physical chemistry - Why are activities of solids and liquids taken to be unity?

First, I should mention that the question was already asked here: Why is active mass of a pure solid or liquid always taken as unity? However, the treatment there is qualitative and I do not really understand the reasoning for the disclusion of solids and liquids from the reaction qoutient that is presented there. I have attempted to derive this fact mathematically and failed, which why I ask this question again.

Let us look at an example reaction which involves solids and gases. The treatment for other reactions or reactions with liquids shouldn't differ significantly. A very simple reaction for this matter will be the combustion of graphite

$$\ce{C(s) + O2(g) -> CO2(g)}$$

Reaction quotients are derived from the equation for change in Gibbs free energy at nonstandard pressures, so let's find that. Suppose the reaction is carried out in nonstandard pressure, and $\Delta G^o_r$ is known. Let's (I) compress the system isothermally to standard pressure, (II) carry out the reaction and then (III) return it isothermally to the original pressure.

I. $\Delta G$ for the oxygen here is the well known $RT\ln (1/(P/P_o))$. For the carbon: In any isothermal process $dG=VdP$, let's assume for our ideal solid that the volume is constant (I have also tried this in a model where the volume changes with pressure with bulk moduli, the result does not differ significantly), so $\Delta G=V(P-P_o)$.

II. Here simply $\Delta G=\Delta G^o_r$.

III. Here it's just gas compressing so again $\Delta G=RT\ln (P/P_o)$

Adding the contributions from all three stages we have $$ \Delta G=RT\ln (1/(P/P_o))+V(P-P_o)+\Delta G^o_r+RT\ln (P/P_o)=\Delta G^o_r+V(P-P_o) $$

This result is different from what a model of an activity of unity for the carbon would predict. What went wrong? And why are activities of solids and liquids taken to be unity? Any help will be appreciated.

Answer

Nothing "went wrong". Hf for CO2 is ~393 kJ/mole. 1 latm = 9.87 kJ, so lets see what a mole of C does Gibbs energy wise going from 1 atm to 11. OK 1 mole is 0.0053 l and multiplying it by 10 atm we get 0.053 latm ~ 0.15 kJ or 0.15% of the heat of formation for the CO2. If you want to get down to this level of accuracy, you gotta throw out the Ideal Gas Law (the simple approximations it is based on aren't that accurate) and all sorts of consequences follow. Then, we have to consider how you actually change pressure "reversibly" (hint:you don't). The point of this level of treatment of thermodynamics is to show you how useful it can be to get an accurate idea of what is going on. It is accurate, unless your idea of "accurate" is errors in the parts per thousand range. (Ideal Gas Law is generally good to between 1 and 3% (at near STP), but it can be off over 10% for some not-so-uncommon gasses (and let's not even consider about gasses near their boiling points, or at cryo-temperatures or at mega-pressures...). A quick search for activity of graphite vs pressure indicates that at 100 bar, its activity coefficient is 1.01. That is, assuming an activity of 1.00 is a great approximation near STP. Liquids (but NOT solutions!) will be similar (but not quite as well 'behaved'). So, pragmatically, the answer to your question is:"Because it gives us an accurate estimate of the system's behavior." Another way to put it is: "Because it is useful." One other factor to consider is what the alternatives are, that is, if we wanted to get more accurate predictions, could we formulate a generally applicable means to do so? The answer is, as far as I know:no. There is no generally applicable (call this a "powerful") method; to get more accurate requires case-by-case considerations: how is heat gained or lost by system, what kind of container walls is it in contact with, how was complete combustion insured, what were the losses due to friction during compression and expansion, etc. etc.

history - Is it possible for mesorah to be discovered?

Suppose there was a previously unknown collection of midrash or sayings or halachah or commentary that was discovered in an archaeological excavation and is hundreds or thousands of years old.

Suppose that this text is not only readable but it is also somehow clear (to archaeologists or historians) that it is teaching from a well known sage of that era.

Since that text was not handed down from teacher to student over the generations, is it ineligible to be incorporated into official Jewish "canon"? Or could it still be considered in some sense authoritative simply on the basis of the ancient sage who wrote it? Would it depend on the type or genre of material or era?

Has anything analagous to this occurred before, and if so, what was the result?

Answer

There is a teshuva of the Rema in which he writes that if you find a Teshuva of the Gaonim, you could follow its opinion.

