Suppose that we have a cube with dimensions $25 \times 25 \times 25$ centimeters containing a single hydrogen molecule. How can we calculate the pressure within the cube?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
4
Pressure would not be defined in this case since it is a statistic measure of the force a big number of molecules apply on surfaces when they randomly collide with it. The property of pressure depends indeed on the assumption of a big number of particles involved such that all the macroscopic values converge on the average/expected value and the velocities of singles molecules becomes irrelevant. So in this case what we call pressure would be the force the hydrogen apply on the cube faces when it collides with them, this force would be discontinuous and would not have any pressure-like symmetry, well the answer I would say is indeed that pressure is not defined in this case since it is only a macroscopic statistical phenomenon.
-
$\begingroup$ So how can we calculate this "macroscopic statistical phenomenon"? $\endgroup$– user147133May 13, 2017 at 14:16
-
$\begingroup$ If pressure is force per unit area, wouldn't a single molecule would be easier than the statistical case? $\endgroup$– danielMay 13, 2017 at 14:26
-
2$\begingroup$ You could just calculate the momentum of the molecule and you would already have all the information that pressure gives and much more. It's just not meaningful to talk about pressure in this case. It is like asking what is the sum of the values that comes out of the the rolling of 10^10 dice? Well that's 3.5*(10^10) and you can be wrong of very very little (very little), this is what you do when you calculate pressure. Now you 're asking me what is the sum of the rolling of a single dice? You can still say the expected value is 3.5, but it's a much much weaker information. $\endgroup$ May 13, 2017 at 14:46
-
$\begingroup$ How can you derive the speed of the molecule? $\endgroup$– user147133May 13, 2017 at 16:34
$\begingroup$
$\endgroup$
4
I have drawn the velocity vs time diagram for a single molecule bouncing back and forth in a 1D box. Assume the particle keeps moving with some constant speed $v$ between bounce. Since the wall imparts some force on the molecule to change its direction, that is the only place where the velocity changes. In the above image, $T$ is the time between consecutive bounces and $t_c$ is the finite time required to change the velocity from $v$ to $-v$.
Thus the 'average' force from the wall on the molecule would be $$F=\dfrac{2mv}{t_c}.$$ Thus to answer your question, we cannot calculate force if you do not know $t_c$; and this is something that you haven't provided in your question. Note that technically there is no link between $T$ and $t_c$. This is because $t_c$ depends on the walls of the container and the molecule in question, while $T$ depends on the length of the container and the speed of the particle. These two quantities are independent.
Thus your question is not well-posed ergo expecting an answer is wrong.
-
$\begingroup$ There is no problem in 1 dimension: $F=mv^2/L$: galileo.phys.virginia.edu/classes/252/kinetic_theory.html $\endgroup$– user137289Mar 4, 2020 at 22:02
-
$\begingroup$ @Pieter Appreciate the comment. The problem with the reference you cited, is that you are dividing by the wrong time interval. I am using the definition of the force as the derivative in order to obtain my answer. Please feel free to go through my answer and correct what you think is wrong. $\endgroup$ Mar 5, 2020 at 0:43
-
$\begingroup$ No, I cannot fix this by editing, I disagree with your analysis (but I give you credit for the drawing that shows your thoughts clearly). The "pressure" (in one dimension there is no area) is the average force over a long time, totally analogous to ordinary kinetic theory. $\endgroup$– user137289Mar 5, 2020 at 8:37
-
$\begingroup$ @Pieter If you disagree with my analysis, then you should be able to tell me exactly what I am doing wrong. Let me first clarify that I am not calculating the pressure, just the force. My point is that this analysis should not be mixed with the standard analysis of the Kinetic Theory of Gases, that's why I drew the figure. Anyway, I appreciate you taking the time to comment. Please feel free to drop a comment correcting me if you figure out in the future what I did wrong. $\endgroup$ Mar 5, 2020 at 13:07