Intermediate Thermodynamics Questions & Answers  
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 $R=k_B/m?$
03.03.14
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.
03.04.14
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.
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.
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 $\xi$.
In class, we found that $\xi=N \bar{q}_x$. We used dimensions to give us a hint only. Ultimately, the number of particules hitting the wall per unit time per unit area is equal to the number of particules per unit volume ($N$) times the average velocity of the particules perpendicular to the wall. The velocity perpendicular to the wall corresponds to the average velocity of the molecules along one direction ($\bar{q}_x$, or $\bar{q}_y$, but not $\bar{q}$..). I chose the positive $x$ direction for illustrative purposes — I could have chosen the negative $x$ direction, or the positive $y$ direction, and we would have obtained the same answer. I'll give you 0.5 point for this question.
03.09.14
Question by Student 201127134
According to your lecture, when we calcuated $$ \overline {q_x}=\frac{1}{4}\overline{q}~~~~~~$$ this was because of the characteristic of molecules which can move in every possible directions. However you gave me an exaple of rectangle to siplify the process of calculating integration which means every possbile directions. Does this have something related with the shape of SPHERE? Actually ,the area of the sphere is $ 4 \pi r^2$.
03.11.14
You're on the right track. Consider one molecule. It can move in any direction with equal probability. Thus, to find the average velocity along one direction, you need to integrate the molecule's velocity in spherical coordinates (on the surface of a sphere). I'll give you 0.5 point for this question. I would have given more if you would have asked it the first time without using an attached image and if the post would be free of spelling mistakes.
Question by Student 201327103
Professor, Question is in the jpg file
question.jpg    
No questions in attached files are allowed. Please edit your post and type your question directly.
Question by Student 201027111
professor, I'd like to ask about today(3.17) problem when solve B part( find heat needed to go from b2 to b3) prof said 1st law thermo can't use because unknown veriable is two with pdv and Q but in process V is constant so isn't that dv=0 ? set dv=0 than d(me)=Q than intergral and we can get (me)_b3 - (me)_b2 = Q and assume calorically perfect gas than e=CvT so Q=mCv(T(b3)-T(b2)) we can get solve this way is wrong?
03.17.14
This is a good question. But before I answer, please edit your post and write the mathematics properly (with correct subscripts and same notation as in class).
Question by Student 200927141
Professor. When we derive the h=CpT, we assume that P is constant. But, today when we solve the problem about 3) heat needed to obtain state B3 from B2, Why can we apply h=CpT despite of Pressure is changed from 2.49MPa to 3.29MPa?
Please use the right notation when asking a question. Cp is not the same as $C_P$, h=CpT is not the same as $h=C_P T$, etc. After you edit your question and change the notation, I'll answer it. Also, don't ask me two questions at the same time. Please wait before I answer the question, and then ask a new one. Otherwise I'll delete the second question.
Question by Student 201127101
I don't know proof is right, but I'll try it. $v$ is magnitude of any velocity. $\theta$ is an angle that is between the x-axis and the straight line of the plane xy. $\alpha$ is an angle that is between the straight line of v and plane xy. When considering only half of the x-axis, the total velocity is shown below. $$v_{total} = \int_{- \pi / 2}^{\pi / 2} \int_{- \pi / 2}^{\pi / 2} v cos \alpha cos \theta d \theta d \alpha $$ $$v_{total} = 4v $$ $v$ is magnitude of any velocity, so v can be seen as the x-axis velocity. $$v_x = \frac {1}{4} v_{total}$$
It's getting there, but not close enough yet. Think about it more ;)
Question by Student 200927141
Professor. When we derive the $$ h=C_PT$$ we assume that P is constant. But, today when we solve the problem about 3) heat needed to obtain state B3 from B2, Why can we apply $$ h=C_PT$$ despite of Pressure is changed from 2.49MPa to 3.29MPa?
No, we did not assume that $P$ is constant when deriving the calorically perfect gas relationship $h=C_P T$. Rather we assumed that $C_P$ is a constant. What may be confusing you here is that $C_P$ is the so-called specific heat at constant pressure, which corresponds to: $$ C_P =\left(\frac{\partial h}{\partial T}\right)_P $$ The latter is true in the general case when the enthalpy depends on both pressure and temperature. But $h$ doesn't depend on pressure for a thermally perfect gas. Thus, for a thermally perfect gas $C_P$ becomes: $$ C_P =\frac{d h}{d T} $$ Thus, assuming constant $C_P$, we can then use $h=C_P T$ and this does not entail that $P$ should be constant. I'll give you one point for this question. I would have given more if you would have asked it with no typesetting mistake the first time around.
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