I asked R' Zvi Berkowitz about this and he said this was restricted specifically to the period of the Gaonim, because the Rishonim themselves (on whom much of our codification is based) would have taken the position of the Gaon into account and may very well have changed their opinion accordingly. However, if it would be a "discovered" Rishon, it would not have this authority, because the existing body of Rishonim would not have been impacted by the opinion of yet another Rishon.

Therefore, manuscripts can be accepted, but their impact into Halacha would depend on the authority they would have had in the Halachic process had they been known.

I know that there are others who accept the Meiri's Beis Habechira (written in 13th century and rediscovered in the 20th century) as having the authority of any other Rishon. I was told by a student of R' Avraham Yehoshua Soloveitchik (Rosh Yeshiva of Brisk) that the Brisker tradition treats the Meiri as having the authority of an Acharon, and can be used but not against the opinion of a Rishon.

halacha - How assur is mixed dancing?

What is the level of the issur on mixed dancing? Minhag? D'Rabanan, D'oraita, etc.?

What about separate dancing but without a mehitza?

Answer

According to R' Gil Student, touching affectionately during a slow dance would be forbidden biblically (according to the Rambam) or rabbinicaly (according to the Ramban).

He says that there is also a ban on mixed dancing (even without touching) going back to the Maharam miRottenburg, and he says that R' Yehuda Henkin says that this ban is still in place he says that the fact that there is a ban is no proof that it would have been permitted without one, as many bans from the Rishonim's times were bans which re-banned already existing prohibitions).

He also mentions a Sdei Chemed that one is not even allowed to be present where women are dancing.

DC blocking IIR filter clipping?

I'm processing audio data for voice input from a mic. The data arrives in 32 bit floats [-1 ~ +1].

My first filter is to remove DC:

// x = new input value, y = filtered output value
m_x += ( 0.01 * ( x - m_x ) );
y = ( x - m_x );

When I feed it audio that is close to clipping (but not actually hitting -1 or +1), I'll get values back that actually do go above the [-1 ~ +1] limit - sometimes way above. I find this behavior curious.

Can anyone explain why this happens?

Also, what's the best way to "fix" this? Do a simple clamp for the returned y value? Pre-scale the input down via (x * 0.7071) first?

Thanks!

Answer

Your DC blocker transfer function is

$H(z) = (1-a)\frac{1-z^{-1}}{1-(1-a)z^{-1}}$

and an alternative (equivalent) difference equation is

$y_n = (1-a)(x_n - x_{n-1}+y_{n-1})$

Although this filter provides no gain it is possible to find inputs that exceed +1. For instance this input [1 -1 -1 -1 1].

Because you seem to work with a floating point realization you should not need to scale down before filtering.

Edit: The output of your filter is bounded this way: $|y_n|<2(1-a)$. So an input gain of 0.5 guarantees that the output of the filter can't exceed +1 if the input is bounded between -1 and +1. Can you come up with an input that produces an output value that exceeds +1.5 or an input that produces an output values that is less than -1.5?

Edit2: Filters can provide amplification as well as attenuation, otherwise they wouldn't be that interesting. Your filter does not provide amplification in the sense that a sinusoid passing through your filter will not have an increased amplitude. You can see that by plotting the amplitude/frequency response of your filter. However, for more complex inputs the non-linear phase response of your filter can cause 'overshoots' in the output. An allpass filter for instance, even though it has a completely flat amplitude response can also provide 'overshoots' for fullscale inputs.

How to deal with the spikes depends very much on what the other modules in your signal path are doing. Maybe some of the modules create headroom maybe some of them consume headroom. I don't know. Considering your DC blocker in isolation then you can apply a headroom gain of say 0.7 or 0.8 and then saturate your output. Although this setting is likely to perform some saturations my guess is that they will be completely inaudible. You will have to confirm this by experiment.

words - In what type of social settings could one say "百姓{ひゃくしょ う}"?

My understanding is that " 百姓 " is a very impolite way to refer to modern-day traditional Japanese farmers. A good translation might be " country bumpkin ".

As best I can remember:
I learned that word while relaxing with friends at an 居酒屋{いざかや} in Tokyo. Subsequent to that, I tried using it in daily conversations with not very close friends. Each time, I was cautioned about saying " 百姓 ". Those who refer to farmers as " 百姓 " imply that they, themselves, are cultured and refined, while the " 百姓 " lead uneducated, unrefined, bumpkin-ish lifestyles.

1. Is my recollection of meaning and usage correct?


2. In what type of settings would a non-native Japanese speaker be able to say " 百姓 " and not be judged as an elitist jerk? In fact, I'd hope to be judged as being someone who knows clever slang.

Answer

1.Is my recollection of meaning and usage correct?

