-
We want to integrate. So let's take the integral of both sides, going from the initial point to the final point.
分别从初态,到末态做积分，消去微分。
-
The reason for inexact doesn't mean it's a crummy measurement, t means that it's path dependent, and so the value of this integral depends on how you get from one to two.
这是因为它是,与积分路径有关的，因此这里的积分值，取决于从一端到二端的,具体路径。
-
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。
-
So we can put that in here and do this integral.
因此把压强带入到这里，算出积分。
-
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。
-
du So, we can also write delta u as integral from 1 to 2 of du.
我们也可以将Δu写成,从1到2的积分。
-
V2 p1 times V2 minus V1. What that turns out to be, 0 this area right here. It's V1 minus V2 times p1.
第二步做的功是-pdV，积分从V2到，这部分积分出来是多少？应该是。
-
The ideal gas constant doesn't change, temperature doesn't change, and so v we just have minus nRT integral V1, V2, dV over V.
理想气体常数不变，温度也不变，因此,是负的nRT，积分从v1到v2，dv除以。
-
So, the work for this process is the integral, pdv or minus the integral, V1, V2, p dV.
此过程中的做功等于负的积分，从v1到v2，外部。
-
V1 This first integral is zero V1 to V1, then I get minus p2 times V2 minus V1 or p2 times V1 minus V2. Again, 0 a positive number.
先保持体积为，对pdV积分1,然后对p2dV从V1到V2积分，第一部分积分为。
-
/T We've got Cv integral from T1 to T2, dT over T is equal to minus R from V1 to V2 dV over V.
左边是Cv乘以，从T1到T2对dT积分。
-
So, if you're doing physics, x2 work is the integral of force, dx, x1 to x2.
如果你正在干活，功就是力的积分，从x1，到。
-
So it's nR integral from V1 to V2 dV over V.
即nR除以V乘以dV从V1到V2的积分。
-
That's equivalent to doing the integral, and so, what we end up getting is that the reversible work v2 pdv is equal to minus integral V1, V2, p dV.
这与刚才的积分过程效果相同，最后，我们得到的结论是可逆过程的功,是负的积分，从v1到。
-
V2 That's minus p1 times V2 minus V1.
做的功是-p1dV，积分从V1到。
应用推荐