Respuesta :
Using the ideal gas equation pV = nRT, where R is the ideal gas constant, we can show that p=nRT/V. Since n, R and T are all constants, p2/p1 = V1/V2 where p1 and p2 are the start and final pressures respectively and V1 and V2 are the start and final volumes respectively. For if p1 = 3*p2, the pressure would have fallen to one third of its original value, and it follows that V2 = 3*V1. Therefore, for the pressure to fall to a third of its original value, the volume must increase by a factor of 3.
Considering the Boyle's law, the pressure to fall to a third of its original value, the volume must increase by a factor of 3.
As the volume increases, the gas particles (atoms or molecules) take longer to reach the walls of the container and therefore collide with them less times per unit of time. This means that the pressure will be lower because it represents the frequency of collisions of the gas against the walls.
In this way, Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant.
Boyle's law is expressed mathematically as:
P× V= k
In this law then two variables are related: pressure and volume, so it is assumed that the temperature of the gas and the number of molecules of the gas are constant.
Studying two different states, an initial state 1 and a final state 2, it is satisfied:
P1× V1= P2× V2
In this case, you know that the pressure of an enclosed gas is be reduced to one-third of its original value. This is that P2= [tex]\frac{1}{3}[/tex]× P1. Then, replacing:
P1× V1= [tex]\frac{1}{3}[/tex]× P1× V2
Solving:
V1= [tex]\frac{1}{3}[/tex]× V2
3× V1= V2
Finally, the pressure to fall to a third of its original value, the volume must increase by a factor of 3.
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