# Why does frequency of collision increase as volume decreases?

Today I was presented with the following scenario: A syringe with one end sealed and with a frictionless piston is dipped into cold water. What is the change in frequency of collision(ie, more or less)?

In my opinion, it should be lesser. When temperature decreases, average velocity of the gas particles decreases as $T\propto KE$. Hence, since velocity decreases, frequency of collision should decrease too.

However, the correct answer was more. The argument is as follows: when temperature decreases, speed decreases. By ideal gas law and that the piston is frictionless, the piston would slide down to maintain the pressure. My teacher mentioned that pressure is made up of 2 components: force and frequency. Since speed is lower, force is lower, and frequency must increase accordingly such that pressure is the same.

I am rather skeptical of this answer. How does speed correspond to force? I would also like to know why my line is reasoning is flawed.

• Speed corresponds to force because on each bounce, the change in momentum of the particle depends on its speed. But I also can't follow the argument presented in the question. – Jasper Aug 17 '18 at 12:48

Here is a simple answer: the collision rate can be said to be proportional to the ratio between velocity and the total surface area of the piston.

$PV=nRT$. Hence, $V\propto T$. The total volume is $\pi r^2h$, so we can say $V\propto h$, where $h$ is the height of the column of gas.

Next, $KE=\frac{3}{2}kT$, where $k$ is Boltzmann’s constant and $T$ the temperature. Thus we can say $T\propto v^2$.

Hence, $h\propto v^2$, and from this relation, the collision rate $Z\propto \frac{h}{v}$. we notice that if $h$ decreases by a factor of $2$, $v$ would decrease by a factor of $\sqrt 2$, and the new ratio $Z\propto\sqrt 2\times \frac{h}{v}$. From which, we can see that collision frequency would increase.

As to why there is a force, upon collision, there is change in momentum: impulse, which is the cross product of force and time.

We need to know what pressure is. Pressure in this case is the force applied by the gases particule per surface unit. It is in fact a mean value like many macroscopic value in thermodynamic

All the computations are done without the numerical constants

To compute it let's say that the variation of momentum exchanged by the collision of a particule with a wall is $\delta p = v \delta m$ (in fact : $2v\delta m$, you can derive it through a momentum budget on the particule before and after the colision).

Then from Newton's second law we have $$\frac{dp}{dt} = F$$ We can take the mean on a small interval of time $\delta t$ $$\frac{1}{\delta t}\int_{t}^{t+\delta t} \frac{dp}{dt} \, dt =\frac{1}{\delta t}\int_{t}^{t+\delta t} F \, dt$$ $$\frac{\Delta p}{\delta t} = \overline{F}$$ And we can find $\Delta p$ : it is the sum of all the small $\delta p$ created by all the particles. All the particles that can reach the wall of surface $S$ are at the maximum distance from it of $v\times \delta t$. So they are all in the volume $S\times v \delta t$. The number of particles that reache the wall is then $N = C\times S v \delta t$ where C is the concentration of particules. (In fact it is $\frac{1}{6} N$ because whe shall only count the particles moving toward the wall)

$$\Delta p = N\times \delta p$$

So we have $$\frac{ C S v \delta t m v}{\delta t} = \overline{F}$$ $$C v \delta m v = f \times \delta m v= P$$

And here we can see the frequency of collision appearing : $f=Cv$ What happens is that the temperature goes down but the pressure remains the same. Therefore the value $f \times \delta m v$ must stay the same. Yet we know that v decreases with the temperature therefore f has to increase. With the equations we wrote you can see like yout teacher said that pressure equals speed times frequency with a mass factor : $f \times \delta m v= P$

What happens is that though the speed is decreasing in the term $f = Cv$ the concentration is rising due to the diminution of the volume with conservation of the mass and it counterbalances the decreasing of v

• I think you mean delta v m and not v delta m. Mass is constant but velocity changes. – Jasper Aug 17 '18 at 18:16
• @Jasper ; I use the $\delta$ to mean that it is a very small quantity : the mass of a particle. It is not an infinitely small variation, I would have used dm instead. I'm sorry I have not been clear. In what I have defined v is constant. It is a mean value (RMS) and it corresponds to the mean kinetic ernergy corresponding to température. In other word if I have $m$ the mass of a particle then due to the equipartition theorem $\frac{3}{2}k_BT=\frac{1}{2}m\overline{v}^2$. Hence $v_{rms}=\sqrt{\frac{3kBT}{m}}$. All I did was naming $m$ as $\delta m$. I hope it is clearer now. – Q.Reindeerson Aug 20 '18 at 13:48