# Doesn't the gas molecule hit the opposite wall during the time $\Delta{t} = \frac{2L}{v_x}$?

We should include only those accelerations that are caused by the piston, not those caused by the opposite side. The best way to accomplish this is to take $\Delta{t}$, exactly the time it takes for the molecule to undergo a round trip: $$\Delta{t} = \dfrac{2L}{v_x}$$

Wouldn't the molecule collide with the opposite wall during that time-interval? It is moving to the left after being rebounded from the piston. So, wouldn't the molecule during the one-round trip collide with the wall on the opposite side?

And why does the author want to exclude the acceleration imparted by the opposite wall? If he is doing so, how can he measure the total pressure as he is ignoring the force of the molecule on the opposite wall? Please help.

• Can you include a picture? So we can see exactly what $L$ is and how a one-round trip is defined. – Steeven Jan 9 '15 at 13:54

Willn't the molecule collide with the opposite wall during that time-interval? It is moving to the left after being rebounded from the piston. So,willn't during the one-round trip, the molecule collide with the wall on the opposite side?

If I am right that $L$ is the cylinder lenght, then $\Delta t$ is the time from a molecule leaves the piston to it reaches it again. And we consider only one molecule know.

So the answer is yes. The molecule will hit the opposite wall and bounce back.

Assuming the molecule has horizontal speed $v_x$ (for more molecules $v_x$ is thus considered an average speed), during $\Delta t$ the molecule is sure to hit the piston exactly once. That might be the point of the text.

And why does the author want to exclude the acceleration imparted by the opposite wall??

Actually he is not "totally" excluding it - the rebouncing is necessary for the setup to make sense. If the opposite wall was not changing the direction of the molecule (rebouncing in an assumably elastic collision), which means changing the horizontal speed from $-v_x$ to $+v_x$, the piston would not necessarily be hit by the molecule during $\Delta t$.

He is just not including it in his calculations. To find pressure you don't need to consider more than one surface. By including as little as possible the model is kept simple.

If he is doing so, how can he measure the total pressure as he is ignoring the force of the molecule on the opposite wall??

Remember that pressure is per area, $p=F/A$. So ignoring the force on the opposite wall is irrelevant. You can say, you only need to consider one square meter - or just one surface of which you know the exact area. In this case the piston.

• He knows that exactly one molecule hits the piston during $\Delta t$ (that is, it is being accelerated by having the direction changed, $+v_x$ to $-v_x$). If you had $n$ molecules you then know that exactly $n$ collisions will happen during $\Delta t$.

• He knows the mass $m$ (I assume he knows what gas molecule it is).

• So at some point during $\Delta t$ the total mass $M$ (or $M=n*m$ for many molecules) is being accelerated by the piston area $A$. That force would be $F=Ma$ (or $F=Ma=nma$ for more molecules).

• The pressure is just $p=F/A$ (or $p=F/A=nma/A$ for more molecules). It is like an average force per square meter during that time $\Delta t$.

The speed $v_x$ (which would be average speed for more molecules) depends on temperature e.g. If that $v_x$ is known, he can find the acceleration also and find the pressure.

• I always see that everything is done using the x-component. So,how can one conclude that the pressure is same in the y and z direction? – user36790 Jan 9 '15 at 18:06
• Sir,you are correct that only one surface is required for measuring pressure. But this is possible when in all the directions ie. in all the walls,pressure is same. Now,how can I say so? – user36790 Jan 9 '15 at 18:09
• Another thing: In his lectures by Feynman & in Peter Atkins' book, they write at any instant,half of the molecules move left while half move in the right. How did they come to this conclusion? – user36790 Jan 9 '15 at 18:12
• @user36790 I always see that everything is done using the x-component That's a good point. But I do think that the point is that $v_x$ is average speed in the x-direction (in the case where you have many more molecules than just one). Since all motion is random in all directions, what is average in one direction is also average in the other two directions. So average speeds perpendicular to each direction are equal. – Steeven Jan 9 '15 at 22:53
• @user36790 Sir,you are correct that only one surface is required for measuring pressure. But this is possible when in all the directions ie. in all the walls,pressure is same. Now,how can I say so? If we agree that all average speeds in any direction are equal, then in any direction you can look at a piece of the wall. (At a curved surface like the cylinder, you might need to look at only an infinitisemal piece of the wall to have a plane area.) In any direction you can then perform the same procedure as above. In any direction the result for the pressure $p$ will be the same. – Steeven Jan 9 '15 at 22:59