al zami Posted June 10, 2013 Posted June 10, 2013 why adiabatic process is called isentropic. we know disorder or any change makes entropy to change.then how adiabatic process keeps entropy constant .a clear explanation will be appreciated.
EdEarl Posted June 10, 2013 Posted June 10, 2013 no change in entropy See: http://en.wikipedia.org/wiki/Isentropic_process
studiot Posted June 10, 2013 Posted June 10, 2013 Well the definition of entropy change is (assume constant temperature for simplicity of explanation) dS = q/T But the definition of adiabatic is q = 0 So dS = 0 In other words there is zero entropy change. Does this help?
daniton Posted June 10, 2013 Posted June 10, 2013 why adiabatic process is called isentropic. we know disorder or any change makes entropy to change.then how adiabatic process keeps entropy constant .a clear explanation will be appreciated. let me ask you what is Entropy and who it is defined?
al zami Posted June 13, 2013 Author Posted June 13, 2013 let me ask you what is Entropy and who it is defined? actually how can i physically define entropy??? i know that change in position.disorder etc means there is a change in entropy,isn't it?
EdEarl Posted June 13, 2013 Posted June 13, 2013 It is hard to beat the Wikipedia definition. I think of the following example to help me understand increasing entropy: Imagine floating in space with a balloon full of smoke, you pop the balloon, and watch what happens to the smoke. It spreads out in all directions, thinner and thinner and thinner.
al zami Posted June 13, 2013 Author Posted June 13, 2013 Imagine floating in space with a balloon full of smoke, you pop the balloon, and watch what happens to the smoke. It spreads out in all directions, thinner and thinner and thinner. what is meant by this example???it is not clear to me ?can you explain this to me?? Well the definition of entropy change is (assume constant temperature for simplicity of explanation) dS = q/T But the definition of adiabatic is q = 0 So dS = 0 In other words there is zero entropy change. Does this help? thanks!it clears the mathemetical part.i wanted to know the actual meaning of increasing and decreasing entropy?what happens when entropy increase or decrease?how it is related to heat content and temperature??
EdEarl Posted June 13, 2013 Posted June 13, 2013 Heat flows from a high temperature source to the lower temperature environment, as the smoke disperses in space=>increasing entropy.
studiot Posted June 13, 2013 Posted June 13, 2013 I note you are studying engineering. You should look up indicator diagrams. They are much used by engineers. There are several pairs of quantities that when multiplied together have the units of energy or work. Force x distance Pressure times volume Surface tension times area EMF times charge Magnetic field times magnetic moment. If you plot one of these pairs on the x axis and the other on the y axis of a graph the area enclosed or between a line and a curve will 'indicate' the work done or energy transferred. But you will note there is something missing from my list Heat energy. Entropy was introduced to pair with temperature in the same way so heat transferred, q = TdS and the area enclosed by a line or curve on a T-S indicator diagram 'indicates' the heat transferred. There are many published tables and diagrams of entropy v temperature as it makes a very convenient form of calculation. The link to statistical mechanics was established later in history and has proved to be one of the most intriguing correspondences in science. Does this help? 1
al zami Posted June 14, 2013 Author Posted June 14, 2013 (edited) I note you are studying engineering. You should look up indicator diagrams. They are much used by engineers. There are several pairs of quantities that when multiplied together have the units of energy or work. Force x distance Pressure times volume Surface tension times area EMF times charge Magnetic field times magnetic moment. If you plot one of these pairs on the x axis and the other on the y axis of a graph the area enclosed or between a line and a curve will 'indicate' the work done or energy transferred. But you will note there is something missing from my list Heat energy. Entropy was introduced to pair with temperature in the same way so heat transferred, q = TdS and the area enclosed by a line or curve on a T-S indicator diagram 'indicates' the heat transferred. There are many published tables and diagrams of entropy v temperature as it makes a very convenient form of calculation. The link to statistical mechanics was established later in history and has proved to be one of the most intriguing correspondences in science. Does this help? thanks !but for w=F*x means because of F force a particle covers x distance .here i can measure the force and distance x . in here q=Tds because of temperature T .......... what can i say in the dash to make it a perfect definition for q as i did it for work done w. and if so , what is the difference between entropy and enthalpy Edited June 14, 2013 by al zami
studiot Posted June 14, 2013 Posted June 14, 2013 (edited) Yes force and distance are 'observables'; you can directly measure them. We take these observables for granted. From this we deduce that there is a physical quantity we call work that we can calculate from the observables. You cannot directly measure entropy, but you can directly measure both (edit: the components of) work done and heat transferred and temperature. So in post 3 we 'deduce' a physical quantity we call entropy that we can calculate from observables. You can't work the definition given in post#3 both ways round. The heat flow is used to define entropy, not entropy to define heat flow. That would be be a circular argument! I have not asked before but if temperature varies we have to perform an integration to get the entropy. Are you OK with calculus? We deduce another physical quantity we call enthalpy from these same observables, can you recall the equation? edit: It contains more than just heat transferred. You started this thread asking why adiabatic expansion is also called isentropic. Is there an enthalpy change during an adiabatic expansion? Edited June 14, 2013 by studiot
