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In my textbook, the proof for this formula is as follows:

$$F \propto \text{Area}_{\text{wall}} \cdot \text{(number of density of particles I.e here m is const) I.e} \frac{n}{V} *\text{Average Kinetic energy of particles}$$ Then, we get $$\frac{F}{A} = R(\text{Prop const}) \cdot \frac{n}{v} \cdot T$$ Since $T$ is $\propto$ average kinetic energy.

Let the above equation be called equation 1

What I didn’t get:

  1. How did we get them to be proportional to each other I.e How is equation 1 true ? Like I know F = ma. Then , if I shift A from 2nd expression to the right. $A*R(\text{Prop const}) \cdot \frac{n}{v} \cdot T$. Here, can I say this expression is equal to m*a ?

  2. Why is $K.E = T$ I.e How are we able to replace T and Kinetic energy? In my text book , they say K.E average is difficult to measure . Why is that ?

The main point is that just by explaining that certain quantities are proportional to one another. It doesn’t make a formula. Who knows if we missed some variables.

I hope I have added enough clarity to the Q to understand it.

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  • $\begingroup$ Can you write it like in your textbook? The second line about average kinetic energy seems alone, where is the connection in this proof. $\endgroup$
    – Mark_Bell
    Commented Apr 20, 2021 at 11:35
  • $\begingroup$ @Mark_Bell I have edited it now. Pls check $\endgroup$
    – S.M.T
    Commented Apr 20, 2021 at 11:52
  • $\begingroup$ In the second expression, $\frac{F}{A}=R\frac{n}{V}T$ not $v$ in the denominator of RHS. For your 2nd question, average Kinetic energy$\neq T$ but equal to $constant\times T$ which is included in $R$ in you last expression. I am not able to understand your 1st question. Can you restate it more explicitly? $\endgroup$
    – Iti
    Commented Apr 20, 2021 at 12:21
  • $\begingroup$ @Iti Thank you for your answer. I have edited the Q to make it more clear. $\endgroup$
    – S.M.T
    Commented Apr 20, 2021 at 12:24
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    $\begingroup$ 1. The proportionalities are established by experiments. 2 physics.stackexchange.com/questions/45785/…. $\endgroup$
    – Ryder Rude
    Commented Apr 20, 2021 at 12:28

1 Answer 1

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Here is my attempt. First of all, this seems like a rough derivation, not a formal one, just understood by setting some things proportional to others. Say you have a balloon with some amount of gas inside it. The gas is made up of many particles that bounce off the walls. Now with this setup I can try to give the intuitive (not rigurous) answer to your questions.

  1. In this balloon, as I said, the particles move around and constantly bounce off the walls, exerting some force on it each time they do so. Step by step, we can see the factors that affect this force. First off, if the balloon is bigger there will be a larger surface for the particles to collide and thus the net force on the balloon is proportional to its surface area, $F \propto A$. We also see that if you have more particles then the force on the walls will be greater as well: more particles (per unit volume, I'll develop this later) $\rightarrow$ more bounces per unit time $\rightarrow$ more force on the area. And so $F \propto$ (Number of particles per unit volume, or particle density). Finally, if the particles have a very high kinetic energy they will also have a higher velocity to bump into the walls, and so the force will also be larger. Therefore $F \propto$ (Average KE of the particles). Combining these you get the relation you cited. $F=ma$ relates the net force on a particle to its acceleration. Here we only wanted to see what the force on the walls is proportional to in terms of the properties of the system.
  2. Relating average kinetic energy to temperature just comes from what we mean by temperature: it is a measure of the average kinetic energy of the particles. Therefore average kinetic energy can be related to temperature for this case. Measuring the average kinetic energy of the particles could be simple in some cases if you know how it relates to temperature, but measuring it by its definition (add kinetic energy of each particle, then divide by the number of particles) would be impossible on practice as gases consist of many particles of tiny size.
  3. Forget about the Ideal Gas Law for a minute and think about what it means to say "particle density". We mean the number of particles in a certain volume, for example. Say you have two moles of particles in a volume of 1 Liter. Instead of considering them separately, it makes more sense for practical applications to use their ratio, $2\frac{mol}{L}$, as you can take a system with the same density of particles but with double de volume and still use the same value for the density (of course it is still useful to know the moles and the volume as well). In our case before, we don't really care about the volume of the balloon. Visualize the neighboring area of one little fraction of the wall. Say you have 10 particles in that region. Some of them will move and collide with the wall, some will not. However, if you had 20 particles, then (about) twice as many as before would bounce off the wall, thus doubling the net force on the wall (this approximation gets better when you consider the system as a whole). And with all this, how can we get a general formula for the density of particles in the gas? Just divide the amount of particles by the volume of our gas, i.e. take the ratio $\frac{n}{V}$

Hope this helped in some way and wasn't too tedious if you were looking for short answers.

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