- StudyBlue
- Alabama
- University of Alabama - Tuscaloosa
- Chemistry
- Chemistry 342
- Gupta
- Ch%2001%20notes.pdf
Ch%2001%20notes.pdf
Chemistry 342 with Gupta at University of Alabama - Tuscaloosa
About this note
By: Gannon Parker
Textbook:
Physical Chemistry: AND Physical Chemistry Calculations
Created: 2010-04-25
File Size: 19 page(s)
Views: 7
Textbook:
Physical Chemistry: AND Physical Chemistry CalculationsCreated: 2010-04-25
File Size: 19 page(s)
Views: 7
About StudyBlue
STUDYBLUE makes things that make you better at school.
Things like online flashcards with photos and audio.
Things like personalized quizzes and friendly reminders about when (and what) to study next.
Think of it as a digital backpack™: access to all of your study materials online and on your phone.
STUDYBLUE exists to make studying efficient and effective for every student, for free. Join us.
“Simply amazing. The flash cards are smooth, there are many different types of studying tools, and there is a great search engine. I praise you on the awesomeness.”
Dennis
Dennis
Sign up (free) to study this.
Thermodynamics ? Thermodynamics is a branch of science and engineering (chemistry, physics, metallurgical, chemical, materials?) which deals with the energy, heat and work of a system (macroscopic the scale). ? Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiments - statistical thermodynamics uses atomic and molecular properties to calculate the macroscopic pp properties of matter. ? There are three principal laws of thermodynamics (four including th th l )e zeroe aw . ? Each law leads to the definition of thermodynamic properties which help us to understand and predict the operation of a pp physical system. ? N +3H? 2NH Examples of Applications of Thermodynamics 2 + 3H 2 3 How to improve yield: temperature and/or pressure? (ammonia yield in equilibrium will be higher if the temperature is yield if the is decreased and the pressure is increased; Chapter 6). CH OH (l) + 3/2O (g) ? CO (g)+2HO(l)? 3 2 2 ) + 2H 2 O ( Internal combustion engine versus electrochemical fuel cell, what are the relative efficiencies of the two propulsion systems? the of the propulsion (electrochemical fuel cell is more efficient; Chapter 5). ? Will the use of a catalyst promote to increase the yield of a reaction? (th i i ld t d d ilib i diti te max mum y eld expec e un er equ r um conditions mus be calculated first. If the yield is insufficient, a catalyst is useless; Chapter 6). Basic Definitions Needed to Describe Thermodynamic Systemsyy ? A thermodynamic system consists of all the materials involved in the process of study. Examples: tt f bk tii t- contents o an open ea er con a n ng reagen s - electrolyte solution within an electrochemical cell - contents of a cylinder and movable piston assembly in an ieng ne. ? The rest of the universe is referred to as the surroundings - if a system can exchange matter with the surrounding, it is called an open system - if not, it is a closed system - both open and closed systems can exchange energy with the surroundings - systems that can exchange neither matter nor energy with the surroundings are called isolated systems. Basic Definitions Needed to Describe Thermodynamic Systems?Continuedyy ? The interface between the system and its surroundings is called the boundary - boundary determines if energy and mass can be transferred ygy between the system and the surroundings: leads to distinction between open, closed, and isolated systems - considering the earth?s ocean as a system and the rest of the gy universe the surroundings: solid-liquid interface and water-air interface are the system-surroundings boundary - for an open beaker in which the system is the contents: an the contents: inner wall of the beaker and the open top of the beaker are boundary surfaces. ? The portion of the boundary formed by the beaker in the above example is called a wall - walls are always boundaries but boundaries need not be a wall always , wall - walls can be rigid or movable and permeable or non- permeable: e.g., surface of a balloon is a movable wall. Equilibrium ? The exchange of energy and matter across the boundary between exchange across the system and surrounding is central to the important concept of equilibrium system and surroundings can be in equilibrium with respect to one- be in to or more of several different system variables such as Pressure (P), Temperature (T), and concentration th d i ilib i ft ditihih- ermo ynam c equ r um refers to a condition n whic equilibrium exists with respect to P, T, and concentration - equilibrium