important for storage) Gas particles are widely separated and move at very

fast speeds Most gases have relatively low densities Gas have relatively low viscosity (resistance to

movement) allowing them move freely through pipes and even small orifices

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The Gas State Chemical behavior of gases depends on composition Physical behavior of all gases is similar Gases are miscible mixing together in any proportion Behavior of gases described by ideal gas law and

kinetic-molecular theory, the cornerstone of this chapter

Gas volume changes greatly with pressureGas volume changes greatly with temperatureGas volume is a function of the amount of gas

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Cylinders of Gas

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The Empirical Gas Laws Gas behavior can be described by pressure,

temperature, volume, and molar amount

Holding any two constant allows relations between the other two

Boyle’s Law : The volume of a sample of gas at a given temperature varies inversely with the applied

pressure

Vα 1/P

PV= constant

P1V1=P2V204/21/234

The Empirical Gas LawsBoyle’s LawGas Pressure-Volume Relationship

5

Practice Problem

Boyle’s LawA 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? using P1V1=P2V2

V2 = P1V1/P2 = (1.0 atm) × (1.8 L) / (4.0 atm) V2= 0.45 L

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The Empirical Gas LawsCharles’s LawThe volume occupied by any sample of gas at constant pressure is directly proportional to its absolute temperature.

Vα T absAssumes constant moles and pressure

Temperature on absolute scale (oC + 273.15)

V/T = constant V1/T1 = V2/T2

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(Tabs (K( = oC + 273.15(

The Empirical Gas LawsCharles’s Law: Linear Relationship of Gas Volume and Temperature at Constant Pressure

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V = at + b0 = a(-273.15) + b

b = 273.15bV = at + 273.15a = a(t +

273.15)

Practice Problem

Charles’s LawA 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.using V1/T1 = V2/T2

Convert temperature (°C ) to absolute (kelvin)50 °C = 5.0 +273.15 = 278.15 K86.0 °C = 86.0 + 273.15 = 359.15 KV2 = (T2/T1) V1 = (359.15 K /278.15 K ) × 3.8 L

V2 = 4.9 L04/21/239

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The Combined Gas LawCombined Gas Law: In the event that all three parameters, P, V, and T, are changing, their combined relationship is defined as follows

assuming the mass of the gas (number of moles) is constant .

V1 / V2 = T1 / T2 (fixed P,n(

P1V1 = P2V2

)fixed T,n(Boyle’s Law

Charles’ Law

V x P = const

V / T = const

1662

1787

P1V1/T1 = P2V2/T2

Avagadro’s LawAvogadro’s Law

The volume of a sample of gas is directly proportional to the number of moles of gas, n

Vα nV/n = constant

Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules (mol)

V1/n1 = V2/n2

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Avagadro’s Law

Avogadro’s Law 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 (273.15 oK) and 1 atm pressure At STP, the molar volume, Vm, that is, the volume

occupied by one mole of any gas, is 22.4 L/mol

VSTP/NSTP = Vm = 22.4 L ( at STP )

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The Ideal Gas LawFrom the empirical gas laws, we see that volume varies

in proportion to pressure, absolute temperature, and the mass of gas (moles) present

Boyle law Vα 1/P V= constant x 1/P

Charles law Vα T abs V = constant x T

Avogadros law Vα n V = constant x n

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The Ideal Gas LawThis implies that there must exist a proportionality

constant governing these relationships

Combining the three proportionalities, we can obtain the following relationship.

V m = “R” x n x (T abs/P)where “R” is the proportionality constant referred to as the Ideal Gas Constant, which relates Molar Volume (V) to the ratio of Temperature to Pressure T/P

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The Ideal Gas LawThe ideal gas equation is usually expressed in the following form:

PV = nRT

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P = Pressure (in atm)V = Volume (in liters) n = Number of atoms (in moles)R = Universal gas constant 0.0821 L.atm/mol.KT = Temperature (in 0 Kelvin = °C + 273.15)

The Ideal Gas Law

What is R, universal gas constant?

the R is independent of the particular gas studied

PVR

nT (1atm)(22.414L)

(1.00 mol)(273.15 K)

-1-1 K mol atm L 0.082057 R

PV = nRT

Practice ProblemA steel tank has a volume of 438 L and is filled with 0.885 kg of O2.

