A student collects 400 mL of oxygen at 9.80 kPa. If the temperature remains
constant, what volume would this gas occupy at 9.40 kPa?
V1/T1 = V2/T2
T1 and T2 must be in Kelvin.
417m
To solve this problem, we can apply Boyle's Law, which states that for a given amount of gas at a constant temperature, the pressure and volume are inversely proportional. Boyle's Law can be mathematically represented as P1V1 = P2V2, where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the final pressure and volume.
In this case, we are given the initial pressure (P1 = 9.80 kPa) and volume (V1 = 400 mL) of the gas. We need to find the final volume (V2) when the pressure changes to 9.40 kPa while keeping the temperature constant.
Let's plug in the known values into the Boyle's Law equation:
P1V1 = P2V2
(9.80 kPa)(400 mL) = (9.40 kPa)(V2)
Now, we can solve for V2 by rearranging the equation and dividing both sides by 9.40 kPa:
V2 = (9.80 kPa)(400 mL) / (9.40 kPa)
V2 = (9.80 kPa * mL) / (9.40 kPa)
V2 = 416.6 mL
Therefore, the gas would occupy a volume of 416.6 mL at 9.40 kPa, assuming the temperature remains constant.