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sgstudent
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When we use Boyle's Law it means that as pressure increases the volume of the container would decrease. In my notes they stated that if I were to halve the volume there would be two times the number of collisions per second which implies that the force exerted is doubled. But why is this so? I agree that the pressure would be doubled but since now V= k/P so if I were to simplify in terms of forces and area it would be V=kA/F so if I were to halve it, the force would not be doubled as the surface area would not remain constant. So how can we say that the number of collisions is doubled per unit time?
The force would definitely increase by it won't be a by a factor of 2 since the surface area also increases so are my notes inaccurate?
But also reading this http://library.thinkquest.org/12596/boyles.html they state that the force should remain the same. Is this true? I'm imagining a 5 by 1 by 10m cuboid so the surface area is 50m^3 and the surface area is 50m^2. So if I were to set the constant to be 1, the force would be 1N. So if I were to halve the volume at the 10m mark, the volume would now be 25m^3 while the volume is 30m^2. So the new force would be 1.2N.
What's wrong with my concept here?
Thanks for the help
The force would definitely increase by it won't be a by a factor of 2 since the surface area also increases so are my notes inaccurate?
But also reading this http://library.thinkquest.org/12596/boyles.html they state that the force should remain the same. Is this true? I'm imagining a 5 by 1 by 10m cuboid so the surface area is 50m^3 and the surface area is 50m^2. So if I were to set the constant to be 1, the force would be 1N. So if I were to halve the volume at the 10m mark, the volume would now be 25m^3 while the volume is 30m^2. So the new force would be 1.2N.
What's wrong with my concept here?
Thanks for the help
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