CH%20301%20Chapter%206%20notes%20part%201v2

CH%20301%20Chapter%206%20notes%20part%201v2 - CH301 Notes...

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CH301 Notes Chapter 6 part 1 Thermodynamics - study of the changes or transfers in energy accompanying chemical and physical processes. Will two (or more) substances react when mixed under specified conditions? What energy changes and transfers are associated with this? How far does the reaction occur? Thermodynamic Terms system: region of interest, or, substances involved in the chemical and physical changes. In lab, its the chemicals inside the beaker. surroundings: environment around the system - where observations are made. system + surroundings = universe . Types of Systems: depends on how energy and/or matter is exchanged with surroundings: OPEN: exchange energy & matter (Example: human body) CLOSED: exchange energy only. (Example: sealed ice pack) ISOLATED: NO exchange (energy or matter) (Example: Thermos flask with rigid lid) WORK: Motion against an opposing force. WORK = FORCE x DISTANCE UNITS: Joule, J 1J = 1kg m 2 s -2 Types: (there are others but these are important for CH301) Expansion (expansion against an external force) Nonexpansion (work is done but there is no expansion against an external force) INTERNAL ENERGY, U: Total energy inside system (can be used to do work) We NEVER know the absolute value of U – dont know exact energy of all component particles We CAN find the CHANGE in U : Δ U = U (final) - U (initial) ANY CHANGE in a quantity X is always given by: Δ X = X (final) - X (initial) NOTE FOR HW SERVER Some other books use E rather than U for Internal Energy. You may see this in questions . Internal Energy and Work: If work done on system: ( energy in form of work transfers to system from surroundings ) w is positive, so is U : system’s internal energy increases by: Δ U = w Assumes NO other type of energy transfer occurred! If work done by system on surroundings: w is negative, Δ U is negative , system lost energy. Expansion Work against a Constant External Pressure System: Piston, area A: pushes against external (surrounding) pressure P ext ; experiences opposing force f = P ex A (remember pressure P ext is a force per unit area, we are stating the total force f) Move piston out a distance d : Work done by system on surroundings: system loses energy: Work = d P ex A But d A = Δ V so w = - P ex Δ V P ex MUST BE CONSTANT! Expansion Work Against a Vaccum P ex = zero so w = zero! FREE EXPANSION- No expansion work done by system- no energy lost. Getting Units to Work Out: If P is in Pa (1Pa = 1kg m -1 s -2 ) and Δ V is in m 3 , we get w in J. But if Δ V is in liters, and P is in atm always use the conversion factor (will be given on exams) 1 L. atm = 10 -3 m 3 x 101325Pa = 101.325 Pa.m 3 = 101.325 J Example: Ideal gas is allowed to expand from a volume of 10L to a volume of 40L against a constant opposing pressure of 0.75atm. Find the work done.
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This note was uploaded on 09/09/2009 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas.

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CH%20301%20Chapter%206%20notes%20part%201v2 - CH301 Notes...

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