properties of gases 1. key objectives: be able to state (i.e. define) and use in practical...

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Properties of Gases

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Key objectives: Be able to state (i.e. define) and use in practical calculations, the following.

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1. Concept of force and pressure.

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1. Concept of force and pressure.2. Boyle’s Law

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1. Concept of force and pressure.2. Boyle’s Law3. Charles’ Law

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1. Concept of force and pressure.2. Boyle’s Law3. Charles’ Law4. Avogadro’s Hypothesis

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1. Concept of force and pressure.2. Boyle’s Law3. Charles’ Law4. Avogadro’s Hypothesis 5. Ideal gas law

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1. Concept of force and pressure.2. Boyle’s Law3. Charles’ Law4. Avogadro’s Hypothesis 5. Ideal gas law6. Dalton’s law of partial pressures

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1. Concept of force and pressure.2. Boyle’s Law3. Charles’ Law4. Avogadro’s Hypothesis 5. Ideal gas law6. Dalton’s law of partial pressures7. Graham’s law of effusion

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1. Concept of force and pressure.2. Boyle’s Law3. Charles’ Law4. Avogadro’s Hypothesis 5. Ideal gas law6. Dalton’s law of partial pressures7. Graham’s law of effusion8. Kinetic theory and real gases

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Gas: Refers to a substance that is entirely gaseous at ordinary temperatures and pressures.

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The 4 important measurable properties of gases are:

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1. pressure

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1. pressure 2. volume

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1. pressure 2. volume 3. temperature

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1. pressure 2. volume 3. temperature 4. mass (or moles)

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Pressure: Gases exert pressure on any surface with which they come into contact. That is, all gases expand uniformly to occupy whatever space is available.

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Pressure is define by the equation:

force pressure = area

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Pressure is define by the equation:

force pressure = area or F P = A

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Units of pressure

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Units of pressure

distance velocity = time

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Units of pressure

distance velocity = time velocity acceleration = time

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Units of pressure

distance velocity = time velocity acceleration = time

i.e. distance acceleration = time2

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To find the SI units of force:

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To find the SI units of force: force = mass x acceleration

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distance force = mass x time2

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Now plug in the SI units for each quantity on the right-hand side of the equation. The SI unit of force is the newton (named after Newton) and abbreviated N.

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Now plug in the SI units for each quantity on the right-hand side of the equation. The SI unit of force is the newton (named after Newton) and abbreviated N.

1 N = 1 kg m s-2

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The SI units of pressure are: Nm-2

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The SI unit of pressure is called the pascal (named after Pascal) and is abbreviated Pa.

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1 Pa = 1 Nm-2 = 1 kg m-1 s-2

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1 Pa = 1 Nm-2 = 1 kg m-1 s-2

The most common unit of pressure is the non-SI unit, the atmosphere, abbreviated atm.

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1 atm = 76 cm Hg

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1 atm = 76 cm Hg = 760 mm Hg

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1 atm = 76 cm Hg = 760 mm Hg = 760 torr (named after Torricelli)

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1 atm = 76 cm Hg = 760 mm Hg = 760 torr (named after Torricelli) = 101325 Pa (by definition)

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1 bar = 10 5 Pa

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1 bar = 10 5 Pa

Hence 1atm = 1.01325 bar

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Atmospheric pressure is measured by a barometer.

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The difference between a gas and a vapor:

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The difference between a gas and a vapor: A gas is a substance normally in the gaseous

state at ordinary temperatures and pressures.

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state at ordinary temperatures and pressures. A vapor is the gaseous form of any substance

that is a liquid or a solid at normal temperatures and pressures.

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state at ordinary temperatures and pressures. A vapor is the gaseous form of any substance

that is a liquid or a solid at normal temperatures and pressures.

Thus, at room temperature and 1 atm, we speak of water vapor and helium gas.

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Temperature: For the gas equations, the temperature is measured on the Kelvin scale – the units are K (kelvin).

