Fundamental Concepts in Thermodynamics

Doba Jackson, Ph.D.Associate Professor of Chemistry & Biochemistry

Huntingdon College

Outline of Chapter

• What is thermodynamics and what is useful about it?

• Macroscopic variables: Volume, Temperature, Pressure

• Basic definitions of Thermodynamics

• Equations of State

What is Thermodynamics?

Chemistry- the study of matter and changes that it may undergo.

Physical Chemistry- the branch of chemistry that establishes and develops the principles of the subject using the underlying concepts of physics and mathematics.

Thermodynamics- The branch of science that describes the behavior of matter on the macroscopic scale (visible scale)

How to Study for this class

Time: For every 1 hour class, you should spend 2 hrs of studying.

Taking Notes:Unlike most classes, you don’t need to write down my conversation. Occasionally I will give you something you need to remember but not often. Also the PowerPoint's will be made available.

How to Study for this classIn Class Problems: You should pay particular attention to problems we work on in class. These problems will be very similar to test problems. Try to follow in class and review them after class and prior to the test.

MasteringChemistry problems:No problems this semester

End of Chapter Problems:You must work on the end of chapter problems outlined in the syllabus. The selected problems are similar to the problem we discuss in class.

What is Thermodynamics?Macroscopic scale- (also called bulk scale)- visible scale and properties.

Macroscopic properties: - Physical state (gas, liquid, solid) - Volume - Temperature - Pressure - Amount (mass, moles) - Heat Capacity - Color - Boiling, melting points - Density (mass/volume) - Bulk Energies (KE, PE, H, S, G, U) - Electrical conductor, producer

Microscopic properties: - Absorption/emission of energy - Dipole moment/ charges present - Atoms present (type, amount) - Bonds (covalent, noncovalent) - Orbitals occupied - Molecular movements (T, R, V) - Atomic movements (spin, orbits) - Atomic energies - Solubility/Miscibility

Thermodynamics (PCHEM 1)

Quantum Mechanics (PCHEM 2)

Properties of Matter (Definitions to review)

Property: Any characteristic that can be used to describe or identify matter.

Intensive Properties: Does not depend on the amount of sample. Ex: Temperature, Melting point, Density

Extensive Properties: Does depend on the amount of sample. Ex: Length, Volume, Mass, Moles

Extensive properties can be converted to intensive properties

Density: is mass divided by volume (extensive) to produce density

Molar Volume: is volume divided by moles

Macroscopic Variables: Volume, Temperature, and

Pressure

Temperature

(°F - 32 °F)5 °C

9 °F°C =

°C + 32 °F9 °F

5 °C°F =

K = °C + 273.15

Temperature: A measurement of direction and magnitude of energy flow in the form of heat.

Derived Units: Quantities based on other quantities

Volume measurement: one liter is one cubic decimeter

Measurement of Pressure• Evangelista Torricelli, in 1863 first

devised a method for measuring the pressure of an atmosphere using a mercury barometer.

Aneroid Barometer

Mercury Barometer

Water Barometer

Mercury Barometer

(exact)

Conversions

1 torr = 1 mm Hg

1 atm = 101 325 Pa

(exact)1 atm = 760 mm Hg

(exact)1 bar = 1 x 105 Pa

Pressure:Unit area

Force

Gases and Gas Pressure

Basic Thermodynamic definitions

Basic Thermodynamic Definitions

• System- Part of the world of interest

• Surroundings- Region outside the system

• Open system- Allows matter and energy to pass

• Closed system- Cannot allow matter to pass

• Isolated system- Cannot allow matter or energy to pass

If “A” is in thermal equilibrium with “B” and “B” is in thermal equilibrium with “C”, then “A” should be

in thermal equilibrium with “C.”

