-
pV=RT dT here because the pressure is constant, dV=RdT/p so dV is equal to R over p dT.
因为对1摩尔气体有，于是。
-
P dV is equal to R dT. pV = RT for 1 mole, pdV=RdT so I just take dV here.
对等压过程，那么。
-
Now you have to know from elementary calculus that v times dv/dt is really d by dt of v square over 2.
根据初等微积分知识你们得知道,v乘以dv/dt其实就是d/2)/dt
-
So if you get these two guys together you get CvdT=-pdV Cv dT is minus p dV.
把它们联系,起来。
-
I know I only need 2, so I can relate dV dV to dp through the ideal gas law.
我只需要两个就够了，因此可以用,理想气体状态方程消去。
-
And in particular let's look at, for example, du/dV du/dV at constant temperature.
更特殊一点考察，恒定温度下的。
-
It's constant pressure. OK, so now, last time you looked at the Joule expansion to teach you how to relate derivatives like du/dV.
这是恒压的，好，上节课你们,学习了焦耳定律,以及怎样进行导数间的变换。
-
dS/dV And that, now, we know must equal dS/dV, with a positive sign. At constant temperature.
我们知道这个等于恒定温度下的，符号为正。
-
And so just like here, w2 now q2 is minus w2, that's integral going from three to four p dV.
就跟这儿一样，现在是q2等于负，等于从第三点到第四点过程的pdv的积分。
-
dV we're still going to write that it's an ideal gas.
理想气体的条件,也依然成立。
-
V+dV And this is going to be here V plus dV.
这里是。
-
TdS It comes from the fact that dq reversible is T dS, pdV and dw reversible is minus p dV.
这个结论来自于：可逆过程下dq等于，做功dw等负的。
-
du external dV minus T surroundings dS is less than zero.
加上压强p，du，plus，p,乘以dV减去环境温度T乘以dS小于零。
-
dV OK, so we've got dT here dV here.
等式的左边是dT，右边是。
-
And I know pdV what those turn out to be. It's minus S dT minus p dV.
我们知道,这最终就是负SdT减去。
-
pdv It's an isothermal expansion, so dw is just negative p dV.
因此dw等于负，这就是。
-
v dv We have minus V1, V2, nRT over V dV.
负的v1，v2，nRT除以。
-
Now, for an ideal gas, du/dV under =0 constant temperature is equal to zero.
对于理想气体，温度一定,时偏U偏V等于零。
-
pdV Minus S dT, that's the p dV term that's left, minus p dV.
应该是负SdT，留下的应该是pdV项负的。
-
pdV So, du is T dS minus p dV.
即du等于TdS减去。
-
So there's p v work and it's given by the integral minus p external dv or just the integral from one to two of dw.
我们知道对于pv系统来说,功w可以写成黑板上的两种不同形式。
-
p dA/dV, at constant T, must be negative p.
在恒定温度下，dA/dV等于。
-
That's where that term comes from, du/dV dV/dT.
乘以偏V偏T，p恒定，这项的来源。
-
You can take a derivative of the derivative and you can get the acceleration vector, will be d^2r over dt^2, and you can also write it as dv over dt.
你可以对导数再求一次导,你就可以得到加速度矢量,也就是 d^2 r / dt^2,你也可以写成 dv / dt
-
It's going to be the same temperature V+dV as before but the volume is V plus dV now.
将升温到跟路径1的结果一样,但是现在的体积是。
-
I'm going from V2 to V2 dv - what do you think this integral is? Right, W1 so this is easy part, zero here.
这样才保证了过程的可逆性,下面来计算，路径1的功。
-
p Well, it's not just p dS/dV because there's some dS/dV at constant T.
它不是简单的,因为式子中还包含，恒定温度下的。
-
dV dT Is equal to minus dV/dT at constant pressure.
它等于，负的恒定压强下的。
-
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。
-
So it's, this is just the integral pdv And it's an ideal gas, isothermal, right.
从一点到二点的,的积分。，from，one，to，two，of，p，dV。,这是理想气体，恒温过程，好的。
应用推荐