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The absolute lowest temperature is zero degrees kelvin.

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The absolute lowest temperature is zero degrees kelvin.

The temperature in K is determined from the temperature measured in oC by:

degrees K = degrees oC + 273.15

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The Gas Laws

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The Gas Laws

The first systematic and quantitative study of gas behavior was carried out by Boyle.

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The Gas Laws

The first systematic and quantitative study of gas behavior was carried out by Boyle.

Boyle noticed that when the temperature and

the amount of gas are held constant, the volume of a gas is inversely proportional to the applied pressure acting on the gas.

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(const. T and n)

This is a statement of Boyle’s Law. (V is the volume, P is the pressure, T is the

temperature, and n is the moles of gas).

P1V

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(const. T and n)


temperature, and n is the moles of gas). C V = (const. T and n) P

P1V

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(const. T and n)



(C is the proportionality constant).

P1V

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(const. T and n)



(C is the proportionality constant). This is also a statement of Boyle’s Law.

P1V

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We can write Boyle’s law as:

PV = C (const. T and n)

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If the initial values of pressure and volume are Pi and Vi and if the conditions are changed to a final pressure Pf and final volume Vf then we can write:

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initial conditions PiVi = C

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initial conditions PiVi = C

final conditions PfVf = C

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Hence: PiVi = PfVf (const. T and n)

This is also a statement of Boyle’s Law.

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Hence: PiVi = PfVf (const. T and n)

This is also a statement of Boyle’s Law.

Problem Example: If 0.100 liters of a gas, originally at 760.0 torr, is compressed to a pressure of 800.0 torr, at a constant temperature, what is the final volume of the gas?

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No mention is made about the moles of gas – so make the assumption that n remains constant. Since temperature is constant, we can use Boyle’s Law.

PiVi = PfVf

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PiVi = PfVf

This can be rearranged to read: Pi Vf = Vi Pf

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PiVi = PfVf

This can be rearranged to read: Pi Vf = Vi Pf 760.0 torr Vf = x 0.100 l 800.0 torr Vf = 0.0950 l

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Charles and Gay-Lussac’s Law

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Charles and Gay-Lussac’s Law Charles (1787) and Gay-Lussac (1802) were the

first investigators to study the effect of temperature on gas volume. Their studies showed that at constant pressure, and a fixed number of moles of gas, that the volume of a gas is directly proportional to the temperature of the gas.

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Charles and Gay-Lussac’s Law Charles (1787) and Gay-Lussac (1802) were the

first investigators to study the effect of temperature on gas volume. Their studies showed that at constant pressure, and a fixed number of moles of gas, that the volume of a gas is directly proportional to the temperature of the gas.

(const. P and n) This is a statement of Charles’ Law.

TV

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The proportionality can be turned into an equality:

V = C’ T (const. P and n)

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The proportionality can be turned into an equality:

V = C’ T (const. P and n) This is a statement of Charles’ Law. (The prime is used to signify that this constant

is different to the constant appearing in Boyle’s law).

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If the initial values of temperature and volume are Ti and Vi and if the conditions are changed to a final temperature Tf and final volume Vf then we can write:

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Vi initial conditions = C’ Ti

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Vi initial conditions = C’ Ti Vf final conditions = C’ Tf

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Vi initial conditions = C’ Ti Vf final conditions = C’ Tf

Hence: Vi Vf = (const. P, n) Ti Tf

This is a statement of Charles’ Law.

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Problem Example: A gas occupying 4.50 x 102 ml is heated from 22.0 oC to 187 oC at constant pressure. What is the final volume of the gas?

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There is no mention of the moles of gas changing, so assume that n is constant. Since pressure is constant, apply Charles’ law:

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Vi Vf = Ti Tf

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Vi Vf = Ti Tf

Note that it is necessary to convert the temperatures given to units of K.