Calculate the pressure of oxygen in the tank at 21oC use PV = nRTV = 438 L, R = 0.0821 L.atm/mol.K, T = 21 + 273.15 = 294.15 K , n = 885/32 = 27.7 mol.So the pressure = nRT/V = 27.7 mol x 0.0821 L.atm/mol.K x 294.15 K/483 L= 1.53 atm

Mixtures of Gases Dalton’s Law of Partial Pressures

The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.

V

RTn P

V

RTnnnPPP P

V

RTnP ,

V

RTnP ,

V

RTnP

Totaltotal

321321total

33

22

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The Ideal Gas Law is an empirical relationship based on experimental observations.Boyle, Charles and Avogadro.

Kinetic Molecular Theory is a simple model that attempts to explain the behavior of gases.

The Kinetic Molecular Theory of Gases

The Kinetic Molecular Theory of Gases1 .A pure gas consists of a large number of identical molecules separated

by distances that are large compared with their size. The volumes of the individual particles can be assumed to be negligible (zero).

2 .The molecules of a gas are constantly moving in random directions with a distribution of speeds. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas.

3 .The molecules of a gas exert no forces on one another except during collisions, so that between collisions they move in straight lines with constant velocities. The gases are assumed to neither attract or repel each other. The collisions of the molecules with each other and with the walls of the container are elastic; no energy is lost during a collision.

4 .The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

Diffusion :

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

Effusion:

Is the process in which individual molecules flow through a hole without collisions between molecules.

According to Graham's law, the rate at which gases effuse (i.e., how many molecules pass through the hole per second) is dependent on their molecular weight; gases with a lower molecular weight effuse more quickly than

gases with a higher molecular weight .

The equation for effusion is given as

1 2

2 1

Rate of effusion of gas M

Rate of effusion of gas M

Where M1 and M2 are molecular masses of gases 1 and 2.

Real Gases: Deviations from Ideality

Real gases behave ideally at ordinary temperatures and pressures.

At low temperatures and high pressures real gases do not behave ideally.

The reasons for the deviations from ideality are:1. The molecules are very close to one another, thus

their volume is important.2. The molecular interactions also become important.

Real Gases:Deviations from Ideality

van der Waals’ equation accounts for the behavior of real gases at low temperatures and high pressures.

P + n a

VV nb nRT

2

2

The van der Waals constants a and b take into account two things: a accounts for intermolecular attraction

For nonpolar gases the attractive forces are London Forces For polar gases the attractive forces are dipole-dipole attractions or hydrogen bonds.b accounts for volume of gas molecules

At large volumes a and b are relatively small and van der Waal’s equation reduces to ideal gas law at high temperatures and low pressures.

Real Gases:Deviations from Ideality

Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the ideal gas law.

PV = nRTP = nRT/V n = 84.0g * 1mol/17 g T = 200 + 273P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K)

( 5 L)P = 38.3 atm

Real Gases: Deviations from Ideality Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L

container at 200. oC using the van der Waal’s equation. The van der Waal's constants for ammonia are: a = 4.17 atm L2 mol-2 b =3.71x10-2 L

mol-1

P + n a

VV nb nRT

2

2

n = 84.0g * 1mol/17 g T = 200 + 273P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K) (4.94 mol)2*4.17 atm L2 mol-2

5 L – (4.94 mol*3.71E-2 L mol-1) (5 L)2

P = 39.81 atm – 4.07 atm = 35.74P = 38.3 atm

7% error

2

2

V

an -

nb-V

nRTP