A 5-kg piston in a cylinder with a diameter of 100 mm is loaded with a linear spring and the ou

A 5-kg piston in a cylinder with a diameter of 100 mm is...

A $5-\mathrm{kg}$ piston in a cylinder with a diameter of $100 \mathrm{~mm}$ is loaded with a linear spring and the outside atmospheric pressure is $100 \mathrm{kPa}$, as shown in Fig. $\mathrm{P} 1.101 .$ The spring exerts no force on the piston when it is at the bottom of the cylinder, and for the state shown, the pressure is $400 \mathrm{kPa}$ with volume $0.4 \mathrm{~L}$. The valve is opened to let some air in, causing the piston to rise $2 \mathrm{~cm}$. Find the new pressure.

Key Concepts

Geometric Relationships in Cylindrical Systems

In a cylindrical piston system, the geometry—specifically the relation between the piston’s cross-sectional area and its displacement—is important for connecting changes in volume with the physical movement of the piston. Knowing the area and the displacement facilitates the calculation of the change in system volume, which together with the gas laws, helps determine the resulting pressure change in the cylinder.

Hooke’s Law

Hooke’s Law describes how the force exerted by a linear spring is proportional to its displacement from the natural (unstressed) length. When the piston moves away from its initial position, the spring compresses or extends, thereby adding an extra force that must be accounted for in the equilibrium analysis. The linear relationship between spring force and displacement is key to quantifying the additional load the spring introduces as the piston moves.

Ideal Gas Law

The ideal gas law relates the pressure, volume, and temperature of a gas (PV = nRT) and is a fundamental concept when dealing with systems where the gas state changes. In problems involving pistons and change in volume, assuming a particular thermodynamic process (such as isothermal) allows for the prediction of how the pressure will adjust when the volume changes due to piston movement, especially when additional gas is allowed into the system.

Static Force Equilibrium

This concept involves setting the net force on the piston to zero when the system is in balance. In such a system, the upward force from the gas pressure (acting over the piston’s area) balances the downward forces from the atmospheric pressure, the weight of the piston, and any additional forces (such as from a compressed or extended spring). Recognizing and applying this balance is essential in solving problems where the piston position changes due to adjustments in pressure or added forces.

Pressure?Area Relationship

Understanding that force exerted by a fluid is related to its pressure and the area over which it acts is crucial. The force is computed as the product of the pressure difference across the piston and the piston’s cross?sectional area. This relationship directly ties pressure changes in the fluid to the mechanical response of the piston, facilitating the analysis of how variations in pressure will affect the piston’s position.

Transcript

00:01 So we want to write an expression that relates pressure to volume so we can find out by how much the pressure is changed when we move this piston up and down by pushing an air to this volume here.

00:15 So because we have a linear spring here and we know that the force will be proportional to the displacement, we can write that the equilibrium pressure, so in the piston, and it's at rest and we have a pressure inside this volume here.

00:37 We'll similarly be linearly proportional to the volume.

00:41 So we have an expression like this.

00:43 So if you think about graphing the pressure with respect to volume, we want to fill out this equation so that we find a value for a, which is the intersect, and for b, which would simply be the slope of this line.

01:01 So to solve for a first, what we're going to do is we're going to balance the forces when the piston is at rest on the ground, on the bottom of the cylinder, so that would mean that the volume is zero.

01:13 So this would mean that we have an expression that looks like this.

01:26 And we get this expression by setting up the forces and then solving for p.

01:32 So when we read the forces in terms of the pressure on either side of the piston and the gravitational force acting on the piston.

01:41 And for the area, we're given the diameter in millimeters, but we're going to go ahead convert to meters.

01:48 And then what we're going to do is use this expression to solve for the area, pi times d squared over 4, and that gives us 0 .0075 meters squared.

02:05 So that's our value for a.

02:07 And we plug in and solve this calculation right here.

02:12 What we get is 106 .2 kilopascals.

02:20 So if we look back on our graph, this point here would be 106 .2 on the y -axis...

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