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Ti = 22.0 + 273.15 = 295.2 K

Tf = 187 + 273.15 = 4.60 x 102 K

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Ti = 22.0 + 273.15 = 295.2 K

Tf = 187 + 273.15 = 4.60 x 102 K

Tf Vf = Vi Ti 4.60 x 102 K = 4.50 x 102 ml 2.952 x 102 K = 701 ml

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Avogadro’s Law

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Avogadro’s Law In 1811 Avogadro published a hypothesis

which stated that at the same temperature and pressure, equal volumes of gases contain the same number of molecules (or atoms if the gas is monatomic).

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Avogadro’s Law In 1811 Avogadro published a hypothesis

which stated that at the same temperature and pressure, equal volumes of gases contain the same number of molecules (or atoms if the gas is monatomic).

It follows that the volume of any given gas must be proportional to the number of molecules present.

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It is more convenient to work in terms of moles. A mole contains 6.02214 x 1023 items.

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For example, one mole of argon contains 6.02214 x 1023 atoms of argon, a mole of dinitrogen contains 6.02214 x 1023 molecules of N2.

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A mole of apples is really, really, ...really big!

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A mole of apples is really, really, ...really big!

The mole is a convenient “lab-sized” unit for amount of substance.

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Avogadro’s law can be written as:

(const. P and T)

nV

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(const. P and T) The proportionality can be replaced by an

equality: V = C’’n (const. P and T)

nV

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equality: V = C’’n (const. P and T) This is a statement of Avogadro’s Law.

nV

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equality: V = C’’n (const. P and T) This is a statement of Avogadro’s Law. (The double prime is used to signify that this

constant is different to the constants appearing in Boyle’s law and Charles’ law).

nV

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If the initial values of volume and moles are Vi and ni and if the conditions are changed to a final volume Vf and final moles nf then we can write:

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Vi initial conditions = C’’ ni

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Vi initial conditions = C’’ ni Vf final conditions = C’’ nf

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Hence: Vi Vf = (const. P, T) ni nf

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Hence: Vi Vf = (const. P, T) ni nf

This is a statement of Avogadro’s Law.

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Avogadro’s law means that, when two gases react with each other, their volumes have a simple ratio to each other. If the product is a gas, its volume is related to the volume of the reactants by a simple ratio.

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Example: 3 H2(g) + N2(g) 2 NH3(g)

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Example: 3 H2(g) + N2(g) 2 NH3(g)

3 volumes 1 volume 2 volumes

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Ideal Gas Law

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Ideal Gas Law

Summary so far:

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Ideal Gas Law

Summary so far:

1 Boyle’s law: V (const. T and n) P

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Ideal Gas Law

Summary so far:

1 Boyle’s law: V (const. T and n) P Charles’ law V T (const. P and n)

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Ideal Gas Law

Summary so far:

1 Boyle’s law: V (const. T and n) P Charles’ law V T (const. P and n)

Avogadro’s law V n (const. P and T)

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Math Aside:

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Math Aside:

If f X

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Math Aside:

If f X and f Y

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Math Aside:

If f X and f Y and f Z-1

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Math Aside:

If f X and f Y and f Z-1

Then f X Y Z-1

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Using the result from the math aside, we can put the expressions for Boyle’s law, Charles’ law, and Avogadro’s law together, so that:

nT V P

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nT V P The proportionality can be replaced by an

equation by inserting a proportionality constant. In this case the constant is represented by the symbol R, and is called the gas constant. Hence:

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nT V P The proportionality can be replaced by an

equation by inserting a proportionality constant. In this case the constant is represented by the symbol R, and is called the gas constant. Hence:

PV = n R T Ideal gas equation

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Ideal gas: Is defined to be a gas that satisfies the ideal gas equation.

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Evaluation of the gas constant: Experiments show that 1 mol of any ideal gas at 0 OC and 1 atm pressure occupies 22.414 liters. The conditions 0 OC and 1 atm are called standard temperature and pressure, abbreviated STP.

