Solve Pressure Problem: Find Plug Position

[bfr]

[SOLVED] Pressure Problem

SOLVED

Homework Statement

A vertical cylindrical tank is 42 cm tall, 2.00 cm in radius, and is open at the top. Atmospheric pressure is 1.013 x 10^5 N / m^2. A close-fitting cylindrical plug of mass 5 kg is inserted at the top, and is let fall inside. If the temperature of the trapped air does not change, how far from the top of the cylinder is the base of the plug when it comes to rest?

(This is the exact problem, word for word)

Homework Equations

(p1*v1)/t1=(p2*v2)/t2

The Attempt at a Solution

Should I compare the pressure caused by the plug to the atmospheric pressure? ((5*9.8*x)/((.02)^2 * pi))((.02)^2 * pi * x)=1.013 * 10^5 * (.02)^2 * pi * .42 ...or, in terms of variables... m*g*h/(pi*r^2) * (pi*r^2*h)=atmospheric pressure * (pi*r^2*height_of_cylinder)

(pi*r^2*height_of_cylinder) is the volume of the cylinder, (pi*r^2*h) is the volume under the plug.

EDIT: I think this is right. Thanks!

 

[Andrew Mason]

Should I compare the pressure caused by the plug to the atmospheric pressure? ((5*9.8*x)/((.02)^2 * pi))((.02)^2 * pi * x)=1.013 * 10^5 * (.02)^2 * pi * .42 ...or, in terms of variables... m*g*h/(pi*r^2) * (pi*r^2*h)=atmospheric pressure * (pi*r^2*height_of_cylinder)

(pi*r^2*height_of_cylinder) is the volume of the cylinder, (pi*r^2*h) is the volume under the plug.

What is the downward force on the plug? What is the upward force on the plug? How are the two related? (ie. does the plug move?). Translate the downward force into pressure and apply the ideal gas law: P1V1 = nRT1 = nRT2 = P2V2

AM

 

[Moetasim]

Based on the given information, the pressure problem can be solved using the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and the number of moles of gas, and inversely proportional to its volume. In this case, the trapped air in the cylinder does not change temperature and is assumed to be an ideal gas.

Using the ideal gas law, we can set up the following equation:

P1*V1 = P2*V2

Where P1 and V1 are the initial pressure and volume of the trapped air, and P2 and V2 are the final pressure and volume of the air after the plug is inserted.

We know that the initial pressure, P1, is equal to the atmospheric pressure, which is given as 1.013 x 10^5 N/m^2. The initial volume, V1, can be calculated using the formula for the volume of a cylinder, V = πr^2h, where r is the radius of the cylinder (0.02 m) and h is the height of the cylinder (0.42 m). Thus, V1 = π(0.02)^2(0.42) = 0.0000264 m^3.

For the final pressure, P2, we can use the same atmospheric pressure, since the air is still trapped in the cylinder. However, the final volume, V2, will change due to the presence of the plug. The volume of the air under the plug can be calculated as the difference between the volume of the cylinder and the volume of the plug. The volume of the plug can be calculated using the formula for the volume of a cylinder, V = πr^2h, where r is the radius of the plug (0.02 m) and h is the height of the plug (x m).

Thus, V2 = V1 - Vplug = π(0.02)^2(x) = 0.0004x m^3.

Substituting these values into the ideal gas law equation, we get:

P1*V1 = P2*V2
(1.013 x 10^5 N/m^2)(0.0000264 m^3) = (1.013 x 10^5 N/m^2)(0.0004x m^3)
0.002698 N = 0.0004x