First Law of Thermodynamics

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First Law of Thermodynamics Created by: Marlon Flores Sacedon Physics section, DMPS June 2010

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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 changes

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Work Done by the system

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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 V

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Work Done by the system Signs of work done

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Work Done by the system Paths Between Thermodynamics States

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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 energy

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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 = 100J

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The First Law of Thermodynamics

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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?

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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)

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Kinds of Thermodynamic Process

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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.

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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 volume

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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 Gases

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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 gas

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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 process

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Adiabatic Process for an Ideal Gas No heat transfer, Q = 0

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Adiabatic Process for an Ideal Gas Adiabatic process, ideal gas Adiabatic process, ideal gas Adiabatic process, ideal gas