Zeroith Law of Thermodynamics

Equations of state and Ideal gas law

Ideal Gas Law: PV= nRT

PV/nT = constant (R)

Ideal Gas Constant (R)

*R is used in other thermodynamic equations

**

Equations of State are equations that relate the major macroscopic variables of a

physical state

Volume is a decreasing function of pressure

Boyle’s Law: PV = const (T,n)

y = 1/xPinitialVinitial = PfinalVfinal

Volume is an increasing function of temperature

Charles’ Law: V/T = const (P, n)

Absolute zero (-273.15 ºC)

=Tfinal

Vfinal

Tinitial

Vinitial

Avogadro’s Law

(constant T and P)

= kn

V

=nfinal

Vfinal

ninitial

Vinitial

V ∞ n

Two major types of problems that can be solved using gas laws

ONE STATE PROBLEM:- Using known variables in one state, find the unknown variable in that same state.

- Given T, P, n; find V

MULTI-STATE PROBLEM:- Using known variables in one state, find an unknown variable in another state assuming some variables remain constant.

P1V1 = P2V2 ; assumes n,T are constant

Example of a Single-State Problem

The reaction used in the deployment of automobile airbags is the high-temperature decomposition of sodium azide, NaN3, to produce N2 gas. How many liters of N2 at 1.15 atm and 30.0 °C are produced by decomposition of 45.0 g NaN3?

2Na(s) + 3N2(g)2NaN3(s)

Stoichiometric Relationships with Gases

2Na(s) + 3N2(g)2NaN3(s)

45.0 g NaN3

65.0 g NaN3

1 mol NaN3

2 mol NaN3

3 mol N2

x x

Volume of N2 produced:

= 1.04 mol N2

Moles of N2 produced:

= 22.5 LV =P

nRT=

(1.15 atm)

(1.04 mol) 0.082058K mol

L atm(303.2 K)

Calculate the volume that .65 moles of ammonia gas occupies at 37*C and 600 torr.

Problem

Problem 2• Calculate the pressure exerted by 18 g of steam

(H2O) confined to a volume of 18 L at 100*C.

What volume would the water occupy if thesteam were condensed to a liquid water at25*C? The density of liquid water is 1.00 g/mlat 25*C.

Show the approximate level of the movable piston in drawings (a) and (b) after the indicated changes have been made to the initial gas sample.

Multistate problem

Multi-State problems use Boundaries: ex. Temperature Boundaries

Diathermic Boundary: A boundary that allows energy in the form of heat to transfer from one object to the next.

Adiabatic Boundary: (insulating)- Will not allow energy to transfer as heat between two objects in contact

Multi-state problems: consider a plot of all states

Isobar- Constant pressure

Isotherm- Constant temperature

Isochore- Constant volume

Multi-state problems: Changes in state occur usually some

conditions are constant

Muti-state problems: often deviations occur from a standard state

Standard Ambient Temperature & Pressure (SATP)

T = 25ºC or (298.15 K) 5 sig figsP = 1.0 bar (exactly)Vm = 24.79 dm3/mol 4 sig figs

Standard Temperature and Pressure (STP)

T = 0ºC or (273.15 K) 5 sig figsP = 1.0 atm (exactly)Vm = 22.41 dm3/mol 4 sig figs

Example of a Multi-State Problem

In an industrial process, nitrogen is heated to 500 K in a vessel of isochoric conditions (constant volume). If it enters the vessel at 100 atm and 300 K, what pressure would it exert at the working temperature if it behaved as a perfect gas.

• A weather balloon is partially filled with helium at 20*C to a volume of 43.7 L and a pressure of 1.16 atm. The balloon rises to the stratosphere where the temperature and pressure are -23.0*C and 6.00 x 10-3 atm. Calculate the volume of the balloon in the stratosphere.

Problem 3

Dalton’s Law of Partial Pressures

• Dalton’s law- Pressure exerted by a mixture of perfect gases is the sum of the pressure the gases would exert is they were alone in a container at the same temperature.

A B A + B+ =

PA= 5 atm PB= 20 atm PA + PB = PT= 25 atm

NA= 5 moles NB= 20 moles NA + NB = NT= 25 moles

Partial Pressures

A + B

PA + PB = PT

PA= XAPT

PB= XBPT

NA + NB = NT

PA and PB are considered partial pressures

AA

T

NX =

Nmole fractionof A

T

NX =

NB

B mole fractionof B

Problem 1.10b• A gas mixture consists of 320 mg of

methane, 175 mg of argon, and 225 mg of neon. The partial pressure of neon at 300 K is 8.87 kPa. Calculate (a) the volume and (b) the total pressure of the mixture.