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Evaluation of the gas constant: Experiments show that 1 mol of any ideal gas at 0 OC and 1 atm pressure occupies 22.414 liters. The conditions 0 OC and 1 atm are called standard temperature and pressure, abbreviated STP.

(1 atm) (22.414 l) R = ( 1 mol) (273.15 K) = 0.082057 l atm K-1 mol-1

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In the SI unit system,

R = 8.3145 J K-1 mol-1

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R = 8.3145 J K-1 mol-1

Here, J stands for joule, the SI unit of energy.

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R = 8.3145 J K-1 mol-1

Here, J stands for joule, the SI unit of energy. 1 J = 1 Nm = 1 kg m2 s-2

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R = 8.3145 J K-1 mol-1

Here, J stands for joule, the SI unit of energy. 1 J = 1 Nm = 1 kg m2 s-2

Exercise: Try to convert R = 0.082057 l atm K-1 mol-1 to the

value given in SI units. (It’s a factor-label exercise).

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Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO2 at STP.

Assume CO2 can be treated as an ideal gas.

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Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 oC = 273.15 K (exact value)

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Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 oC = 273.15 K (exact value) n R T V = P

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Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 oC = 273.15 K (exact value) n R T V = P (0.168 mol) (0.08206 l atm mol-1 K-1) (273 K)V = (1 atm) = 3.76 l

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Combined Gas Law

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Combined Gas Law If the initial values of the pressure, volume, temperature,

and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write:

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Pi Vi initial conditions = R ni Ti

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Pi Vi initial conditions = R ni Ti Pf Vf final conditions = R nf Tf

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Hence: Pi Vi Pf Vf = ni Ti nf Tf

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Hence: Pi Vi Pf Vf = ni Ti nf Tf

This is the Combined Gas Law.

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Problem Example, Combined Gas Law: A scuba diver carries three tanks of air. Each has a capacity of 7.00 l and is at a pressure of 1.50 x 102 atm at 25.0 oC. What volume of air does this correspond to at STP?

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Given initial data: Pi = 1.50 x 102 atm

Vi = 21. 0 l

Ti = 298.2 K (25.0 + 273.15)

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Given initial data: Pi = 1.50 x 102 atm

Vi = 21. 0 l

Ti = 298.2 K (25.0 + 273.15)

Given final data: Pf = 1 atm

Tf = 273.15 K

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Pi Vi Pf Vf = ni Ti nf Tf

Assume that the number of moles of gas is fixed, so the combined gas law simplifies to:

Pi Vi Pf Vf = Ti Tf

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Pi Vi Pf Vf = ni Ti nf Tf

Assume that the number of moles of gas is fixed, so the combined gas law simplifies to:

Pi Vi Pf Vf = Ti Tf

The equation can be rearranged, so that

Pi Tf Vf = Vi Pf Ti

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(1.50 x 102 atm) (273.15 K) Vf = (21.0 l) (1 atm) (298.2 K)

= 2.89 x 103 l at STP

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Problem Example: Molar mass of a gas. Calculate the molar mass of methane if 279 ml of the gas measured at 31.3 OC and 492 torr has a mass 0f 0.116 g. Two steps: first find moles of gas, then determine the molar mass.

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From PV = nRT,

PV n = RT

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From PV = nRT,

PV n = RT 1 atm 1 l (492 torr)( )( 279 ml)( ) 760 torr 1000 ml = (0.08206 l atm mol-1 K-1)(304.5 K) = 7.23 x 10-3 mol

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The molar mass is given by:

molesgrams in massmass molar

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0.116 g = 7.23 x 10-3 mol


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0.116 g = 7.23 x 10-3 mol

= 16.0 g mol-1


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Dalton’s Law of Partial Pressures

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Dalton’s Law of partial pressures: In a mixture of gases, each component exerts the same pressure as it would, if it were alone and occupied the same volume.

