Gas Laws and Processes PDF
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This document covers various aspects of gas laws and thermodynamic processes, including isothermal, isobaric, and isochoric processes. It also touches on concepts like internal and external irreversibility. The text includes diagrams and equations.
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Ideal Gas Processes In thermodynamics, processes usually involve the transfer of energy such as heating or cooling, compressing or expanding, stirring or pumping. The amount of energy transferred depends on the process as well as the end states (properties). Ideal Gas Processes For example, it is p...
Ideal Gas Processes In thermodynamics, processes usually involve the transfer of energy such as heating or cooling, compressing or expanding, stirring or pumping. The amount of energy transferred depends on the process as well as the end states (properties). Ideal Gas Processes For example, it is possible to transfer different amounts of energy by different processes and still maintain the same initial and end states. The path taken to change a property is known as a process. External irreversibility is some irreversibility external to Another form of external the system, when the system is the irreversibility is due to the flow of working substance; for example, heat through the containing walls, frictions on the moving parts such in as much as an adiabatic wall is a as piston and cylinder walls, and fiction of the mind. between the atmosphere and the rotating members; all absorbed some of the work output of the system. Internal irreversibility is irreversibility caused by fluid Conditions for reversibility: friction within the system, or it is processes controlled through a an internal event which is series of equilibrium states (no irreversible, such as the mixing or fluid friction). diffusion of two or more gases. (whirlpools, eddies and fluid no mechanical friction. friction) no temperature difference during transfer of heat. no diffusion. Constant Volume Process (isometric) Isometric Process or Constant volume process is under the constraints of incompressibility. The specific volume and density remain constant. Isometric Process PV – T relation Change in Enthalpy = 𝐶; if V = constant 𝑑ℎ 𝑷𝟏 𝑷𝟐 𝐶𝑝 = = 𝑑𝑡 𝑻𝟏 𝑻𝟐 ∆𝑯 = 𝒎𝑪𝒑∆𝑻 Change in Internal Energy Work Non – Flow 𝐶 = ; 𝑑𝑢 = 𝐶 𝑑𝑡 𝑊 = 𝑃𝑑𝑉 ; 𝑉 = 𝐶 ∆𝑼 = 𝒎𝑪𝒗 ∆𝑻 𝑾𝒏𝒇 = 𝟎 Isometric Process Work Steady – Flow Considering Ideal Gas Equation; 𝑃𝑣 = 𝑚𝑅𝑇 W = −𝛥𝑃𝐸 − ∆𝐾𝐸 − ∆𝑃𝑉 𝐖𝒔𝒇 = −𝒎𝑹∆𝑻 If 𝛥𝑃𝐸 & ∆𝐾𝐸 is neglected, therefore W = −𝑉 ∆𝑃 W =− vdP −𝑽 ∆𝑷 = − 𝒗𝒅𝑷 For non – flow system with paddle work (𝑊 ) considering figure, Isometric Process Change in Entropy 𝑄 𝑆= ∆𝑇 𝑑𝑄 𝑚𝐶𝑣𝑑𝑇 ∆𝑆 = = ∆𝑇 𝑇 ∆𝑆 = 𝑚𝐶𝑣𝑙𝑛 ; = 𝑄 + 𝑊 = 𝛥𝑈 + 𝑊 𝑷𝟐 𝑄 = 𝛥𝑈 + 𝑊 − 𝑊 ∆𝑺 = 𝒎𝑪𝒗𝒍𝒏 𝑷𝟏 Hence, for constant volume process 𝑸 = ∆𝑼 − 𝑾𝒑 Isometric Process, V=C PV – T relation 𝑷𝟏 𝑷𝟐 = 𝑻𝟏 𝑻𝟐 Change in Internal Energy ∆𝑼 = 𝒎𝑪𝒗 ∆𝑻 Change in Enthalpy ∆𝑯 = 𝒎𝑪𝒑∆𝑻 𝑾𝒏 𝒇 = 𝟎 Work Non- Flow Work Steady – Flow W = −𝛥𝑃𝐸 − ∆𝐾𝐸 − ∆𝑃𝑉 Change in Entropy 𝑷𝟐 ∆𝑺 = 𝒎𝑪𝒗𝒍𝒏 𝑷𝟏 Problem 1 A perfect gas has a value of R= 58.8 and k=1.26. If 20 btu are added to 5lb of this gas at constant volume when the initial temperature is, 90°𝐹. Find (a)T2, (b)∆𝐻, (c)∆𝑆, (d)∆𝑈, and (e) Work for non flow process. Constant Pressure Process (isobaric) Isobaric Process or Constant pressure process is change of state during which the pressure remains constant. It may be reversible or irreversible, non-flow or steady flow. Isobaric Process PV – T relation Change in Enthalpy = 𝐶; if V = constant 𝑽𝟏 = 𝑽𝟐 𝐶 = 𝑻𝟏 𝑻𝟐 ∆𝑯 = 𝒎𝑪𝒑 ∆𝑻 Change in Internal Energy 𝐶 = ; 𝑑𝑢 = 𝐶 𝑑𝑡 ∆𝑼 = 𝒎𝑪𝒗 ∆𝑻 Isobaric Process Work Steady – Flow W = −𝛥𝑃𝐸 − ∆𝐾𝐸 − ∆𝑃𝑉 Work Non – Flow If 𝛥𝑃𝐸 & ∆𝐾𝐸 is neglected, therefore 𝑊 = ∫ 𝑃𝑑𝑉 ; 𝑃 = 𝐶 𝐖𝒏𝒇 = 𝑷∆𝑽 W = −𝑉 ∆𝑃 W = − ∫ vdP Considering Ideal Gas Equation; −𝑉 ∆𝑃 = − ∫ 𝑣𝑑𝑃 𝑃𝑣 = 𝑚𝑅𝑇 If pressure is constant, therefore 𝐖𝒏𝒇 = 𝒎𝑹∆𝑻 𝐖𝒔𝒇 = 𝟎 Isobaric Process Change in Entropy 𝑄 𝑆= ∆𝑇 ∆𝑆 = = ∆ ∆𝑆 = 𝑚𝐶 𝑙𝑛 ; = 𝑽𝟐 ∆𝑺 = 𝒎𝑪𝒗𝒍𝒏 𝑽𝟏 Isobaric Process, P=C PV – T relation 𝑽𝟏 𝑽𝟐 = 𝑻𝟏 𝑻𝟐 Heat 𝑄 =∆𝑯 Change in Internal Energy ∆𝑼 = 𝒎𝑪𝒗 ∆𝑻 Change in Enthalpy ∆𝑯 = 𝒎𝑪𝒑∆𝑻 𝐖𝒏𝒇 = 𝑷∆𝑽 Work Non- Flow Work Steady – Flow W =0 Change in Entropy 𝑽𝟐 ∆𝑺 = 𝒎𝑪𝒗𝒍𝒏 𝑽𝟏 Problem 2 Three pounds of a perfect gas with. R= 38 and k=1.667 have 300Btu oh heat added during a reversible non-flow constant pressure change of state. The initial temperature is 100℉, Determine the (a) final temperature, (b) ∆𝐻, (c) W, (d) ∆𝑈, (e) ∆𝑆.