Delta u, delta H, familiar state functions, q w changes in their values, q, w, heat and work.
U,△H,很熟悉的态函数,它们的值在变化;
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?
U OK, so let's look at delta u. Delta u is q plus w.
好的,让我们看看Δ
u=q+w We have a relationship between them.
它们之间有关系Δ
u=q+w All right, what is delta u? delta u is q plus w.
好,Δu是多少?Δ
w Now, we have u, q and w.
现在我们有u,q和。
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。
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。
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。
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