So, these two are equal to each other as well which tells me that this derivative, Cp dH/dT constant pressure is Cp.
所以这两者也相等,这告诉我们在恒压下微分,等于。
Cv+R=Cp Cv is equal, oh Cv plus R is equal to Cp it's a relationship that we had up here that we wanted to prove.
我们就得到了,我们一开始,想要证明的。
Cp, I forgot to put my little bar on top here because it's per mole Cp dT that's my dq here.
上面的Cv我忘记加上横线了,因为它也是摩尔热容。
dT This is equal to Cp minus R dT.
等于。
Cp This will give us something about C sub p.
而这里可以计算出。
Cp so this is exactly what we want.
正好出现了我们需要的。
but right now you're going to have to take it for granted. So, if the Joule-Thomson coefficient is equal to zero, just like we wrote, du = Cv dT du = Cv dT for an ideal gas, we're going to dH = Cp dT have dH = Cp dT for an ideal gas as well.
但是现在请你们应该把它看成理所当然的,所以,如果焦耳-汤姆逊系数等于零,就像我们写的,对于理想气体,我们也可以得到对于理想气体。
Cv So, for Cp and Cv, these are often quantities that are measured as a function of temperature, and one could, in fact, calculate this integral.
一般Cp和,都是温度的函数,因此实际上,我们可以将这个积分计算出来。
By definition I'm going to define gamma by Cp over Cv by definition.
把Cp/Cv记作γ,这完全是定义。
T So we know that T dS/dT at constant volume is Cv over T, T and dS/dT at constant pressure is Cp, over T.
在恒定压强下定压比热容Cp乘以dT除以,所以在恒定体积下dS/dT等于Cv除以,在恒定压强下dS/dT等于Cp除以。
So we already know that. So now we can write CpdT or differential dH as Cp dT plus dH/dp, pdp constant temperature, dp.
我们已经知道了这个,所以我们现在,可以写出H的微分式:dH等于,加上恒温时的偏H偏。
T It's Cp dT over T at constant pressure.
定容比热容Cv乘以dT除以。
We already did that. OK, dH/dT constant pressure is Cp. That was easy one.
我们已经做过这个计算了,好的,在恒压,状态下的偏H偏T就是Cp,这个很简单。
UC So we can immediately write delta u C is Cv times T1 minus T2. Delta Hc C is Cp times T1 minus T2, right?
所以我们可以直接写出△,是Cv乘以,△HC是Cp乘以,对吧?
Cp And delta T is given by the heat, which has to do with how much of the candle burnt, divided by the constant pressure heat capacity.
T等于热量q除以恒定的等压热容,其中热量与,蜡烛燃烧的多少相关。
Over here, we have dq=Cp dT, the heat, the proportionality between heat - and temperature rise is given by this, the constant pressure heat capacity.
这里我有dq=CpdT,这是热量,这是联系热量,和温度变化的系数,恒压热容。
This is only true for an ideal gas. Since it's true for an ideal gas, then we can go ahead and replace this with Cv, and then we have Cp=Cv+R Cp with Cv plus R, which is what we were after.
常犯的一个错误,这只对理想气体成立,因为对理想气体成立,所以我们可以继续,用Cv代替,这项,最后得到。
There I have it Cp is equal to Cv plus R, right?
所以Cp=Cv+R,对吗?
Cv What do we know about the Cp and Cv?
关于Cp和?
Cv The only difference is it'll be Cp instead of Cv, B but there it is for pathway B. There it is for C a pathway C. So the state functions that we're familiar with are doing what we expect they ought to be doing, right? If you go around in a cycle, starting and ending at the same place the state functions have to stay the same.
是Cp而不是,这是路径,这是路径,所以我们熟悉的态函数的行为,正与我们预期的相同,对吧?,如果你沿着循环走一圈,开始和结束于同一个位置。
应用推荐