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Dalton’s Law of partial pressures: In a mixture of gases, each component exerts the same pressure as it would, if it were alone and occupied the same volume.

Consider a simple case: A mixture of two gases A and B in a container of volume V and temperature T.

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The pressure exerted by gas A – called the partial pressure of gas A – is given by:

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nART PA = V

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nART PA = V

Similarly for gas B,

nBRT PB = V

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Now the total pressure PTotal is

nTotal RT PTotal = V

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nTotal RT PTotal = V Now nTotal = nA + nB so that:

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(nA + nB )RT nA RT nB RT PTotal = = + V V V

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(nA + nB )RT nA RT nB RT PTotal = = + V V V Hence, PTotal = PA + PB

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(nA + nB )RT nA RT nB RT PTotal = = + V V V Hence, PTotal = PA + PB

This is Dalton’s Law of partial pressures: The total pressure is the sum of the partial pressures.

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The preceding result can be generalized to any number of components:

PTotal = PA + PB + PC + PD + …

where PA, PB, PC, PD, etc. are the partial pressures of the individual gases.

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Problem Example: Assume that 1.00 moles of air contain 0.78 moles of dinitrogen, 0.21 moles of dioxygen, and 0.01 moles of argon. Calculate the partial pressures of the three gases when the air pressure is at 1.0 atm.

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VRTnnnV

nRTP ArONair 22)(

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VRTnnnV

nRTP ArONair 22)(

VRTnP 2

2

NN

156

)()( ArON

N

ArON

N

air

Nnnn

n

VRTnnn

VRTn

PP

22

2

22

22

157

airArON

NN P

nnnnP

22

22 )(

)()( ArON

N

ArON

N

air

Nnnn

n

VRTnnn

VRTn

PP

22

2

22

22

158

But

airArON

NN P

nnnnP

22

22 )(

)()( ArON

N

ArON

N

air

Nnnn

n

VRTnnn

VRTn

PP

22

2

22

22

mol1.00nnn ArON 22 )(

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Similar calculations give and

atm 0.78atm 1.0mol1.00mol0.78P

2N

atm 0.21P2O atm 0.01PAr

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Dalton’s law has a practical application when calculating the volume of gases collected over water. For a gas collected over water, the measured pressure is given by

where denotes the vapor pressure of water.

OHgastotal 2P PP

OH2P

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(p. 217)

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Problem example: O2 generated in the decomposition of KClO3 is collected over water. The volume of the gas collected at 24 oC and at an atmospheric pressure of 762 torr is 128 ml. Calculate the number of moles of O2 obtained. The vapor pressure of H2O at 24 oC is 22.4 torr.

First step: Calculate the partial pressure of O2.

(Dalton’s law)OHgasTotal 2

PPP

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= 762 torr – 22.4 torr = 739.6 torr (extra sig. fig)

= 0.973 atm

OHTotalgas2

PPP

torr760atm1 torr739.6

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From the ideal gas equation PV = nRT,

PV n = RT

(0.973 atm) (0.128 l) = (0.08206 l atm mol-1K-1)(297 K)

= 0.00511 mols

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Graham’s Law of Effusion

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Diffusion: A process by which one gas gradually mixes with another. The term is also used for solutes mixing with a solvent.

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Diffusion: A process by which one gas gradually mixes with another. The term is also used for solutes mixing with a solvent.

Effusion: The process by which a gas under pressure escapes from one compartment of a container to another by passing through a small opening.

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Graham’s law: The rate of effusion of a gas is inversely proportional to the square root of its density when the pressure and temperature are held constant.

Effusion rate: abbreviated R (don’t get this confused with the gas constant).

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Gas density: abbreviated d.

