Intermediate Thermodynamics Questions & Answers  




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.




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). 



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.




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.




No questions in attached files are allowed. Please edit your post and type your question directly.




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).




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.




It's getting there, but not close enough yet. Think about it more ;)




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 socalled 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.



$\pi$ 