Ne- 20.18g/molAr- 39.95g/mol

Question 1: Air at 25.0*C and .998 atm has a density of 1.21 g/dm3. Assuming air consists of only N2

and O2. Calculate the partial pressure of N2

and O2

P = .998 atmT = 298.15 K (25.0 *C)V = 1 dm3

n = ?Mass = 1.21 gMMO2 = 32.00 g/molMMN2 = 28.00 g/mol

3

3

.998 1.0408

.08206 298.15

atm dmPVn

RT dm atm Kmol K

.0408n moles total

2 2 .0408T O Nn n n moles

2 2 1.21T O Nm m m g 2 2 2O O Om n MM

2 2 2 2 1.21T O O N Nm n MM n MM g

2 2 .0408T O Nn n n moles

2 2 1.21T O Nm m m g

2 2 2 2.0408 1.21T N O N Nm n MM n MM g

2 2 2 2 2.0408 1.21T O N O N Nm MM n MM n MM g

2 2 2 2.0408 1.21T O N N Om MM n MM MM g

2 2 2 21.21 .0408N N O On MM MM g MM

2

22 2

1.21 .0408 ON

N O

g MMn

MM MM

0239.00.4

0956.

00.3200.28

00.320408.21.1

=

414.586.1;586.0408.

0239.22

ON atmatmPP

atmatmPP

TOO

TNN

413.998.414.

585.998.586.

22

22

Section 1.5: Introduction to Real Gases

Why are real gases not Ideal?

Ideal Gas Model Assumptions

• The size of the molecules is negligible because the diameters are much smaller than the distance traveled between collisions.

• The molecules do not interact with each other outside of brief, infrequent and elastic collisions

Problems with Boyle’s Law

Problem 1: At high pressures and low molar volumes, inter-molecular forces between molecules become an important factor to consider.

Problem 2: At high pressures and low molar volumes, the volume occupied by the molecules themselves becomean important factor to consider.

Deviations

Other Equations of state for gases

Outline of Chapter

• 1) The Difference between Real and Ideal gases

• 2) Equations of State for Real gases– Van Der Waals Equation– Virial Equation

• 3) Compression Factor

• 4) Law of Corresponding States

Van Der Waals Equation

• Molecules do occupy space and their volume must be excluded from the ideal volume. This reduces any repulsive interactions. The “b” term.

• Attractive interactions between molecules are proportional to the square of the density (or molar concentration) of the gas. The “a” term.

Videal = V – nb “nb- corrects volume”

Pideal = P + a(n/V)2 “a(n/V)2- corrects pressure”

Discovered by Dutch physicist Johannes Diderik van der Waals (1837-1923) who won the Nobel prize in 1910.

Van Der Waals Equation

2m m

RT aP = -

V - b V

RT b– VV

1aP

nRT nb– VV

naP

nRTVP

m

2

m

2

idealideal

Standard Form

Ideal Gas

Van Der Waals terms

Using Molar Volume

The excluded volume term “b” of the van der Waals equation

343V r

34 23V r

3 34V = 2 πr = 8V3 mol

Excluded Volume

“b” terms can be calculated bytaking the volume of the moleculeand multiplying by 8

Van der Waals Equation

Pressure-Volume (CO2)

Problem 1.3• Calculate the pressure exerted by Ar for a molar volume

of 1.31 L/mol at 426 K using the van der Waals equation of state. The van der Waals parameters are 1.355 bar*dm6/mol2, and .0320 dm3/mol. Is the attractive or repulsive portion of the potential dominant.

Problem 2.1• Calculate the pressure exerted by 1.0 mol of C2H6

behaving as a perfect gas and a van der Waals gas. The gas is confined under the following conditions: Condition1: 1000 K, 100 L. (The van der Waals constants are: a=18.57 atm*L2/mol2 and b= .1193 L/mol), R is .08206 L*atm/mol*K.

