Air trapped in a cylinder fitted with a piston occupies 162.2 mL at 1.08 atm pressure. What is the new volume when the piston is depressed, increasing the pressure by 25%?
assuming a constant temperature,
PV = kT is constant.
So, with 5/4 the pressure, you get 4/5 the volume
i dont understand
You can do also by using
P1V1 = P2V2
P1 = 1.08 atm and P2 = 1.08+0.25*1.08 = ?
V1 is given. Solve for V2
116.56 ml
To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature and the amount of gas are constant:
P₁V₁ = P₂V₂
where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.
In this case, we are given P₁ (1.08 atm), V₁ (162.2 mL), and the change (increase) in pressure (25%). We need to find V₂, the new volume.
First, let's convert the initial volume from mL to liters, as the pressure is given in atm:
V₁ = 162.2 mL = 162.2/1000 L = 0.1622 L
Next, let's calculate the final pressure:
P₂ = P₁ + (P₁ * 25%) = 1.08 atm + (1.08 atm * 0.25) = 1.08 atm + 0.27 atm = 1.35 atm
Now, we can substitute the values into Boyle's Law equation:
P₁V₁ = P₂V₂
(1.08 atm) * (0.1622 L) = (1.35 atm) * V₂
0.174936 L atm = 1.35 atm * V₂
Dividing both sides of the equation by 1.35 atm:
V₂ = 0.174936 L atm / 1.35 atm
V₂ ≈ 0.13 L
Therefore, the new volume when the piston is depressed, increasing the pressure by 25%, is approximately 0.13 liters.