Gas pressure and container shape.

  • Thread starter Dragynfyre
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  • #1
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Main Question or Discussion Point

I've been thinking about the ideal gas law lately and there is something I don't understand. Say you have 2 containers filled with an ideal gas. Both containers contain the same moles of gas, the same temperature of gas, and both containers have the same volume. Therefore, according to the ideal gas law the pressure in both containers will be same regardless of the container shape since P = RnT/V and all the variables on the right side are the same for both containers. However, I have thought of a situation where this doesn't seem possible.

Here's an example. One container is a sphere and one container is in the shape of a cube. Now let's say both containers have the same volume of 1m3.

According to the volume formula for a cube the dimensions of the cube will be:
1m*1m*1m

Therefore surface area of a cube is
6m2

According to the volume formula of a sphere the radius of the sphere will be roughly 0.62m and will have a surface area of 4.83m2


As you can see the cube has a larger surface area compared to the sphere even though they both have the same volume. Now since Pressure=Force/Area wouldn't the pressure in the sphere be higher than the pressure in the cube? I would expect if the number of gas molecules in the container are the same and temperature is also the same then the force applied on the container would be the same.

So basically my question is what am I missing here?
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
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If you integrated the absolute force over the walls of the container, the answer would not be the same between the two containers. (If you integrated the force as a vector, the total force in both cases would be 0 since the gas can't propel the container in one direction preferentially when the average velocity of the gas molecules was 0.)

Even though the number of molecules is the same in both cases, the one with more surface area will get more collisions in a given interval of time. If there is more area for the molecules to hit, they will hit more often.
 
  • #3
411
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The number of molecular impacts upon the surface of the cube in one second will indeed be greater than the number of molecular impacts upon the surface of the sphere. However the surface area of the cube is greater than the surface area of the sphere. Divide the number of impacts by the surface area and you get identical numbers of impacts per unit area per second. Ergo, the pressure is the same on both surfaces.
 
Last edited:
  • #4
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Hmm I'm having trouble visualizing why more molecules will hit the surface of the cube.
 
  • #5
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Hmm I'm having trouble visualizing why more molecules will hit the surface of the cube.
Because there is greater surface area: 6.00 square meters versus 4.83 square meters. However, the number of impacts per unit area (one square meter) will be the same.

Let's try it with a different way of visualizing an ideal gas pressure in molecular terms. Let us define pressure as the product of two terms: P=AB. Here, P is the pressure in pascals, A is the frequency of molecular impacts in number per second per square meter, and B is the mean impulse transferred per impulse in newtons.

Given that both geometric containers have the same pressure and temperature, the 6 m3 surface of the cube will experience more impacts in one second than the 4.83 m3 surface of the sphere. The number of impacts per square meter, however, remains the same. Hence, the pressure is the same.
 
  • #6
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Ahh some of my confusion may have arose because I didn't understand how pressure is calculated mathematically when talking about molecular impacts.
 

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