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Gas density: abbreviated d. 1 R (const. P, T)

d

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Gas density: abbreviated d. 1 R (const. P, T) This is Graham’s law of effusion.

d

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The proportionality can be turned into an equality: R = C (const. P and T) This is a statement of Graham’s law of effusion. (The constant is unrelated to any of the previous

constants in Boyle’s law, etc.).

d

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For two gases A and B, the relative rates of effusion can be evaluated as follows:

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For gas A: RA = C

Ad

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For gas A: RA = C

For gas B: RB = C

Ad

Bd

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For gas A: RA = C

For gas B: RB = C

Hence: RA = RB

Ad Bd

Ad

Bd

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For gas A: RA = C

For gas B: RB = C

Hence: RA = RB

RA = (const. P, T) RB

Ad Bd

Ad

Bd

A

Bdd

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For gas A: RA = C

For gas B: RB = C

Hence: RA = RB

RA = (const. P, T) RB

This is a statement of Graham’s law of effusion Law.

Ad Bd

Ad

Bd

A

Bdd

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Problem example, Graham ’s law: Under conditions for which the density of carbon dioxide is 1.96 g/l and that of dinitrogen is 1.25 g/l, which gas will effuse more rapidly? What is the ratio of the rates of effusion of dinitrogen to carbon dioxide?

181

Problem example, Graham ’s law: Under conditions for which the density of carbon dioxide is 1.96 g/l and that of dinitrogen is 1.25 g/l, which gas will effuse more rapidly? What is the ratio of the rates of effusion of dinitrogen to carbon dioxide? Let A be N2 and B be CO2.

182


2

2

2

2

N

CO

CO

Ndd

RR

183


2

2

2

2

N

CO

CO

Ndd

RR

g/lg/l

25.196.1

184


= 1.25

2

2

2

2

N

CO

CO

Ndd

RR

g/lg/l

25.196.1

185


= 1.25

So = 1.25 x , that is, N2 effuses 1.25 times faster than CO2.

2

2

2

2

N

CO

CO

Ndd

RR

g/lg/l

25.196.1

2NR2COR

186

An alternative form can be developed for Graham’s law of effusion, where the ratio of the rates is determined from the molar masses of the two gases.

187


m m d = M = V n

188


m m d = M = V n

so nM d = V

189


m m d = M = V n

so nM d = V

and PV = n RT , so (n/V) = P/(RT)

190


m m d = M = V n

so nM d = V

and PV = n RT , so (n/V) = P/(RT)

PM d = RT

191

Now plug this result for d into the expression for

Graham’s law.

RTPMRT

PM

A

B

BA

RR

192


Graham’s law.

That is (const. P, T)

RTPMRT

PM

A

B

BA

RR

A

B

BA

MM

RR

193


Graham’s law.

That is (const. P, T) This is a statement of Graham’s law of effusion Law.

RTPMRT

PM

A

B

BA

RR

A

B

BA

MM

RR

194

Exercise, Graham ’s law: What is the ratio of the rates of effusion of dinitrogen to carbon dioxide?

195

Exercise, Graham ’s law: What is the ratio of the rates of effusion of dinitrogen to carbon dioxide? Note: In this example, no density information is given, so use the formula involving the molar masses.

196

Kinetic Theory of Gases

197

The kinetic theory of gases provides a “working model” of a gas. It is an attempt to interpret experimental observations at the molecular level.

198


Basic Postulates of Kinetic Theory:

199


Basic Postulates of Kinetic Theory: 1. A gas consists of an extremely large number

of tiny particles that are in constant random motion.

200

2. The particles are separated by distances far greater than their own dimensions. The molecules can be considered as point-like, that is, they posses mass but have negligible volume (compared with the volume of the container).

201

2. The particles are separated by distances far greater than their own dimensions. The molecules can be considered as point-like, that is, they posses mass but have negligible volume (compared with the volume of the container).

3. All molecular collisions are elastic, that is, the sum of the kinetic energy of the colliding molecules remains unchanged before and after the collision.

202

It is found: The average kinetic energy of a collection of gas molecules is directly proportional to the absolute temperature.

