Hi mezza8, thanks for the input and welcome to the forums. Even if we discard completely the D.F. connection to the sample variance, D.F. is still an important concept in statistics. It is applied in the chi-square test for example. A lot of people say that degrees of freedom is an intuitive...
Thank you for replying anyway. I am familiar with the proof you speak of, but some people have said that the n-1 "makes sense because it is the number of degrees of freedom." I rather doubt this claim; In fact, as I said twice, I doubt the entire claim that n-1 is even the number of D.F. to...
Sorry, I forgot to add that this is a common intuitive explanation for why the n-1 creates an unbiased sample variance. I take it it's a bad one? Regardless, n-1 is generally said to be the number of degrees of freedom in the case of n numbers whose residuals must sum to zero. Supposedly, only...
One common explanation of the concept of D.F. is this:
Suppose you have n numbers (a, b, c,...) that make up a sample of a population. You want to estimate the variance of the population with the sample variance. But the sample mean m is being calculated from these numbers, so when...
Oh right. We just need the formula mu=mu^o + RT ln a. Besides, if c^o were 1, then my formula would make mu^o zero, which can't always be right. Thanks for everything.
Ah, I get it. My primary misconception was that the standard state conditions had to include a temperature, such as 298K. There is actually no standard temperature here. Mu^o is the chemical potential when the substance is at standard concentration and the selected T.
Say, this appears to be...
Thank you for taking the time to assist me, DrDu. I'm afraid I still don't understand. It would make sense if mu standard were a function of standard temperature, but the symbol T in my previous post represents the temperature corresponding to nonstandard conditions. Mu standard may very well...
For now, let's focus on ideal solutions please. Let me state my problem more clearly. If we define the standard state concentration as 2 M instead of 1 M, then by
a_i = \gamma_{c,i} c_i/c^{\ominus}\,
we see that the activity is halved, since c^{\ominus}\, has doubled from 1 molar to 2...
I know, and the limit evaluates to ln(1+x). I'm saying if we replace delta x in the first expression with 1, we get the harmonic series summed up until k=x.
Thank you for your help, guys. So let me get this straight. If we double the definition of the standard state concentration, then the activity of a given concentration is halved? Then by this formula (taken from DrDu's link):
\mu_i = \mu_i^{\ominus} + RT\ln{a_i}
and assuming that the...
I can't believe some pig answered my request! Ugh! No seriously, thanks. (Hey, you asked for it.)
The thing that interests me about this formula is that if you replace delta x with 1 instead of -->0, you get the harmonic series up to the x-th term. And you can generalize it by having delta x...
Please show me the derivation for the formula relating Gibbs free energy change and the reaction quotient. I didn't find it on Google, so I decided to turn here. Is it really that hard?
Thanks for any help.
I think I have an explanation, though I thought of it myself so I can't prove it.
Imagine you have a solid, movable barrier (i.e. a piston) that interacts with your liquid and vapor, which are stored in a sealed cylindrical container. On the other side of the piston is your atmospheric...
I don't know where to begin to calculate this limit, but I think I know what the answer is. Could someone please do it with straightforward methods? I suspect the answer is ln(x+1).
Also, I don't know how to write it down in proper notation, so I ask that someone do that too.
limdx-->0...