Air at 1 atm is inside a cylinder 17.9 cm in radius and 17.9 cm in length that sits on a table. The top of the cylinder is sealed with a movable piston. A 24.1-kg block is dropped onto the piston.
a) From what height above the piston must the block be dropped to compress the piston by 1.30 mm?
b) From what height above the piston must the block be dropped to compress the piston by 2.60 mm?
c) From what height above the piston must the block be dropped to compress the piston by 1.30 cm?
To solve these problems, we need to use the principles of Pascal's Law and the ideal gas law.
Pascal's Law states that when pressure is applied to a fluid (in this case, the air inside the cylinder), the pressure is transmitted equally in all directions. Therefore, the pressure applied by the block is transmitted to the air inside the cylinder.
The ideal gas law relates the pressure, volume, and temperature of a gas. In this problem, we assume that the temperature remains constant, so we can use the simplified version of the ideal gas law:
P₁V₁ = P₂V₂
where P₁ is the initial pressure, V₁ is the initial volume, P₂ is the final pressure, and V₂ is the final volume.
Now let's solve each part of the problem:
a) To find the height from which the block must be dropped to compress the piston by 1.30 mm, we need to calculate the initial pressure inside the cylinder.
First, we convert the piston's compression from millimeters to meters:
1.30 mm = 1.30 x 10^(-3) m
Given that the cylinder has a radius of 17.9 cm, we can find the initial volume V₁.
V₁ = πr₁²h₁
where r₁ is the radius and h₁ is the initial height of the air column.
Since the piston is sealed, the height of the air column remains constant, so h₁ = 17.9 cm = 0.179 m.
Now, we can use the ideal gas law to find the initial pressure P₁.
P₁V₁ = P₂V₂
Since the air inside the cylinder is at 1 atm of pressure, P₁ = 1 atm = 1.01 x 10^5 Pa.
We already know the initial volume V₁ and we can find the final volume V₂, knowing that the compression distance is 1.30 x 10^(-3) m.
V₁ = πr₁²h₁
V₂ = πr₂²h₂
where r₂ is the radius and h₂ is the final height of the air column after compression.
Since the radius of the cylinder remains the same, r₂ = r₁.
Using this information, along with the equation for volume and the given compression distance, we can solve for h₂.
V₁ = V₂ + Ah₂
where A is the cross-sectional area of the cylinder.
A = πr₁²
V₁ = πr₁²h₁
V₂ = πr₁²h₂
πr₁²h₁ = πr₁²h₂ + Ah₂
πr₁²h₁ = πr₁²(h₂ + h₂)
πr₁²h₁ = πr₁²(2h₂)
Canceling out πr₁² from both sides:
h₁ = 2h₂
Substituting the values we know:
0.179 m = 2h₂
Solving for h₂:
h₂ = (0.179 m) / 2 = 0.0895 m
So, the block must be dropped from a height of 0.0895 m above the piston to compress it by 1.30 mm.
b) To find the height from which the block must be dropped to compress the piston by 2.60 mm, we can follow similar steps as in part a.
Convert the compression distance from millimeters to meters:
2.60 mm = 2.60 x 10^(-3) m
Using the same formula as before:
h₁ = 2h₂
Substituting the given values:
0.179 m = 2h₂
Solving for h₂:
h₂ = (0.179 m) / 2 = 0.0895 m
So, the block must be dropped from a height of 0.0895 m above the piston to compress it by 2.60 mm.
c) To find the height from which the block must be dropped to compress the piston by 1.30 cm, we use the same steps as in parts a and b.
Convert the compression distance from centimeters to meters:
1.30 cm = 1.30 x 10^(-2) m
Using the formula:
h₁ = 2h₂
Substituting the values:
0.179 m = 2h₂
Solving for h₂:
h₂ = (0.179 m) / 2 = 0.0895 m
So, the block must be dropped from a height of 0.0895 m above the piston to compress it by 1.30 cm.