-
This is just an equality. I have a constant pressure dH process. This term here is equal to zero.
这是一个等式，这是个恒压过程,这项等于零，这意味着。
-
And so, again, we see a temperature increase, and we know the work, and the temperature increase, it's a constant pressure thing.
好，我们看到温度升高了，然后我们有做功量和温度的升高量，这是一个恒定压力下的值。
-
It is taking place inside this thing, and it's a constant pressure, and we'll do it reversibly, right. So that's what we've got.
它是绝热的，在这个内反应,是在恒压下，它是可逆的，对吧？
-
OK, so this, what I've sketched here would be a constant pressure calorimeter. There's a reaction.
好，我画的就是一个恒压量热计,其中进行一个反应。
-
So this is a constant pressure calorimeter.
它做了很好的隔热处理。
-
It's going to take place in there. It's going to be a constant pressure, it might be open to the air, or even if it isn't, there might be plenty of room, and it's a liquid anyway, so the pressure isn't going to change significantly.
也许它是液体，它在这个位置,这是恒压的，它也许是连通大气的，就算不是，它也有，足够的空间，而它是液体,压强不会显著地改变。
-
So that's going to have to be a constant pressure path.
它应当包含,一个等压过程。
-
Path number 2 on my diagram it's a reversible, this path number 2, it's a reversible constant pressure path.
路径，首先是一个，等压过程。
-
It's a state function, so we're at constant temperature and pressure, and now we want to consider some chemical change or a phase transition or you name it.
这就是态函数，我们处于恒定的温度和压强之下，然后考虑某些化学变化或者相变,或者你想考虑的东西。
-
You just change volume to pressure and basically you're looking at enthalpy under a constant -- anything that's done at a constant volume path with energy, there's the same thing happening under constant pressure path for enthalpy.
可以看到这就是把体积换成了压强,一般我们都是在一种恒定状态下,考虑焓的，任何在恒容条件下,能伴随能量变化的东西，也在恒压条件下伴随焓同样地变化,所以你可以经常。
-
This piston is being brought out, so we expect 0 the work to be negative, negative. And we start o V2 ut with zero volume. We end up with a volume p2 of V2, and the external pressure is constant to p2.
所以我们可以想象功是负的,开始的时候体积是,最终的容积是,外界的压力恒为。
-
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。
-
Let's hang on for a little while longer to a set of conditions where we will maintain constant entropy, namely constant entropy and pressure.
我们先继续考虑,保持熵不变的情况，即保持熵和压强不变。
-
I mean maybe up the street we whisper, but here we know it. And, so here is a different kind of system where we have a constant external pressure.
或许在街上你们低声细语，但在这里不应该这样，这是另一个系统,我们有一个恒定的外压强。
-
So I need, well the pressure is constant, but there's a change in volume.
压强不变，体积变化。
-
Could be done, but easier is to just do the whole thing at constant volume, right, and just run the reaction that way and redo the calculation to be a constant volume rather than constant pressure calorimeter, right.
可以进行测量,但是如果在体积恒定的条件下,做这些会容易得多,还是这样进行反应,但是在等体而不是恒压条件下重新计算。
-
I'm pressing on the gas. So I expect that to be a positive number. The pressure is constant 0 p. The V goes from V1 to zero.
我们对气体加压,所以这应该是一个正数,压强是常数，p，V从V1变成。
-
What is dH/dT as a function, keeping pressure constant, what is dH/dp, keeping temperature constant?
恒定时偏H偏T是什么,温度恒定时的偏H偏p又是什么呢？,好的，让我们解决第一个问题？
-
If you want to write a function that describes this line here, it's pressure as a function of volume related to each other with this constant.
如果要写出描述这条等温线的方程,它的压强和体积,就通过常数C相联系起来。
-
Let's take a system. Under constant pressure T1 V1, going to a second -- this is the system, so let me write the system here.
我们建立一个系统1,在恒定的压强T1，V1，下，变成了另一个系统，-这个初始的系统让我把它写在这。
-
Let's say we start from some V1 and p1 here, so high pressure, small volume and we end up with a high volume low pressure, under constant temperature condition.
例如我们要从压强比较高，体积比较小V1，p1出发,到达低压强，大体积的末态，过程中温度不变。
-
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。
-
pV Also A plus pV and G is minimized at equilibrium with constant temperature and pressure.
同时等于亥姆赫兹自由能A加上,同时在恒定的温度和压强下。
应用推荐