What happens to a 4L volume of gas as you change the pressure from 6 atm to 760 torr? What law was used, and show work
you change pressure by a factor of 6 (760torr is one atm).
volume increases by a factor of six.
V1P1=V2P2
V2=V1(P1/P2)=4L*6/1
To understand what happens to a 4L volume of gas as you change the pressure from 6 atm to 760 torr, we can use Boyle's Law. Boyle's Law states that at constant temperature, the pressure and volume of a gas are inversely proportional.
The equation for Boyle's Law is:
P1 * V1 = P2 * V2
where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.
In this case, the initial pressure (P1) is 6 atm, the initial volume (V1) is 4L, the final pressure (P2) is 760 torr, and we need to find the final volume (V2).
To solve for V2, we rearrange the equation:
V2 = (P1 * V1) / P2
Now we can plug in the values:
V2 = (6 atm * 4L) / 760 torr
To be consistent, we need to convert atm to torr:
1 atm = 760 torr, so 6 atm = 6 * 760 torr = 4560 torr
V2 = (4560 torr * 4L) / 760 torr
Simplifying:
V2 = 9120 L / 760 torr
V2 ≈ 12 L
Therefore, as you change the pressure from 6 atm to 760 torr, with a constant volume of 4L, the final volume (V2) would be approximately 12L.
In summary, using Boyle's Law, the volume of the gas increased from 4L to approximately 12L as the pressure changed from 6 atm to 760 torr.