Please ask your questions related to Thermodynamics in this thread, I will answer them as soon as possible. To insert mathematics use . For instance, let's say we wish to insert math within a sentence such as $P$ is equal to $\rho R T$. This can be done by typing \$P\$ is equal to \$\rho R T\$. Or, if you wish to display an equation by itself out of a sentence such as: $$\frac{d}{dt}\int_V \rho dV + \int_S \rho (\vec{v} \cdot \vec{n}) dS =0$$ The latter can be accomplished by typing \$\$\frac{d}{dt}\int_V \rho dV + \int_S \rho (\vec{v} \cdot \vec{n}) dS =0\$\$. You can learn more about on tug.org. If the mathematics don't show up as they should in the text above, use Chrome or Firefox or upgrade MSIE to version 9 or above. Ask your question by scrolling down and clicking on the link “Ask a Question” at the bottom right of the page.
 03.02.14
 Question by Student 201227128 In "Ideal gas law" $$P=\rho RT$$ when professor prove this law, you said that $$R=k_B/m$$ And it can be $$k_B=Rm$$ (here $k_B$ is boltzmann constant, and $m$ is mass of one particle) But I learned the definition of $k_B$ in chemistry is that $$k_B=R/N_A$$ (here $N_A$is Avogadro number or constant) I cannot understand why those two formulas are same. In class, Professor said $R$ is not a universal gas constant. so I thought that in those two formulas 'R's are different. But I couldn't get the answer. why did you said that $R=k_B/m?$
 03.04.14
When asking a question, please use the same notation as used in class: use $k_{\rm B}$ instead of $K$ for the Boltzman constant. And what is $M$? I didn't use this symbol yet. Please edit your question and formulate it with the correct notation.
Now I understand your question better. The Boltzman constant $k_{\rm B}$ is equal to $\bar{R}/N_{\rm A}$ with $\bar{R}$ the universal gas constant, not to $R/N_{\rm A}$ ($R$ is the gas constant which varies from gas to gas). I can give you a bonus boost of 0.5 point for this question.
 Question by Student 201127148 Why is the velocity of molecules in X-axis equal to a quarter of the velocity of molecules? I'm sorry that I'm asking a question which you already explained in last class. But I have not been able to understand your lecture about this question. So I need some additional explanation.
 03.08.14
If all molecules would always have a velocity vector pointing in the positive $x$ direction, then the average velocity in the positive $x$ direction would be $\overline{q}$ (with $\overline{q}$ the molecular speed). But, molecules move in all directions randomly, not just in the positive $x$ direction. When integrating and taking a time average over all molecules in a gas, then it can be shown that the average velocity in the positive $x$ direction is $\frac{1}{4} \overline{q}$. I can give you 0.5 point for this question. For more points, you need to express better what you don't understand.
 03.09.14
 Question by Student 201227128 Question about 'example-(a)'  In class, professor said that $$\overline{q_x}=\frac{1}{4}\overline{q}~~~~~~(i)$$ $$\overline{q_y}=\frac{1}{4}\overline{q}~~~~~~(ii)$$ $$\overline{q_z}=\frac{1}{4}\overline{q}~~~~~~(iii)$$ I have read other student's question's answer about this.(above this question) But could I know the exactly method that how to integrate and take time average over the all molecules in gas? I want to know that because when you derive $P=\rho R T$, you said that $$q^2=u^2+v^2+w^2$$ and from this we got that $$\overline{q^2}=\overline{u^2}+\overline{v^2}+\overline{w^2}$$ I understand to use this formula with (i),(ii),(iii) is not correct. ($\overline{u^2}$ is not equal with $\overline{q_x}^2$) Thank you.
There is a mistake in your question: I didn't say that $\overline{q}^2=\overline{u}^2+\overline{v}^2+\overline{w}^2$ but I said that $\overline{q^2}=\overline{u^2}+\overline{v^2}+\overline{w^2}$. Please edit your question and make sure that you use exactly the same notation as in class.. Also, to put some blank spacing in equations, use the tilde character. I will answer your question after you make these modifications.
To answer your question, no, you cannot substitute $\overline{u^2}$ by $\overline{q_x}^2$. This is because by definition: $$\overline{u^2} \equiv \frac{1}{\Delta t}\int_0^{\Delta t} u^2 dt$$ $$\overline{q_x} \equiv \frac{1}{\Delta t}\int_0^{\Delta t} \max(0,u) dt$$ That is, $\overline{u^2}$ is the average in time of the square of the $x$ component of the velocity while $\overline{q_x}$ is the average in time of the component of the velocity in the positive $x$ direction. Taking the square of $\overline{q_x}$ will give a totally different answer as $\overline{u^2}$: $$\frac{1}{\Delta t}\int_0^{\Delta t} u^2 dt \ne \left( \frac{1}{\Delta t}\int_0^{\Delta t} \max(0,u) dt\right)^2$$ You can find more information about how to integrate the latter integrals in some book on the “Kinetic Theory of Gases” — but this is beyond the scope of this course. I'll give you 1 point for this question — for more points, you need to formulate it correctly the first time with the right notation.

For the second part of your question, please delete it and ask a new question below (only one question per post).
 Question by Student 201227128 This is another question about 'example-(a)'. In class, when professor get $\xi$ which is #of particles striking wall per unit time per unit area, you used N(number density) and $\overline{q_x}$. Here I have a doubtful thing. The logic that we can get $\xi$ is that $$\xi =N \left[\frac{particles}{m^3}\right]\times \overline{q_x}\left[\frac{m}{s}\right]=\frac{ particles}{m^2 s}$$ But the $\xi$ is not about the x- direction area, but is about just unit area. Furthermore, if we use $\overline{q}$ to get $\xi$, there is no problem in dimension.(we can also get $\left[\frac{ particles}{m^2 s}\right]$) So I wonder why did you use $\overline{q_x}$ not $\overline{q}$ to get the answer $\xi$.
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