2
$\begingroup$

In a video about the relation between pressure and volume on Khanacademy, Sal says that on increase in volume the particles hit the wall less often because the volume has increased and the surface area has also increased. In this way the pressure is going to decrease.
In this video at about 9:15

Hence according to this video pressure exerted by the gas does depend upon the surface area of the container. So if we keep the volume constant and increase the surface area can we decrease the pressure exerted by a gas?

$\endgroup$
2
$\begingroup$

As you increase the surface area you must bring parts of the surfaces closer together and this will increase the rate at which the surfaces are hit by the molecules which will compensate for the increase in area.

$\endgroup$
2
$\begingroup$

The argument in the video seems flawed: it is true that the surface area has increased, but that is irrelevant, because when you calculate pressure you look at the number of molecules hitting a unit of area, which depends on the gas density, and here on the volume, but not on the total area. The derivation is made here: https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Pressure_and_kinetic_energy, and you can see that in the end only the volume matters.

And o f course, for a fixed volume, the net force on the walls will be proportional to the area of the container.

$\endgroup$
  • $\begingroup$ do you mean the net pressure will be proportional to the area or the net force? $\endgroup$ – Osheen Sachdev Jun 2 '16 at 17:07
  • 1
    $\begingroup$ at constant volume the pressure will be constant, but the net force different, because F=PA $\endgroup$ – user83548 Jun 2 '16 at 17:08
1
$\begingroup$

It's the volume that determines the pressure (for a given amount of gas at a given temperature), so if you consider a rectangular container, increasing the surface area of two opposite sides and/or increasing the distance between those sides will increase the volume and therefor decrease the pressure. But increasing the total area while keeping the volume the same (in other words: changing the shape, not the volume) will have no effect.

To demonstrate this, assume the original container is a cube of 1 m³ (with each side a square of 1 m * 1 m) , and you change it to a rectangular box with width and length = 2 m and height = 0.25 m (top and bottom side become squares of 2 m * 2 m; left, right, front and back side become rectangles of 2 m * 0.25 m): now the average distance between particles and the top (or bottom) side of the box is reduced by a factor 4, so 4 times more particles hit the surface each second, but the surface area is also four times bigger (4 m²), therefor the pressure stays the same. The same is true for the sides of the box: the average distance has doubled, so the number of particles hitting them each second is halved, but the surface area is also reduced to half the original (0.5 m² instead of 1 m²), resulting in the same pressure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.