exists with respect to variable if it does not change ith ti d it h th l th h t th t dw time an as the same va ue roug ou e sys em an surroundings (e.g., equilibrium with respect to P but not concentration for an idealized bubble). ? Temperature is an abstract quantity that is only measured indirectly. In thermodynamics, temperature is the property of a system that determines if the system is in thermal equilibrium with the other system or the surroundings. Dilute gas for which the ideal gas law describes the Temperature which ideal relationship among P, T, and molar density ? = n/V P = ?RT T = P/?R For the same molar density, a pressure gauge can be used to compare two systems to determine which of T 1 or T 2 is greater. Thermal equilibrium Same molar density P 1 P 2 ? of exists if P 1 = P 2 (for same molar density) For 1 1b the walls (eg coffee in styrofoam cup) are . , walls e.g. referred to as adiabatic (do not conduct heat). Because P1?P2, systems are not in thermal equilibrium and have different temperatures adiabatic P 1 P = different For 1.1c, the walls (e.g. coffee in copper cup are referred to as diathermal (conduct heat). Both 2 () systems achieve the same pressure; thermal equilibrium exists: T 1 = T 2 diathermal ? The zeroth law of thermodynamics begins with a definition of Zeroeth Law of Thermodynamics zeroth law of thermodynamic equilibrium - it is observed that some property of an object, like the pressure in a volume of gas the length of a metal rod or the electrical conductivity of , length of rod, electrical conductivity a wire, can change when the object is heated or cooled. If two of these objects are brought into physical contact (with exchange of heat) there is initially a change in the property of both objects. But, eventually, the yg y, change in property stops and the objects are said to be in thermal, or thermodynamic equilibrium. - thermodynamic equilibrium leads to the large scale definition of yq temperature. When two objects are in thermal equilibrium they are said to have the same temperature. During the process of reaching thermal equilibrium, heat, which is a form of energy, is transferred between the objects. The details of the process of reaching thermal equilibrium are described in the first and second laws of thermodynamics. ? The zeroth law of thermodynamics is an observation. When two objects are separately in thermodynamic equilibrium with a third object, they are in equilibrium with each other. Zeroeth Law of Thermodynamics http://www.grc.nasa.gov/WWW/K-12/airplane/thermo0.html Thermometry t(x)=a+bxt(x) = a + bx Equation defines a temperature scale in terms of specific thermometric properties (volume of a liquid, electrical resistance of metal or semiconductor, electromotive force generated at junction of dissimilar metals, etc.) The constant ?a? determines the zero of the temperature scale and constant a zero of the constant ?b? determines the size of a unit of temperature (degree). Fahrenheit Celsius Kelvin Boiling Point of Water 212°F 100°C (99.975°C) 373 K Water Freezing Point of Water 32°F 0°C (0.01° C) 273 K Water Absolute Zero -459°F -273°C 0 K Gas Thermometer The gas thermometer is a practical gp thermometer with which the absolute temperature can be measured (international standard for thermometry for at very low temperature). Charles? law: P = c + dt for a constant volume, V ?t? is the temperature in the Celcius scale Pressure exerted by 5.00 x 10 -3 mol of dilute gas as a function of temperature (Charles? Law). . Kelvin scale: T (K) = T (ºC) + 273.15 T (K) = 273.16 P/P tp , where P tp is the pressure corresponding to the triple point of water (the temperature is (the temperature is independent of pressure and of the gas when extrapolated to zero pressure). Temperature Equations of State and the Ideal Gas Law Equation of state PV = ?T, where T is the absolute temperature ?= nR, where n is the number of moles and R is a constant (ideal gas constant). The ideal gas equation of state: PV = NkT = nRT h k ithBlt t t d N ith 100 K differential w ere is the Boltzmann cons an an is the number of molecules. Any three variables (out of the four) is sufficient to three four) is sufficient completely describe the ideal gas. The total number of moles ? not the moles of the Relationship between P and V of total moles individual gas components of a gas mixture ? appear in the ideal gas law. 0.010 mol of He for fixed values of gas temperature (Boyle?s Law). Equations of State and the Ideal Gas Law?contd. Of the four variables, P and T are independent of the amount of