• It's not constant pressure, because we have a delta p going on. It's not constant volume either.

    也不是恒容,这个限制,是这个实验的限制。

    麻省理工公开课 - 热力学与动力学课程节选

  • It's tabulated in books, and this we can measure p in the experiment. Delta p here is the change in pressure from the left side to the right side, and we can put a thermometer, measure the temperature before the experiment and measure the temperature after the experiment.

    这列在书上,这个量我们在,实验中也可以测量,在这里Δ,是从左边到右边的压强变化,我们可以放一个温度计,去测量实验前的温度,再去测量试验后的温度。

    麻省理工公开课 - 热力学与动力学课程节选

  • And so they defined them, p after many experiments, the limit of this 0 delta T delta p and the limit of delta p goes to zero as the Joule-Thomson coefficient.

    他们定义了这些量,以及它们的范围,ΔT比Δ,Δp的极限趋近于,叫做焦耳-汤姆逊系数。

    麻省理工公开课 - 热力学与动力学课程节选

  • STUDENT: from the T delta V p to the delta p here?

    学生:,从TΔV到这里的Δ

    麻省理工公开课 - 热力学与动力学课程节选

  • So in this experiment here, delta p is less than zero. You need to have this whole thing greater than zero. So delta T is less than zero as well. So if you're below the inversion temperature and you do the Joule-Thomson experiment, you're going to end up with something that's colder on this side than that side.

    所以在这个实验中,Δp小于零,这全部都大于零,因此ΔT也小于零,所以如果在低于转变,温度的情况下做焦耳-汤姆孙实验,最后的结果是,这边的温度比这边低。

    麻省理工公开课 - 热力学与动力学课程节选

  • Adiabatic q equal to zero. It's also delta H 0 which is zero. The two didn't necessarily follow because remember, delta H is dq so p is only true for a reversible constant pressure process.

    在这个过程中ΔH等于,绝热的所以q等于0,而ΔH也等于,这两个也不一定有因果关系,因为,记住,ΔH等于dq只有在恒压。

    麻省理工公开课 - 热力学与动力学课程节选

  • H=qp The u plus p V. Delta H is equal to q V.

    括号里面的就是H,等于u+pv,Δ

    麻省理工公开课 - 热力学与动力学课程节选

  • T Remember, we're trying to get delta H, p we're trying to get dH/dT constant pressure and dH/dp constant temperature. OK, these are the two things were trying to get here.

    想要得到在恒压状态下的偏H偏,和在恒温状态下的偏H偏,好的,这是两个我们,在这里想要得到的东西。

    麻省理工公开课 - 热力学与动力学课程节选

  • V Minus p2, V2. Minus p delta V. So the total work is the work from the left hand side plus the work on the right hand side, p1 V1- p2 V2 which is p1 V1 minus p2 V2.

    p2V2【w=-p2v2】,即负pΔ,所以全部的功就是,左边的功加上右边的功,等于。

    麻省理工公开课 - 热力学与动力学课程节选

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