Yes, it is correct and that is because 「[百姓]{ひゃくしょう}」 is closer to "peasant" than to "farmer" in feeling.

2.In what type of settings would a non-native Japanese speaker be able to say " 百姓 " and not be judged as an elitist jerk? In fact, I'd hope to be judged as being someone who knows clever slang.

Whether you are a Japanese-learner or native speaker, there are basically no situations where you can call a farmer a 百姓 without offending someone. The only exception would be when a farmer calls himself a 百姓, which happens quite often.

We have the term 「お百姓さん」, which sounds more "correct" than the plain 「百姓」 but it still would not be used much in public by the more careful speakers -- in particular, by the media. In private conversations, though, 「お百姓さん」 is still heard fairly often.

The safest word choices would be 「農家{のうか}」 or 「農家の人」.

grammar - What does こと refer to in だけのことはある

I'm trying to figure out what "こと” is referring to in this idiom/grammar point.

In the following example sentence:

彼は若いころから日本に住んでいただけのことはあって、さすが日本語がうまいんだね。

What does koto refer to here?

Additionally, is this perfectly interchangeable with だけあって

Compare the following 2 sentences

彼は若いころから日本に住んでいただけのことはあって、さすが日本語がうまいんだね。
彼は若いころから日本に住んでいただけあって、さすが日本語がうまいんだね。

Are these different? If so, why?

atoms - Rutherford's Alpha Ray Scattering Experiment

I understood the result of this experiment that the nucleus is nearly empty and things like that. But what I have on mind is that when an alpha particle goes nearer to the thin gold foil why couldn't it capture electrons from gold and convert to very stable helium gas? It could also happen right? All the textbooks only mention the deviation of alpha particles. Why couldn't the above happen or is there anything wrong about my question?

Answer

Your assumption is correct. For alpha particles, the main contribution to the total stopping power can be attributed to the electronic stopping power, i.e. inelastic collisions with electrons. Only a small contribution comes from the nuclear stopping power, i.e. elastic Coulomb collisions in which recoil energy is imparted to atoms.

The stopping power of a material is defined as the average energy loss per path length that the alpha particle suffers when travelling through the material.

According to the International Commission on Radiation Units and Measurements (ICRU) Report 49 Stopping Powers and Ranges for Protons and Alpha Particles (1993), the contributions to the total stopping power for alpha particles in gold are as follows.

Typical low-energy alpha particles with $E=1\ \mathrm{MeV}$:

• Electronic stopping power: $3.887\times10^2\ \mathrm{MeV\ cm^2\ g^{-1}}$


• Nuclear stopping power: $8.394\times10^{-1}\ \mathrm{MeV\ cm^2\ g^{-1}}$

Typical high-energy alpha particles with $E=10\ \mathrm{MeV}$:

• Electronic stopping power: $1.650\times10^2\ \mathrm{MeV\ cm^2\ g^{-1}}$


• Nuclear stopping power: $1.315\times10^{-1}\ \mathrm{MeV\ cm^2\ g^{-1}}$

Since the energy that is required for excitation or ionization is only a few $\mathrm{eV}$, an alpha particle with an initial energy of a few $\mathrm{MeV}$ can liberate many electrons on its path. When the alpha particle has sufficiently slowed down due to the stopping interactions, it can finally catch two electrons to form a neutral helium atom. Hence, you can release and detect helium when you dissolve old minerals of uranium or thorium.

In the Rutherford experiment, however, the used gold foil was very thin ($<1\ \mathrm{\mu m}$; note that gold is very malleable and can be beaten thin enough to become semi-transparent). Even for low-energy alpha particles of only $E=1\ \mathrm{MeV}$, the range in gold is still $r/\rho=3.974\times10^{-3}\ \mathrm{g\ cm^{-2}}$, which can be calculated using the continuous-slowing-down approximation (CSDA), i.e. by integrating the reciprocal of the total stopping power with respect to energy. Considering a density for gold of $\rho=19.3\ \mathrm{g\ cm^{-3}}$, the range $r$ can be calculated as

$$\begin{align} r&=\frac{3.974\times10^{-3}\ \mathrm{g\ cm^{-2}}}{19.3\ \mathrm{g\ cm^{-3}}}\\[6pt] &=2.06\times10^{-4}\ \mathrm{cm}\\[6pt] &=2.06\ \mathrm{\mu m} \end{align}$$

I.e., the experiment was deliberately designed so that the alpha particles were not completely stopped and converted to neutral helium atoms within the gold foil.