"His name is John Barovetto, B-a-r-o-v-e-t-t-o.
VOA: standard.2010.05.31
pV=RT p plus a over v bar squared times v bar minus b equals r t. All right if you take a equal to zero, these are the two parameters, a and b. If you take those two equal to zero you have p v is equal to r t.
我们就回到,也就是理想气体,状态方程,下面我们来看看,这个方程。
and the same for favour, they spell f-a-v-o-r, we put u in it,
同样地,对于发烧这个单词,他们拼写成f-a-v-o-r,我们要加一个u在里面,
Were going to make it for a mole of gas, T1 so it's R times T1, V and then we'll have dV over V.
假设是有一摩尔气体,那么就是R乘以,然后有dv除以。
When a particle moves in a circle, it has an acceleration towards the center of this size, v^2 over R.
当一个质点在圆周上运动时,它有一个指向圆心的加速度,v^2 / R
If you think of two towns down here, Brive,b-r-i-v-e,famous for its rugby, and Tulle,t-u-l-l-e,which is a capital.
如果你可以想起来这儿的两个城镇,布瑞福,b-r-i-v-e,以英式橄榄球运动而负有盛名,图勒,t-u-l-l-e,是一个省的首府
And then we can take the derivative with respect to temperature, it's just R over molar volume minus b.
这样我们求,压强对温度的偏导数,结果等于R除以摩尔体积V杠减去b的差。
RT/V this expression becomes Cv dT over T is equal to CvdT/T=-RdV/V minus R dV over V.
这样,or,RT,over,V,bar。,So,now,这个等式就,可以化成。
v^2 / R is the acceleration directed towards the center of that circle.
^2 / R 就是指向那个圆心的加速度
Whenever you see a particle moving in a circle, even if it's at a constant speed, it has an acceleration, v square over r directed towards the center.
只要看到质点做圆周运动,即使是匀速圆周运动,也存在一个加速度,大小为 v^2 / r,方向指向圆心
But now, I can also write it as v^2 over R.
但现在我也可以写成 v^2 / R
The size of that is v^2 over R.
它的大小是 v^2 / R
It's true for any gas, and if I remove this limit here, r t is equal to p v bar, I'm going to call that an ideal gas.
这样的气体被称作理想气体,这就是理想气体的性质,理想气体的涵义是什么?
So instead of v bar, we write p v bar minus b, equal r t.
现在考虑,这些气体分子之间。
take the derivative of this, get the velocity vector and you notice his magnitude is a constant Whichever way you do it, you can then rewrite this as v square over r.
对这个式子求一次导,就能得到速度矢量,你会发现其模长是常数,不管用什么方法,加速度也可以写成 v^2 / r
If it's going in a circle, you will say from now on, that it, indeed, has an acceleration, even though no one's stepping on the accelerator, of amount v^2 over R.
如果它在一个圆周上运动,你会说从现在起它其实有加速度,即使没有人去踩油门,加速度大小为 v^2 / R
It just turns out that if you put the v^2 and you put the r, and r is 93 million miles, you will find the acceleration is small enough for us to ignore.
这样的话,如果你把 v^2 和 r 代进去,r 是 9300 万英里,你会发现加速度小得足以被忽略
So it's minus R T1 dV over V, right?
那么这是-RT1,dV/V,对吧?
The magnitude is v square over r.
模长为 v^2 / r
/T We've got Cv integral from T1 to T2, dT over T is equal to minus R from V1 to V2 dV over V.
左边是Cv乘以,从T1到T2对dT积分。
We want a relationship in p-V space, not in T-V space. So we're going to have to do something about that. But first, it turns out that now we have this R over Cv.
我们想要p-V空间中的结果,而不是T-V空间中的,因此需要做一些变换,先来看现在的关系,它跟R/Cv有关。
If you want to write it as a vector equation, you want to write it as v square over r minus-- I want to say that it's pointing in the direction toward the center.
如果要写出矢量方程,就应该写成 -v^2 / r,也就是说,方向就向圆心
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