gas, whereas Vou ad ou and n are proportional to the amount of gas. ? A variable that is independent of the size of the system ( example, P and T) is referred to as an intensive variable. ? One that is proportional to the size of the system (example, V) is referred to as an extensive variable. ? The gas law can be written exclusively in terms of intensive variables: P = ?RT Fidl it RTPV ? For an idea gas m x ure: n i i = ? ? RTn i For non-interacting gases where P i is the partial pressure of each gas (independent of other gases in the it ) ++=== ii PPPP V P i ...321 mixture (mole fraction) i ii x n n P P == Units of Pressure, Volume and Temperature -1 In International System of Units (SI): Pressure is measured in Pascal (Pa). Volume is measured in cubic meters (1 m 3 = 10 3 L and 1L = 1dm 3 = 10 -3 m 3 . Tti diKliTemperature s measure n e v n. Gas constant R = 8.314 JK -1 mol -1 Example Problem 1.1 Solution Because the number of moles is constant fiiffii TVPVPVP i f f fi TV P TT == ; )02815273( 5 5 + ifii VTVP Pa1050.3 )00.515.273( .. 03.1 Pa1021.3 ×= ? ×××== ii f f VTV P This pressure is within 10% of the recommended pressure. Example Problem 1.2 T = 298K Solution Th b f l f H N d X i i be num er o mo es o e, e, an e s g ven y: mol121.0 K298molKbarL108.314 L2.00bar50.1 1- 1- 2- = ×× × == RT PV n He 1.00L)1.00bar;(mol0403.0 === VPn Xe 3.00L)2.50bar;(mol303.0 === VPn eN 0.0860 0.653;0.261;fractionsmole;464.0 =====++= Xe Ne He He Xe Ne He xx n xnnnn n bar92.1 L00.6 K298molbar KL108.3145mol464.0)( 1-1-2- = ××× = ++ = V RTnnn P XeNeHe The partial pressures are: bar165.0;bar25.1 bar;501.0bar92.1261.0 ===×== Xe Ne HeHe PPPxP Introduction to Real Gases ? Ideal gas law only holds for gases at low densities - the equation is accurate to higher values pressure and lower values of temperature for He than for NH 3 (why?) ? Atoms and molecules of an ideal gas are assumed not to interact with one another and are treated as point masses ? Three regions: (1) potential energy is essentially zero (r > r transition ); (2) negative (attractive interaction) (attractive (r transition > r > r V=0 ); (3) positive (repulsive interaction) (r < r V=0 ). |V (r transition )| ? kT ? In the attractive region, P is lower than that attractive than calculated using the ideal gas law (attraction). At sufficiently higher densities P is higher than that calculated using ideal gas law (repulsion). Finite gg volume even when P ?? ? van der Walls equation of state: ? The parameters b and a take the finite size of the Potential energy of 2 2 V an nbV nRT P ? ? = finite size of the molecules and the strength of the of the attractive interaction into account, respectively. interaction of two molecules or atoms Example Problem 1.3 Van der Walls parameters are generally tabulated with eitherVa are either of two sets of units: a: Pa m 6 mol -2 or bar dm 6 mol -2 b: m 3 mol -1 or dm 3 mol -1 Determine the conversion factor to convert one system to the conversion other. Note that 1 dm 3 = 10 -3 m 3 = 1 L Solution 26 6 66 5 2-6 moldmbar10 m dm10 Pa10 bar molmPa ? =×× 1 33 3 3 3 1- 3 moldm10 m dm10 molm ? =× Example Problem 1.4 Solution bar1098.9 L250 K300molbar KL108.314mol1 2 1-1-2- ? ×= ××× == V nRT P Similarly, for V = 0.100 L, P = 249 bar For van der Walls equation of state: 2 2 V an nbV nRT P ? ? = 2 262 1-3 1-1-2- L)250( moldmbar370.1)mol1( moldm0.0387mol1-L250 K300molbar KL108.314mol1 ? × ? × ××× = b10989 2? For V = 0.100 L, P = 270 bar Because P real > P ideal (for V=0.1 L), we conclude that the repulsive ar. ×= interaction is more important than the attractive interaction for this specific value of molar volume and temperature. agupta Microsoft PowerPoint - Ch01_PChem modified [Compatibility Mode]
Back
Next
About this note
By: Gannon Parker
Textbook:
Physical Chemistry: AND Physical Chemistry Calculations
Created: 2010-04-25
File Size: 19 page(s)
Views: 7
Textbook:
Physical Chemistry: AND Physical Chemistry CalculationsCreated: 2010-04-25
File Size: 19 page(s)
Views: 7
About StudyBlue
STUDYBLUE makes things that make you better at school.
Things like online flashcards with photos and audio.
Things like personalized quizzes and friendly reminders about when (and what) to study next.
Think of it as a digital backpack™: access to all of your study materials online and on your phone.
STUDYBLUE exists to make studying efficient and effective for every student, for free. Join us.
“Simply amazing. The flash cards are smooth, there are many different types of studying tools, and there is a great search engine. I praise you on the awesomeness.”
Dennis
Dennis