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Thread: hobby scientist

  1. #1 hobby scientist 
    zw3 is offline
    New Member
    Join Date
    Jul 2014
    Hello everybody
    I like to study math and science on free time. I get the text books and take the tests, and I enjoy especially the math and righting out and figuring derivatives and integrals.
    Also like playing learning music and being creative.
    I used to be really diligent in creating stop motion animations.

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  3. #2  
    Forum Freshman
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    Jul 2014
    I have a problem or at least a question: Thermodynamics uses a concept :"inexact differential" to describe internal energy. The symbol used either d with a t dash through or the small Greek delta. dU = delta Q + delta W. It states that internal energy U increases as the sum of heat addition Q and work addition W. How much heat and work are added individually depends on the path taken. thermodynamics states also that, if no work is added, dQ = T dS, temperature times entropy differential.
    When a system, a large number of atoms, is heated, the last entropy added requires TdS heat. Earlier entropy additions required less heat, since the temperature was then lower. If no work was exchanged, dU = Integral(0 to S) T dS. When a gas has zero mechanical energy it has infinite volume. When work is applied dU = -P dV. I can produce a three dimensional plot, a plot on which steam tables are based, plotting U versus S and V with T and P directional slopes dU/dS and dU/dV. By Maxwell I can state:
    (del X / del V)s = (del P / del S)v and due to the third law of thermodynamics, which states that T(s=0)= 0 , PdV = integral (0 to S) ((dT)s)) dS or
    P = integral (0 to S) ( dT / dV)s dS. I can say that Heat addition causes entropy to be added while mechanical energy addition causes the heat requirement of of the entropy differentials to be increased.
    The question is: Does the assumption of inexact differentials of heat and work allow other conclusions?

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