Answer:
The volume that the same gas will occupy at 101.3 kPa if the temperature is kept constant is 5.27 L.
Explanation:
As the volume increases, the particles (atoms or molecules) of the gas 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 pressure and volume are related, determining Boyle's law that says:
"The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure"
Boyle's law is expressed mathematically as:
P * V = k
Now it is possible to assume that you have a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and it will be fulfilled:
P1 * V1 = P2 * V2
In this case, you have:
Replacing:
92 kPa* 5.80 L= 101.3 kPa* V2
and solving, you get:
[tex]V2=\frac{92 kPa* 5.80 L}{101.3 kPa}[/tex]
V2= 5.27 L
The volume that the same gas will occupy at 101.3 kPa if the temperature is kept constant is 5.27 L.