al zami Posted June 15, 2013 Author Posted June 15, 2013 (edited) We deduce another physical quantity we call enthalpy from these same observables, can you recall the equation? edit: It contains more than just heat transferred. you've said It contains more than just heat transferred but if it is constant pressure process then we can say h=u+pdv =>h=q-w+pdv =>h=q-pdv+pdv =>h=q so here enthalpy change is equal to heat transferred,isn't it??,But in adiabatic process we cant say enthalpy depends only on heat added. consider two system where one has volume two tymes than the other one but of same temperature then we can say the larger system has more heat content than the other one .so we can say the larger system has more entropy than the small one??? but can we say the larger one has more enthalpy than the small one????so if we can or can't then what is the difference between these two terms ???? and what is the difference between enthalpy change and entropy change?and what physical phenomena these two terms describe about a system .? Edited June 15, 2013 by al zami
studiot Posted June 15, 2013 Posted June 15, 2013 (edited) Yes you can devise processes eg const pressure, where the enthalpy change equals the heat transferred. This fact is much used by chemists investigating bond energies in calorimetry. For a particular amount of substance or a particular system Enthalpy has the units of energy. Entropy has the units of energy per degree K. Edited June 15, 2013 by studiot
studiot Posted June 15, 2013 Posted June 15, 2013 To help you understand the difference between entropy and enthalpy I suggest you perform the following calculations for yourself. 1kg of pure water as is sitting in an open container as ice. The ice is heated until it is all melted and then heated until its temperature is raised to 20C Calculate the entropy and enthalpy changes for the melting and for the heating of the liquid. Ignore volume changes and specific heat variation as insignificant. What is the difference in each case?
al zami Posted June 15, 2013 Author Posted June 15, 2013 (edited) i dont understand the question.is it water and ice are in same container or in seperate container?? and what is the mass of ice .as i need it to calculate latent heat m*lf Edited June 15, 2013 by al zami
studiot Posted June 15, 2013 Posted June 15, 2013 There is 1kg of ice that you melt. Yes you need the latent heat of fusion so you can calculate the heat flow to melt it. This gives you entropy of fusion. Can you say why? Then you heat it up to 20C by adding more heat. Again calculate the enthalpy and entropy of this second change.
al zami Posted June 16, 2013 Author Posted June 16, 2013 (edited) This gives you entropy of fusion. Can you say why? it confuses me.this is the amount of heat taken for melting.here entropy changes because the molecules of ice gets disordered and gets mixed in water,and we cant get the actual amount ice back from melted ice and water mixture,isn't it?its irreversible but if we apply this equation for entropy change s=q/T then we get s=q/0 [temp at o degree] s=undetermined doesn't making any sense. Edited June 16, 2013 by al zami
studiot Posted June 16, 2013 Posted June 16, 2013 You really do need to understand this calculation to make progress, because this is about the simplest possible case for a thermodynamic calculation. You also need to make sure you are correctly quoting formulae. Engineering is about attention to detail. s=q/0 [temp at o degree] Compare post#3 with your statement, what formula are you quoting? What temperature scale do we normally use for formulae in thermodynamics? So what is the thermodynamic temperature of melting ice? Now try the calculation again. You should be able to get the entropy change for melting or fusion. Why on earth do you think this process is irreversible? What happens if you cool it down again? This is about the simplest possible case for a thermodynamic calculation.
al zami Posted June 16, 2013 Author Posted June 16, 2013 (edited) temp for melting q1=m*lf =1*2100 temp for taking the temp upto 20 degree =m*c*t =1*4200*20 total heat q=q1+q2 here q is the enthalpy entropy S=intrigating q/T within 0 to 20 and s is the entropy isn't that right? actually Why on earth do you think this process is irreversible? What happens if you cool it down again? i tried to say if i put a piece of ice in water and let it melt .we cant get the exact piece of ice back from the water.if we freeze the water then the the whole water will turn into ice .but the exact piece can't be brought back,in that case it is irreversible s=q/dT means heat added per degree of temperature change.how this is related to the terms "disorder" or with "arrow of time" such types of concepts ?????? Edited June 16, 2013 by al zami
studiot Posted June 16, 2013 Posted June 16, 2013 Phase changes such as melting are the nearest we can get to a reversible process in real life. And this is pretty close. Melt 1kg of ice you get 1kg of water. Freeze 1kg of water you get 1kg of ice. The blocks of ice may not look the same but that is not a thermodynamic issue. To the best of normal measurements the heat used to melt the ice = the heat evolved (extracted) when it freezes. I am going to take this as a reversible process for all practical purposes. The latent heat of fusion for water is 333,000 Joules per kilogram So first to calculate the enthalpy change dH, note that the pressure remains constant since the container is open to atmosphere. We are given to ignore the small change in volume so no work is done w = d(PV) = 0 Thus dH = dU = 333000J q + w = 333000J q = 333000J Thus dS = q/T = 333000/273.2 = 1218.9 J per kg per degree K I have asked several time why we are justified in using this equation ? That is the first part and yes integration is needed for the second part. I did ask if you are OK with calculus.
