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 class In 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 Quantum Mechanics (PCHEM 2) Thermodynamics (PCHEM 1)

Properties of Matter (Definitions to review) Chapter 1: Atoms: Matter and Measurement 4/25/2017 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 Copyright © 2010 Pearson Prentice Hall, Inc.

Macroscopic Variables: Volume, Temperature, and Pressure

Chapter 1: Atoms: Matter and Measurement 4/25/2017 Temperature Temperature: A measurement of direction and magnitude of energy flow in the form of heat. °C + 32 °F 9 °F 5 °C °F = The 273.15 offset from Kelvin to Celsius is exact. (°F - 32 °F) 5 °C 9 °F °C = K = °C + 273.15 Copyright © 2010 Pearson Prentice Hall, Inc.

Derived Units: Quantities based on other quantities Chapter 1: Atoms: Matter and Measurement 4/25/2017 Derived Units: Quantities based on other quantities Copyright © 2010 Pearson Prentice Hall, Inc.

Chapter 1: Atoms: Matter and Measurement Volume measurement: one liter is one cubic decimeter 4/25/2017 Copyright © 2010 Pearson Prentice Hall, Inc.

Measurement of Pressure Evangelista Torricelli, in 1863 first devised a method for measuring the pressure of an atmosphere using a mercury barometer. Mercury Barometer Aneroid Barometer Water Barometer

Mercury Barometer

Gases and Gas Pressure (exact) Conversions 1 torr = 1 mm Hg 1 atm = 101 325 Pa 1 atm = 760 mm Hg 1 bar = 1 x 105 Pa Pressure: Unit area Force

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

Zeroith Law of Thermodynamics 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.”

Equations of state and Ideal gas law

Equations of State are equations that relate the major macroscopic variables of a physical state Ideal Gas Law: PV= nRT PV/nT = constant (R) Ideal Gas Constant (R) * * *R is used in other thermodynamic equations

Volume is a decreasing function of pressure Boyle’s Law: PV = const (T,n) PinitialVinitial = PfinalVfinal y = 1/x

Volume is an increasing function of temperature Charles’ Law: V/T = const (P, n) = Tfinal Vfinal Tinitial Vinitial Absolute zero (-273.15 ºC)

Avogadro’s Law (constant T and P) V ∞ n = k n V = nfinal Vfinal ninitial Vinitial

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) Moles of N2 produced: 45.0 g NaN3 65.0 g NaN3 1 mol NaN3 2 mol NaN3 3 mol N2 x x = 1.04 mol N2 Volume of N2 produced: (1.15 atm) (1.04 mol) 0.082058 K mol L atm (303.2 K) P nRT V = = = 22.5 L

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

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 the steam were condensed to a liquid water at 25*C? The density of liquid water is 1.00 g/ml at 25*C.

Multistate problem 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.

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

Multi-state problems: Changes in state occur usually some conditions are constant Isobar- Constant pressure Isotherm- Constant temperature Isochore- Constant volume

Muti-state problems: often deviations occur from a standard state Standard Ambient Temperature & Pressure (SATP) T = 25ºC or (298.15 K) 5 sig figs P = 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 figs P = 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.

Problem 3 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.

Dalton’s Law of Partial Pressures B A + B = + PB= 20 atm PA= 5 atm PA + PB = PT= 25 atm NA= 5 moles NB= 20 moles NA + NB = NT= 25 moles 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.

Partial Pressures A + B PA and PB are considered partial pressures PA + PB = PT NA + NB = NT PA and PB are considered partial pressures PA= XAPT PB= XBPT

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/mol Ar- 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 atm T = 298.15 K (25.0 *C) V = 1 dm3 n = ? Mass = 1.21 g MMO2 = 32.00 g/mol MMN2 = 28.00 g/mol

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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 become an 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 Videal = V – nb “nb- corrects volume” Discovered by Dutch physicist Johannes Diderik van der Waals (1837-1923) who won the Nobel prize in 1910. 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”

Van Der Waals Equation Ideal Gas Van Der Waals terms Using Molar Volume Standard Form

The excluded volume term “b” of the van der Waals equation “b” terms can be calculated by taking the volume of the molecule and 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) 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 Z>1 Compression Factor No forces Ideal Z≈1 Z<1

Real gases show deviations from the ideal gas law mainly because of molecular interactions Compression Factor Z>1 No forces Ideal Z≈1 Z<1

Van der Waals constants are used to solve the Second Virial Coeficient Geometric series Van der Waals Equation Virial Expansion The van der Waals equation can be expanded to form a series Second Virial Coeficient

Problem 1.15: A gas at 250 K and 15 atm has a molar volume 12% smaller than that calculated from the perfect gas law. Calculate (a) the compression factor 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.

The Law of Corresponding States

Critical behavior of certain substances Super- critical Fluid Gas Gas Pressure Liquid Liquid Temperature Critical Point- temperature at which a liquid phase no longer exists and a phase intermediate of a liquid and gas exist.

Critical constants from Van der Waals constants 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 First Second

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. Let BT = 0

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: Pc, Tc, and Vc are critical pressure, temperature and volumes