A fixed amount of CH4 gas occupies a volume of 136 mL at a pressure of 567 mm Hg and a temperature of 19oC. What will be the volume, in mL, occupied by this sample of CH4 gas when the pressure is decreased to 421 mm Hg while holding the temperature constant?
P1V1 = P2V2. You can use this because T is constant. However, since T is given, you can use
(P1V1/T1) = (P2V2/T2) and come out with the same answer.
(567*136/19c)=(421*183.16/19c)
=4,058.5 mL
Is this correct?
No. I see P1 of 567 and V1 of 136. On the right I see 421 and all of those are in the right place. Where did the 183 come from? That's what you're solving for and is the answer. The way you have it set up there is no unknown and I have no idea how you ended up with 4,058.
It should be
(567*136) = (421*V2)
Solve for V2.
V2 = 183 mL.
To solve this problem, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure, as long as the temperature remains constant. Boyle's Law can be written as:
P1 * V1 = P2 * V2
where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
In this problem, we are given the initial pressure (P1 = 567 mm Hg), initial volume (V1 = 136 mL), final pressure (P2 = 421 mm Hg), and we need to find the final volume (V2).
Let's plug the given values into the equation and solve for V2:
567 mm Hg * 136 mL = 421 mm Hg * V2
(567 mm Hg * 136 mL) / 421 mm Hg = V2
V2 ≈ 183.6 mL
Therefore, the volume occupied by the CH4 gas when the pressure is decreased to 421 mm Hg, while holding the temperature constant, is approximately 183.6 mL.