al zami Posted June 17, 2013 Author Posted June 17, 2013 actually i am not worried about calculations.i am trying to grab the concept.entropy of each thermodynamic process is always equal or greater than zero.if i melt ice that will increase entropy .and if in regain ice from water that should increase entropy too.how this happen ??
studiot Posted June 17, 2013 Posted June 17, 2013 A few calculations will help understanding and promote thinking in a themodynamic way. Concepts are no good to you if you only half understand them or get them wrong and calculations help test and improve your understanding. I asked questions associated with the calculation intended to do exactly this and you have ignored them. The whole point of my example is that it forms a very good platform for discussion of thermodynamic issues as we work through it. For example I have asked several time why we are justified in using this equation ? Instead you offer an incorrect statement, misapplied from somewhere else. entropy of each thermodynamic process is always equal or greater than zero. In order to freeze the water, q is negative thus entropy is negative. Are you trying to tell me I can't freeze water?
al zami Posted June 17, 2013 Author Posted June 17, 2013 (edited) I have asked several time why we are justified in using this equation ? i didnt get the question.sorry for that.i dont know the answer.if i dont give answer to your questions i am sorry for that.entropy is not in my syllabus.i know maths will lead to better understanding.but classes are going in a full swing and its hard to make time for out of the syllabus topics.the basic thing which i wanted to know from where the formula dQ/T came from and what it signifies???what leads the scientists to introduce a concept like ENTROPY??? In order to freeze the water, q is negative thus entropy is negative. Are you trying to tell me I can't freeze water? dQ/T is always greater than or equal zero .it is said to be the fundamental law of the universe.entropy was not in my syllabus.i am trying to understand the topic because i have admiration towards astronomy and trying to understand the facts.if we freeze water entropy is negative.why this shows contradiction to dQ/T>=0; Edited June 17, 2013 by al zami
studiot Posted June 17, 2013 Posted June 17, 2013 i didnt get the question.sorry for that.i dont know the answer.if i dont give answer to your questions i am sorry for that.entropy is not in my syllabus That is useful information. What are you actually studying and at what level. I had thought you were perhaps in the first year of engineering college, so I was trying to pitch my replies to suit this. This is why I my explanations were often in the form of questions, based on what I thought you should already know. It is good to see someone who is sufficiently enthusiastic to read further than the syllabus so I will try help. In post 9 I said that entropy was originally introduced for engineers to calculate energy changes, or to look them up in tables. These changes were for steam machinery. Many properties of entropy and other thermodynamic variables were not realised at first, the full story took half a century to develop. So when they introduced entropy S, all the analysis referred to cyclic processes. This stemmed from the fact that machines work in cycles. You input some heat, generate some steam, output some work, release the waste gasses, input some more heat (and water) to end up with more output work and so on and so on. Now the statement about entropy change always being greater than or equal to zero refers to a cyclic process, where what is known as the working fluid is taken from state A to state B and back again to state A. This is not the melting ice in my example which is simple the transition from state A to state B. Now the entropy change round a reversible cyclic process is zero. For an irreversible cycle the entropy change is always greater than zero (ie positive). So if we melt some ice then refreeze it we have a forward transition A to B and a reverse transition B to A, completing the cycle. This, as I said, is about as near to a reversible process as nature allows so the entropy change around the cycle is zero. (SB - SA) = - (SA - SB) So the entropy change is positive in the forward direction and numerically equal but negative in the reverse. Entropy changes for a simple change from state A to state B can be positive or negative. How are we doing now?
al zami Posted July 6, 2013 Author Posted July 6, 2013 (edited) sorry for the delay. been too busy for exam it becomes impossible to make time to visit the forum. ok that gives the intuation . entropy was originally introduced for engineers to calculate energy changes as entropy means energy is used and we are not getting the energy back.so entropy is increasing. previously you compered entropy with voltage difference.as we know voltage difference helps to move electrons from lower to higher voltage.does entropy do similar or such thing.i mean does entropy is responsible for the flow of something.if not then from where this comparison is derived from ???? Edited July 6, 2013 by al zami
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