-
John Q.Public is also a common person - male or female.
VOA: special.2010.02.07
-
For isothermal expansion, that means that delta u does not change, but delta q is equal to delta w?
在等温过程中，是不是内能不变，Δq等于Δw？
-
So q is zero adiabatic. Work is zero.
因为系统绝热，所以q是零。
-
And lower case q sub 2 is the charge on the electron.
小写q加个下标2是指电子所带电量。
-
Sorry, the question was, what happens if I said p is less than q?
抱歉，问题是，如果我这里写p小于q的话？
-
Going from two to three, that's an adiabatic expansion, so q is equal to zero in that step.
从第二点到第三点，是绝热膨胀，因此q等于零。
-
It's well insulated. Heat is not going in or out adiabatic. q is equal to zero.
绝热性很好，热量不会变化,是绝热的，Q等于零。
-
All right, what we've already got that q is zero.
好，现在我们已经知道了q是零。
-
Or I could have a non-adiabatic, I could take the same temperature change, by taking a flame, or a heat source and heating up my substance. So, clearly q is going to depend on the path.
也能改变温度，绝热指的是没有热传递，在非绝热条件下，也同样可以升温,比如用火或者热源加热，这样，q也应当与路径有关。
-
q1 The heat input is just q1, Q and we'll define that as capital Q.
输入的热是，我们把它定义为大写的。
-
u=q+w All right, what is delta u? delta u is q plus w.
好，Δu是多少？Δ
-
w This is just q plus w. There's w, RT1 ln q has to be R T1 log of V2 over V1.
而U等于q加，那是w，q应该是。
-
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
-
q Delta u is w2 prime.
等于零，q，is，zero。,Δu等于w2一撇。
-
So we've already seen that q is zero.
我们已经知道q等于零。
-
U OK, so let's look at delta u. Delta u is q plus w.
好的，让我们看看Δ
-
Q So it's just capital W over capital Q, q1 which is to say it's minus all that stuff over q1.
所以效率就是W除以，等于负的所有这些之和除以。
-
q Work is zero. Delta u is w plus q, work plus heat. This is zero, this isn't.
功是零，△U是w加，功加上热量，这个是零。
-
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。
-
U And then if we can also determine delta u, then we know this, we know delta u is q plus w w, then we can determine work as well, right?
然后，如果我们也能定出△,然后我们知道这个,我们知道△U等于q加？
-
And of course, in either case, w delta u is q plus w, so it's q irreversible plus w irreversible, u being a state function it's the same in either case.
当然了，在这两个过程中，Δu都等于q加，因此现在是q不可逆加w不可逆，同时u在这两个过程中都是态函数。
-
Delta u is q plus w. Delta u isn't zero.
U是q+w，△U不是零。
-
U It's u, because u is to q plus w right, heat and work, but it's adiabatic. So there's no heat, exchange with the environment, and it's constant volume, so there's no p dV work, right.
什么是零？是U，因为,等于q加w，热量和功，但这是绝热的，所以系统与环境间没有热量交换；,同时它是灯体的，所以也没有pdV形式的功。
-
And because there is an explicit relationship between u, delta u, q and w, you can always find the easy way to derive the change in internal energy or the heat or the work.
因为Δu，q和w之间,有明确的关系，一般来说,内能或热或功的变化,易于计算。
应用推荐