logging in or signing up First Law of Thermodynamics mfsacedon Download Post to : URL : Related Presentations : Let's Connect Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Copy embed code: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 822 Category: Entertainment License: All Rights Reserved Like it (0) Dislike it (0) Added: February 18, 2013 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript PowerPoint Presentation: First Law of Thermodynamics Created by: Marlon Flores Sacedon Physics section, DMPS June 2010PowerPoint Presentation: The First Law of Thermodynamics Thermodynamic system is a system that can interact (and exchange energy) with its surroundings, or environment, in at least two ways, one of which is heat transfer. Thermodynamic process is a process in which there are changes in the state thermodynamic system. Work Done during volume changesPowerPoint Presentation: Work Done by the systemPowerPoint Presentation: Work Done by the system Where: W = work done by the system p = pressure dV = differential volume V 1 & V 2 = initial and final volume V 1 V 2 pV-diagram 0 p VPowerPoint Presentation: Work Done by the system Signs of work donePowerPoint Presentation: Work Done by the system Paths Between Thermodynamics StatesPowerPoint Presentation: Internal Energy (U) Internal Energy of a system is the sum of kinetic energies of all of its constituent particles, plus the sum of all the potential energies of interaction among these particles. Where: = change in internal energy U 1 = initial internal energy U 2 = final internal energyPowerPoint Presentation: System Surroundings (environment) The First Law of Thermodynamics = Q-W = +50 J System Surroundings (environment) = Q-W = -50 J System Surroundings (environment) Q = 150J W = 150J = Q-W = 0 Q = -150J W = -100J Where: = change in internal energy (J) W = work done (J) Q = heat quantity (J) Q = 150J W = 100JPowerPoint Presentation: The First Law of ThermodynamicsPowerPoint Presentation: The First Law of Thermodynamics Ex. A gas in a cylinder is held at a constant pressure of 2.30x10 5 Pa and is cooled and compressed from 1.70 m 3 to 1.20 m 3 . The internal energy of the gas decreases by 1.40x10 5 J. a) Find the work done by the gas. b) Find the absolute value of the heat flow into or out of the gas, and state the direction of heat flow. c) Does it matter whether or not the gas is ideal? J, b) 2.55x10 5 J, out of gas, c) no (Ans. a) -1.15x10 5 Ex. A gas in a cylinder is held at a constant pressure of 2.30 x 10 5 Pa and is cooled and compressed from 1.70 m 3 to 1.20 m 3 . The internal energy of the gas decreases by 1.40 x 10 5 J. a) Find the work done by the gas, b) Find the absolute value |Q| of the heat flow into or out of the gas, and state the direction of heat flow, c) Does it matter whether or not the gas if ideal? Why or who not?PowerPoint Presentation: Kinds of Thermodynamic Process 1. Adiabatic Process ( pronounced “ay-dee-ah-bat-ic ”) is defined as one with no heat transfer into or out of a system: Q = 0. (adiabatic process) 2. Isochoric Process ( pronounced “eye-so-kor-ic ”) is a constant-volume process. When the volume of thermodynamic system is constant W=0. (isochoric process) 3. Isobaric Process ( pronounced “eye-so-bear-ic ”) is a constant –pressure process. (Isobaric process) 4. Isothermal Process ( pronounced “eye-so-bear-ic ”) is a constant –temperature process. (Isothermal process)PowerPoint Presentation: Kinds of Thermodynamic ProcessPowerPoint Presentation: Internal Energy of an Ideal Gas Property of Ideal Gas: The internal energy of an ideal gas depends only on its temperature, and not on its pressure and volume.PowerPoint Presentation: Heat Capacity of an Ideal Gas Molar heat capacity at constant volume (C V ) Molar heat capacity at constant pressure (C p ) (First Law) At constant volume (from First Law) (because dQ=dU) or At constant pressure ( from pV=nRT ) (Molar heat capacities of an ideal gas) (ratio of heat capacities) Where: C p = molar specific at constant pressure (J/mol.K) C V = molar specific at constant volume (J/mol.K) R = ideal gas constant initial and final volumePowerPoint Presentation: Type of Gas Gas C V (J/ mol.K ) C p (J/ mol.K ) C p -C V (J/ mol.K ) (J/ mol.K ) Monatomic He 12.47 20.78 8.31 1.67 Ar 12.47 20.78 8.31 1.67 Diatomic H 2 20.42 28.74 8.32 1.41 N 2 20.76 29.07 8.31 1.40 O 2 20.85 29.17 8.31 1.40 CO 20.85 29.16 8.31 1.40 Polyatomic CO 2 28.46 36.94 8.48 1.30 SO 2 31.39 40.37 8.98 1.29 H 2 S 25.95 34.60 8.65 1.33 Molar Heat Capacities of GasesPowerPoint Presentation: Heat Capacity of an Ideal Gas Molar heat capacities for Monatomic ideal gas Molar heat capacities for Diatomic ideal gas Molar heat capacities for Polyatomic ideal gasPowerPoint Presentation: Example . In an experiment to simulate conditions within an automobile engine, 645J of heat is transferred to 0.185 mol of air-conditioned within a cylinder of volume 40.0cm 3 . Initially the nitrogen is at a pressure of 3.00x10 6 Pa and a temperature of 780K. a) If the volume of the cylinder is held fixed, what is the final temperature of the air? Assume that the air is essentially nitrogen gas, use the Table. Draw a pV-diagram for this process. b) Find the final temperature of the air if the pressure remains constant. Draw a pV-diagram for this processPowerPoint Presentation: Adiabatic Process for an Ideal Gas No heat transfer, Q = 0PowerPoint Presentation: Adiabatic Process for an Ideal Gas Adiabatic process, ideal gas Adiabatic process, ideal gas Adiabatic process, ideal gas You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.