systems, properties & equilibrium

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Systems, Properties & Equilibrium System - Part of the world on which we focus attention Surroundings - Everything else System + Surroundings = Universe System + Surroundings = Universe Properties - Examples: temperature, volume, pressure, mass, refractive index density refractive index, density Two types of properties at equilibrium: Intensive - do not change with quantity of matter present (i.e., when system is subdivided) Extensive – scale linearly with quantity of matter present Equilibrium - state when a system’s macroscopic properties do not change with time

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Page 1: Systems, Properties & Equilibrium

Systems, Propertiesy& Equilibrium

System - Part of the world on which we focus attentionSurroundings - Everything elseSystem + Surroundings = UniverseSystem + Surroundings = UniverseProperties - Examples: temperature, volume, pressure, mass, refractive index densityrefractive index, densityTwo types of properties at equilibrium:Intensive - do not change with quantity of matter present(i.e., when system is subdivided)Extensive – scale linearly with quantity of matter present

Equilibrium - state when a system’s macroscopic properties do not change with time

Page 2: Systems, Properties & Equilibrium

Systems, Propertiesy , p& Equilibrium

System - Part of the world on which we focus attentionSurroundings - Everything elseSystem + Surroundings = UniverseSystem + Surroundings = Universe

Page 3: Systems, Properties & Equilibrium

TemperatureTemperature

Zeroth Law of thermodynamicsTwo bodies in thermal equilibrium with a third body

are in equilibrium with each otherare in equilibrium with each other

Thermal Energy kBTk i th B lt t tkB is the Boltzmann constant

T is the Kelvin Temperature Scale (K)T(°C) + 273.15 = T(K)

Measure Temperature based on Thermal Expansion:l l (1 + T + T2)

T( C) + 273.15 T(K)

lt = l0(1 + T + T2) is typically ~10-5 K-1

Page 4: Systems, Properties & Equilibrium

Zeroth Law of ThermodynamicsZeroth Law of Thermodynamics

Page 5: Systems, Properties & Equilibrium

Chapter 1Chapter 1

The Nature of Physical ChemistryThe Nature of Physical Chemistryand the Kinetic Theory of GasesPhysical chemistry is the application

Work

y y ppof the methods of physics to chemical problems

WorkKinetic and Potential Energy

Thermal EquilibriumThermal EquilibriumGas Laws

Kinetic Molecular Theory of GasesKinetic-Molecular Theory of GasesMaxwell Distribution of Molecular Speeds

Page 6: Systems, Properties & Equilibrium

Pressure

Pressure = Force/Area

Units (Nm-2 = kg m-1 s-2 )

SI unit of Pressure is the Pascal

Units (Nm kg m s )

Other Units of Pressure:

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

Other Units of Pressure:

1 bar = 105 Pa = 0.986923 atm

1 atm = 760 mm Hg = 760 Torr

Page 7: Systems, Properties & Equilibrium

Measuring Pressure, P = gh is density (kg/m3), g is acceleration of gravity (m/s2),

h i h i h ( ) P (P )

Measuring Pressure, P gh

h is height (m) P (Pa)Barometer was early

form of pressureform of pressure measuring device

Page 8: Systems, Properties & Equilibrium

Atmospheric Pressurep= 760 mm Hg = 10,300 mm H2O

P = P0exp(-Mgz/RT)

Page 9: Systems, Properties & Equilibrium

Boyle’s Law (empirical)Boyle s Law (empirical)

Describes relationship of pressure (P) and volume (V)Describes relationship of pressure (P) and volume (V) when amount (n) and temperature (T) are constant

SI unit for amount (n) of a substance is the moleSI unit for amount (n) of a substance is the mole

Avogadro’s constant, L = 6.022 x 1023 mol–1

Also known as NA

n = N/LV 1/P or PV = cons’t, Boyle's Law

Th l f fi d t f i i l ith

n = N/L

The volume of a fixed amount of gas varies inversely with the pressure if the temperature is maintained constant.

Page 10: Systems, Properties & Equilibrium

Boyle’s LawBoyle s Law

Page 11: Systems, Properties & Equilibrium

Boyle’s Law (experimental)Boyle s Law (experimental)

PV (h b l )PV = constant (hyperbolae)

T4

Is T > or < T ??

Temp T

Is T4 > or < T1 ??