Virial Equation • Virial Expansion (Expanded Molar Volumes)

2 3mT T T

PV = 1 + B P + C P + D P + .....

RT

2 3m

T T Tm m m

PV 1 1 1 = 1 + B + C + D + .....V V VRT

• Virial Expansion (Expanded Pressures)

• Each term in the Virial equation becomes successively smaller.

• The series does not converge at very high pressures which make molar volumes less than 1.

• Usually the equation is trunicated after the second or third term.

The Compression Factor

Real gases show deviations from the ideal gas law mainly because of molecular interactions

m mom

V PVZ= =

V RTomV = Ideal Molar Volume

Compression Factor

Z = 1: Ideal Gas (no forces)

Z < 1: Attractive forces dominate

Z > 1: Repulsive forces dominate

No forcesIdeal

Z≈1

Z>1

Z<1

Real gases show deviations from the ideal gas law mainly because of molecular interactions

m mom

V PVZ= =

V RTomV = Ideal Molar Volume

Compression Factor

Z = 1: Ideal Gas (no forces)

Z < 1: Attractive forces dominate

Z > 1: Repulsive forces dominate

No forcesIdeal

Z≈1

Z>1

Z<1

Van der Waals constants are used to solve the Second Virial

Coeficient

T T T2 3

m m m

B C DPVZ= =1+ + + +.........

RT V V V

2 3 411 .........

1x x x

x

mm

PV 1 aZ= = -

bRT RTV1- V

2m m

RT aP = -

V - b V

Van der Waals Equation

The van der Waals equation can be expanded to form a series

Geometric series

Virial Expansion

m

m m

PV V aZ= = -

RT V - b RTVm

2 3

m m m m

PV b b b aZ = = 1 + + + + ..... -

RT V V V RTV

2 3

m m m

PV 1 b baZ = = 1 + b - + + + .....RTRT V V V

Second Virial Coeficient

TaB = b - RT

Problem 1.15: A gas at 250 K and 15 atm has a molar volume 12% smaller thanthat calculated from the perfect gas law. Calculate (a) the compressionfactor under these conditions and (b) the molar volume of the gas. Which are dominating in the sample, the attractive or repulsive forces?

12% smaller volume means the real gas is 88% of the actual gas.

m mom

V PVZ= = .88

V RT o

m mV =.88 (V )

om

L atm.08206 250KRT mol KV = = =P 15atm

mV =.88 ( 1.37 L ) = 1.20 L

The Law of Corresponding States

Critical behavior of certain substances

Gas Gas

Liquid Liquid

Super-criticalFluid

Pre

ssu

re

Temperature

Critical Point- temperature at whicha liquid phase no longer exists anda phase intermediate of a liquid andgas exist.

Critical constants from Van der Waals constants

2m m

RT aP = -

V - b V

Van der Waals Equation

The critical Temp, Pressure and Volume will be the point at which the first and second derivative both equal zero. The

inflection point of a cubic equation

First and second derivative

2 3m m m

dP -RT 2a = - 0

dV (V -b) V

2

2 3 4m m

d P 2RT 6a = - 0

dV (V -b) Vm

First

Second

cV = 3bc 2

aP =

27bc

8aT =

27Rb

Van der Waals constants are used to find the Boyle Temperature

Boyle Temperature- the temperature at which the compression factor is the most ideal (Z ≈ 1) over a broad range of pressures and volumes.

T T T2 3

m m m

B C DPVZ= =1+ + + +.........

RT V V V

TZ 1 as B 0

Let BT = 0

TaB = b - 0RTBoyle

BoyleaT = Rb

Principle of corresponding states

• Ideal gas law is independent of the molecular substance.

• Real gases depend on each individual gas.• Find a relative scale for each substance and

use it for all substances.• Use the critical point as a relative scale:

rc

TT =

Tr

c

PP =

P rc

VV =

VPc, Tc, and Vc are critical pressure, temperature and volumes