-
"The tires are larger than you would find on a HUMVEE, so the ground pressure, which is really what matters, isn't all that much different from an up-armored HUMVEE,"
VOA: standard.2009.11.02
-
This depends on the path. It tells you right here the path is constant pressure. These don't depend on the path, right. V doesn't care how you v get there. u doesn't care how you get there.
这由变化的具体路径决定,这个小脚标表明过程是恒压的,这些量都与具体路径无关,即不管是通过什么路径使得体积变化为Δ
-
It's going to take place in there. It's going to be a constant pressure, it might be open to the air, or even if it isn't, there might be plenty of room, and it's a liquid anyway, so the pressure isn't going to change significantly.
也许它是液体，它在这个位置,这是恒压的，它也许是连通大气的，就算不是，它也有，足够的空间，而它是液体,压强不会显著地改变。
-
The pressure points come when the issues are so close to the bone that a variety of people don't feel that they had the luxury to do that, that their interests, their commitments are threatened.
有时也有一些压力,这时话题通常比较敏感，许多人,觉得言论自由成为了一种奢侈品，他们的兴趣和信仰都受到了威胁。
-
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。
-
dT That means that dH is also equal to dH/dT, constant pressure dT. All right, so now I've T ot more dH/dT under constant pressure.
也等于偏H偏T恒压乘以，现在我已经得到了在恒压,状态下的偏H偏。
-
du But here you've got pressure constant. du, T this is du, not H here. du/dT is only equal Cv to Cv when the volume is constant, not when the pressure is constant.
这里是压强横笛，du，这是，不是H，偏U偏,只在体积恒定时等于，而不是压强恒定时。
-
It would be much better if you had a reference point that didn't care where the pressure was.
我们还需要精确地定义气压,你必须知道气压的大小,当然如果我们能找到一个。
-
Because if it's irreversible, it's very likely that I don't know what the pressure inside the system is doing while this is happening.
因为对不可逆过程，系统内部的压强,没有明确的定义,气体不处于平衡态。
-
And so now, instead of using these reference points for the Kelvin scale, we use the absolute zero, which isn't going to care what the pressure is.
就像理想气体温标,与气体的种类无关一样，具有普适性，在开尔文温标中。
-
This is an example where the external pressure here is kept fixed as the volume changes, but it doesn't have to be kept fixed.
在我们举的这个例子中,外界压强不变，气体体积改变。
-
And this volume, temperature and pressure doesn't care how you got there. It is what it is.
另一个状态,也有一组确定的体积。
-
T It's Cp dT over T at constant pressure.
定容比热容Cv乘以dT除以。
-
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。
-
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。
-
I don't know what the pressure is doing in there, doing that expansion.
我不知道这个过程中,内部气体的压强。
-
It doesn't matter what the pressure internal is.
内部的压强是多少无所谓。
-
Negative pressure doesn't make any sense, right?
负的压强没有意义？
-
That is, in real life, the variables that you'd normally control aren't some combination of entropy and these variables, but really their temperature, volume and pressure, any couple of those, might be what you'd really have under experimental control.
在生活中，我们所能控制的,不是熵和其他变量的组合，而是温度，体积，压强，以及其中的两两组合,这些才是试验中所能控制的。
-
In this case, V = /P. Have two quantities and the number of moles gives you another property. You don't need to know the volume. All you need to know is the pressure and temperature and the number of moles to get the volume.
以及气体的摩尔数,就可以得到第三个量,知道压强，温度和气体的,摩尔数就可以推导出气体的体积,这称为状态方程,它建立了状态函数之间的联系。
-
In other words, the order of taking the derivatives with respect to pressure and temperature doesn't matter And what this will show is that dS/dp dS/dp at constant temperature, here we saw how entropy varies with volume, this is going to show us how it varies with pressure.
换句话说,对温度和压强的求导顺序无关紧要,结果会表明,恒定温度下的，对应我们上面看到的，熵如何随着体积变化，这个式子告诉我们，熵如何随着压强变化。
-
Normally I couldn't do that Vdp because this term would have p dV plus V dp, but we've specified the pressure is constant, so the dp part is zero.
一般情况下我不能这么写,因为这一项会包含pdV和，但是我们已经假定压强为常数，所以包含dp的部分等于零。
-
So we have dH/dT keeping pressure constant, is du/dT keeping pressure constant.
等于偏U偏T，p恒定加上，偏pv偏T，p恒定。
-
So we're going to start with a mole of gas, V at some pressure, some volume, T temperature and some mole so V, doing it per mole, and we're going to do two paths here.
假设有1摩尔气体，具有一点的压强p，体积，温度，我们将让它，经过两条不同的路径。
应用推荐