Temp T1

Page 12: Systems, Properties & Equilibrium

ConcepTest #1

One mole of ice is similar in volume to

A. The Lambert GlacierB The iceberg that sank the TitanicB. The iceberg that sank the TitanicC. The ice of the CU skating rinkD An ice cube in a cold drinkD. An ice cube in a cold drink E. A snowflake

Page 13: Systems, Properties & Equilibrium

ConcepTest #1

One mole of ice is similar in volume to

A. The Lambert GlacierB The iceberg that sank the TitanicB. The iceberg that sank the TitanicC. The ice of the CU skating rinkD An ice cube in a cold drinkD. An ice cube in a cold drinkE. A snowflake

Page 14: Systems, Properties & Equilibrium

Charles’s Law

Describes relationship of volume (V) and temperature (T) when amount (n) and pressure (P) are constant( ) p ( )

V T, Charles's Law

The volume of a fixed amount of gasvaries directly with the absolute temperature if thevaries directly with the absolute temperature if the

pressure is maintained constant.

Page 15: Systems, Properties & Equilibrium

Charles’s Law Absolute T ScaleCharles s Law – Absolute T Scale

1 kg of each gas & P=1 atm

Page 16: Systems, Properties & Equilibrium

Avogadro’s LawAvogadro s Law

Describes relationship of volume (V) and amount (n)when pressure (P) and temperature (T) are constantp ( ) p ( )

V n, Avogadro's Law

The volume of a gas varies directly with the amount if the pressure and temperature are constantif the pressure and temperature are constant.

Page 17: Systems, Properties & Equilibrium

Ideal Gas LawIdeal Gas Law

Deduced from Combination of Gas Relationships:

V 1/P, Boyle's LawV 1/P, Boyle s Law

V , Charles's Law

V n, Avogadro's Law

Therefore, V nT/P or PV nT

PV = nRTh R i l t twhere R = universal gas constant

The empirical Equation of State for an Ideal Gas

Page 18: Systems, Properties & Equilibrium

Ideal Gas Equation of StateIdeal Gas Equation of State

Page 19: Systems, Properties & Equilibrium

Ideal Gas LawIdeal Gas Law

PV = nRTwhere R = universal gas constant

R = PV/nT

R = 0 0821 atm L mol–1 K–1

R = 8 314 J mol–1 K–1 (SI unit)

R 0.0821 atm L mol KR = 0.0821 atm dm3 mol–1 K–1

R = 8.314 J mol 1 K 1 (SI unit)

Standard molar volume = 22.4 L mol–1 at 0°C and 1 atm

Real gases approach ideal gas behavior at low P & high T

Page 20: Systems, Properties & Equilibrium

ConcepTest #2ConcepTest #2

A steel vessel contains 1 mole of gas at 100K. 2 moles of gas are added and the temperature is increased to g p200K. How does the pressure change?

A P increases by a factor of 4A. P increases by a factor of 4B. P decreases by a factor of 4C P i b f t f 6C. P increases by a factor of 6D. P decreases by a factor of 6E. P does not change

Page 21: Systems, Properties & Equilibrium

ConcepTest #2ConcepTest #2

A steel vessel contains 1 mole of gas at 100K. 2 moles of gas are added and the temperature is increased to g p200K. How does the pressure change?

A P increases by a factor of 4A. P increases by a factor of 4B. P decreases by a factor of 4C P i b f t f 6C. P increases by a factor of 6D. P decreases by a factor of 6E. P does not change

Page 22: Systems, Properties & Equilibrium

Dalton’s LawDalton s Law

Definition: Partial pressure Pi is the pressure exerted by one component of a gas mixture (at total pressure Pt)p g ( p t)

Pi = xiPt

h i th l f ti ( / )where xi is the mole fraction (xi = ni/ntotal)Dalton’s Law:

P P + P + P + + P

The total pressure is equal to the sum of the

Pt = P1 + P2 + P3 + … + Pi

partial pressures that each individual component gas would exert if it were alone

Page 23: Systems, Properties & Equilibrium

ConcepTest #3ConcepTest #3

A mixture of gases contains 4 g of He and 4 g of H2. The total pressure is 300 Pa. What is the partial p ppressure of helium?

A 100 PaA. 100 PaB. 150 PaC 200 PC. 200 PaD. 250 Pa

Page 24: Systems, Properties & Equilibrium

ConcepTest #3ConcepTest #3

A mixture of gases contains 4 g of He and 4 g of H2. The total pressure is 300 Pa. What is the partial p ppressure of helium?

A 100 PaA. 100 PaB. 150 PaC 200 PC. 200 PaD. 250 Pa

Page 25: Systems, Properties & Equilibrium

Graham’s Law of EffusionGraham s Law of EffusionEffusion is the movement of gasesg

through small passages(e.g., passing through a plug of fine sand)

where essentially all collisions are between gas and sand

This feels like something where the faster you go, the faster you get

Graham observed that for gases A and B

g y g y gthrough the maze, so we might expect an effusion rate velocity.

g

Rate(A)/Rate(B) = [M(B)/M(A)]1/2

The rate of effusion of a gas is inversely proportional to the square root of its molar mass (M)