If I had taken as my interpolation scheme, my white curve here, I could go to infinity and have the equivalent of absolute zero being at infinity, minus infinity.
要注意,如果我们采用,像图中白线这样的插值方案的话,我就可以一直降温下去,相应的绝对零度点。
Well, the energy at infinity is zero -K and the energy in the ground state is minus K.
无限远处的能量值为,而基态能量值为。
SdT This has minus T dS minus S dT, but the dT part is zero because we're at constant temperature.
这一项包含负的Tds和,但是dT的部分等于零,因为温度为常数。
l l So what you can do for a 1 s is just take 1 minus 1 and then l is equal to 0, so you have zero radial nodes.
它等于1减去,是等于0的,所以没有节点,这和我们看到的是相符的。
So when we count, generally, again, we start from zero, we go to N minus 1.
当我们计数时,一般的,再次强调,我们从0开始,到N-1结束。
pV=RT p plus a over v bar squared times v bar minus b equals r t. All right if you take a equal to zero, these are the two parameters, a and b. If you take those two equal to zero you have p v is equal to r t.
我们就回到,也就是理想气体,状态方程,下面我们来看看,这个方程。
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。
w So that means that minus w prime must equal w And w is greater than zero.
也就是说负w一撇等于,在这之中。
Here's heat exchanged in pathway A and in pathway B heat is zero, and in pathway C, Cv here is qC it's Cv T1 minus T2.
这是qA,这是路径A上的热量交换,路径B中的热量交换是零,而在路径C中,这是qC,它是。
That says that minus q1 prime plus q1 is greater than zero.
这就是说1,负q1一撇加q1大于零。
We put the magnesium 2 plus here, and then we put the oxygen here, O 2 minus, and we know that this gives us our R zero, right?
我们把Mg2+放这,然后O2-放这,我们都知道这给我们的是R0,对么?
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。
du/dV So now our du/dV, dp/dT at constant T is just T times dp/dT which is just p over T minus p, it's zero.
现在我们的恒定温度下的,等于T乘以dp/dT,在这里,等于p除以T,最后再减去p,结果是0。
And we wrote something that looks, the energy is equal to minus the Madelung constant times Avogadro's number, 0R0 q1 q2 over 4 pi epsilon zero R zero.
我们写下了,晶格能等于负的马德隆常数,乘以阿伏伽德罗常数,乘以q1q2除以4πε
e The charge on the anion times minus e, so there is the minus e squared, 0R0 and divided by 4 pi epsilon zero r naught, because now I am evaluating this function at r naught, one minus one over n where n is the Born exponent.
阴离子的电荷乘以,因此会有-e的频繁,除以4πε,因为现在我用r圈评估这个函数,1-1/n,n是波恩指数。
1 Here I have m is one, zero, minus one.
再有,m为1,0,或。
I know the energy in this first pair would equal -e^2 That is just going to equal minus e squared over 4 pi epsilon zero r naught.
我们明白第一对的能量将会等于,等于,/4πε0,R圈。
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