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Laws of Thermodynamics - Lecture Slides | MTLE 6030, Study notes of Thermodynamics

Lecture 1 Material Type: Notes; Professor: Keblinski; Class: ADVANCED THERMODYNAMICS; Subject: Materials Science &Engineering; University: Rensselaer Polytechnic Institute;

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2011/2012

Uploaded on 05/15/2012

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Download Laws of Thermodynamics - Lecture Slides | MTLE 6030 and more Study notes Thermodynamics in PDF only on Docsity! Lecture 1 Laws of Thermodynamics  Thermodynamic state - equilibrium  Thermodynamic processes  Laws of thermodynamics  Absolute Temperature  Problems 2.5, 2.6, 2.8  Thermodynamic intensive coordinates are uniform across the whole system (T, P, ) or across each macroscopic phase (e.g., water and ice density density at the melting point.  All thermodynamic coordinates are time independent  Mechanical equilibrium, thermal equilibrium and chemical equilibrium Thermodynamic state - equilibrium Two or more systems in equilibrium do not exhibit heat flow among each other, they are at the same temperature Later we will see that criterion of equilibrium for isolated system, i.e., const E, V, T is the maximum entropy state dS(E, V, N) = 0 0th law of thermodynamics E1, V1, N1 E2, V2, N2 dS = dS1+dS2 = 0  dS1  S1 E1 dE1  S1 V1 dV1  S1 N1 dN1  S1 E1 dE1 Allowing only energy exchange between two isolated systems  dS2  S2 E21 dE2  S2 V2 dV2  S2 N2 dN2  S2 E2 dE2 dS1  dS2  S1 E1 dE1  S2 E2 dE1  dE2  dE1 From conservation of energy  S1 E1  S2 E2 T  E S 1st law of thermodynamics - conservation of energy dE = dQ-dW d indicates inexact differential - depends on the integration path In a cycle, E = 0 net Q in = net W out  Work and heat are not state functions  Energy is a state function dW = Fdx can be PdV, -dl, -it 2nd law of thermodynamics - entropy For a reversible process dQ= TdS Where S is entropy which a state function, and T is an absolute temperature The entropy can by calculated by integrating heat over a reversible path  S f  Si  dQ TR  Problem 2.6 A vessel of volume VB contains n moles of gas at high pressure. Connected to the vessel is a capillary tube trough which the gas may slowly leak out into the atmosphere, where P=P0. Surrounding the vessel and capillary is a water bath, in which is immersed an electric resistor. The gas is allowed to leak slowly trough the capillary into the atmosphere while, at the same time, electrical energy is dissipated in the resistor at such a rate that the temperature of the gas, the vessel, the capillary and the water is kept equal to that of the surrounding air. Show that, after as much gas is leaked as is possible during time , the change of internal energy is where, v0 = molar volume of gas at P=P0,  = potential on the resistor, and i is the current.  E i P0(nv0 VB ) Problem 2.8 The tension in a wire is increased quasi statically and isothermally from 1 to 2. If the length, cross-sectional area and isothermal Young’s modulus (Y) remain practically constant, show that the work done by the wire is:  W   L 2AY ( 2 2  1 2)
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