KE = ½ m v2 T

The bar over KE means average, and over v2 means the average of v2.

203


KE = ½ m v2 T


v2 is called the mean square velocity.

204


KE = ½ m v2 T


v2 is called the mean square velocity.

If there are N molecules v12 + v2

2 + … + vN2

v2 = N

205

Molecules exert neither attractive nor repulsive forces on one another. This ties in directly with postulate 3.

206

Molecules exert neither attractive nor repulsive forces on one another. This ties in directly with postulate 3.

Compressibility: Since gas molecules are separated by large distances (relative to the molecular size), they can be compressed easily to occupy smaller volumes.

207

Pressure – volume connection (p. 215):

208

Pressure – volume connection: Increase the external pressure on the piston at constant temperature, then more molecules strike the container walls, so the force per unit area increases, and hence the pressure of the gas increases.

209

Pressure – volume connection: Increase the external pressure on the piston at constant temperature, then more molecules strike the container walls, so the force per unit area increases, and hence the pressure of the gas increases. That is, a smaller volume implies a larger pressure. Detailed arguments lead to

P V-1.

210

Pressure – temperature connection: An increase in temperature increases the average velocity of the gas molecules.

211


At higher velocities, the gas molecules strike the container walls more frequently and with greater force.

212



If the volume of the container is kept constant, the area being struck is the same, so the force per unit area, that is the pressure, increases.

213



If the volume of the container is kept constant, the area being struck is the same, so the force per unit area, that is the pressure, increases. Expect P T (const. V, n).

214

Temperature – volume connection: Increase the temperature at constant pressure.

215

Temperature – volume connection: Increase the temperature at constant pressure. An increase in temperature increases the average velocity of the gas molecules.

216

Temperature – volume connection: Increase the temperature at constant pressure. An increase in temperature increases the average velocity of the gas molecules. At higher velocities, the gas molecules strike the container walls more frequently and with greater force.

217

Temperature – volume connection: Increase the temperature at constant pressure. An increase in temperature increases the average velocity of the gas molecules. At higher velocities, the gas molecules strike the container walls more frequently and with greater force. This would increase the pressure, but if this is held constant, it would be necessary for the volume of the container to increase to maintain this constant pressure.

218

Temperature – volume connection: Increase the temperature at constant pressure. An increase in temperature increases the average velocity of the gas molecules. At higher velocities, the gas molecules strike the container walls more frequently and with greater force. This would increase the pressure, but if this is held constant, it would be necessary for the volume of the container to increase to maintain this constant pressure.

Expect V T (const. P, n).

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Graham’s law of effusion: Consider two gases (call them A and B) at the same temperature.

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Graham’s law of effusion: Consider two gases (call them A and B) at the same temperature. Kinetic theory indicates that the average kinetic energy for both gases must be the same.

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gas A gas B KEA = ½ mA vA

2 KEB = ½ mB vB2

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gas A gas B KEA = ½ mA vA

2 KEB = ½ mB vB2

If gas B molecules have higher mass, then the only way the average KE can be the same is

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that the average velocity of the B molecules is smaller than gas A.

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that the average velocity of the B molecules is smaller than gas A. The molecules moving more quickly are more likely to hit the opening, and hence the gas with the molecules having the lower mass will effuse faster.

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Maxwell Distribution of Molecular Speeds

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At a given instant, how many molecules are moving at a particular speed?

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At a given instant, how many molecules are moving at a particular speed?

The answer to this question is provided by Maxwell’s distribution of speeds curve.

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Maxwell Distribution of Molecular Speeds (p. 205)

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One important point to note is that although there are always some slow-moving molecules, there are fewer slow-moving molecules at higher temperatures.

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One important point to note is that although there are always some slow-moving molecules, there are fewer slow-moving molecules at higher temperatures. This is important for understanding the rates (i.e. speeds) of gas phase chemical reactions.

properties of gases 1. key objectives: be able to state (i.e. define) and use in practical...

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