Ideal gas in a cylinder with a piston

In summary: For y much smaller than h, y/h is much smaller than 1. Therefore, $$\frac{1}{(1+\frac{y}{h})}\rightarrow 1$$and $$\frac{y}{h+y}\rightarrow...$$
  • #1
Elias Waranoi
45
2

Homework Statement


A vertical cylinder of radius r contains an ideal gas and is fitted with a piston of mass m that is free to move. The piston and the walls of the cylinder are frictionless, and the entire cylinder is placed in a constant-temperature bath. The outside air pressure is p0. In equilibrium, the piston sits at a height h above the bottom of the cylinder. (a) Find the absolute pressure of the gas trapped below the piston when in equilibrium. (b) The piston is pulled up by a small distance and released. Find the net force acting on the piston when its base is a distance h + y above the bottom of the cylinder, where y << h.

Homework Equations


F = pA = ma
pV = nRT (pressure p, volume V, moles in gas n, gas constant R, temperature T)
V = πr2h

The Attempt at a Solution


I managed to solve a) by myself and get p1 = p0 + mg/πr2

Trying to solve b)
Positive force is upwards so ∑F = F - F0 - Fp where F is the force upwards on the piston by the pressure of the gas underneath, F0 is the downward force on the piston by the air pressure p0 above and Fp is the downward force by the weight of the piston.

F0 = p0A = p0πr2
Fp = mg

moles in ideal gas n = p1V1/RT1 (1)
pressure p2 = nRT2/V2 (2)
Because of heat conduction and the constant temperature outside the cylinder, T2 = T1. Inserting (1) in (2) yields:
p2 = p1V1/V2
V1 = πr2h
V2 = πr2(h + y)
p2 = p1h/(h + y)

∑F = p2A - p0A - mg. A = πr2.
Simplifying yields my answer: ∑F = (h/(h + y) - 1)(mg + p0πr2)
Textbook answer: ∑F = -(y/h)(mg + p0πr2)

I tried replacing the variables with values to see the result of both answers. My answer is not correct apparently, what did I do wrong?
 
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  • #2
Elias Waranoi said:

Homework Statement


A vertical cylinder of radius r contains an ideal gas and is fitted with a piston of mass m that is free to move. The piston and the walls of the cylinder are frictionless, and the entire cylinder is placed in a constant-temperature bath. The outside air pressure is p0. In equilibrium, the piston sits at a height h above the bottom of the cylinder. (a) Find the absolute pressure of the gas trapped below the piston when in equilibrium. (b) The piston is pulled up by a small distance and released. Find the net force acting on the piston when its base is a distance h + y above the bottom of the cylinder, where y << h.

Homework Equations


F = pA = ma
pV = nRT (pressure p, volume V, moles in gas n, gas constant R, temperature T)
V = πr2h

The Attempt at a Solution


I managed to solve a) by myself and get p1 = p0 + mg/πr2

Trying to solve b)
Positive force is upwards so ∑F = F - F0 - Fp where F is the force upwards on the piston by the pressure of the gas underneath, F0 is the downward force on the piston by the air pressure p0 above and Fp is the downward force by the weight of the piston.

F0 = p0A = p0πr2
Fp = mg

moles in ideal gas n = p1V1/RT1 (1)
pressure p2 = nRT2/V2 (2)
Because of heat conduction and the constant temperature outside the cylinder, T2 = T1. Inserting (1) in (2) yields:
p2 = p1V1/V2
V1 = πr2h
V2 = πr2(h + y)
p2 = p1h/(h + y)

∑F = p2A - p0A - mg. A = πr2.
Simplifying yields my answer: ∑F = (h/(h + y) - 1)(mg + p0πr2)
Textbook answer: ∑F = -(y/h)(mg + p0πr2)

I tried replacing the variables with values to see the result of both answers. My answer is not correct apparently, what did I do wrong?
Your answer is correct, but simplify the factor h/(h+y)-1, and use that y<<h.
 
  • #3
I don't even know what y<<h is. And simplifying h/(h+y)-1 gives me (h-y)/(h+y)
 
  • #4
Elias Waranoi said:
I don't even know what y<<h is. And simplifying h/(h+y)-1 gives me (h-y)/(h+y)
Check your math. This is wrong.
 
  • #5
h/(h+y) - 1 = h/(h+y) - (h+y)/(h+y) = (h - (h+y))/(h+y) = (h - h - y)/(h+y) = -y/(h+y)
oops, you're right. But what am I supposed to do now?
 
  • #6
Elias Waranoi said:
h/(h+y) - 1 = h/(h+y) - (h+y)/(h+y) = (h - (h+y))/(h+y) = (h - h - y)/(h+y) = -y/(h+y)
oops, you're right. But what am I supposed to do now?
Dividing numerator and denominator by h yields:
$$\frac{y}{h+y}=\frac{y}{h}\frac{1}{(1+\frac{y}{h})}$$For y much smaller than h, y/h is much smaller than 1. Therefore, $$\frac{1}{(1+\frac{y}{h})}\rightarrow 1$$and $$\frac{y}{h+y}\rightarrow \frac{y}{h}$$
 
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Related to Ideal gas in a cylinder with a piston

1. What is an ideal gas?

An ideal gas is a theoretical gas that follows the ideal gas law, which describes the relationship between pressure, volume, temperature, and number of moles of gas. It assumes that the gas particles have no volume and do not interact with each other.

2. What is a cylinder with a piston?

A cylinder with a piston is a container used to hold a gas or fluid. It consists of a cylindrical tube with a movable piston that can be used to change the volume of the gas or fluid inside.

3. How does a piston affect an ideal gas in a cylinder?

A piston can change the volume of the gas in the cylinder by moving up or down. This affects the pressure and temperature of the gas according to the ideal gas law.

4. What happens to an ideal gas when the piston is pushed down?

When the piston is pushed down, the volume of the gas decreases, which leads to an increase in pressure according to the ideal gas law. The temperature of the gas may also increase if there is no heat exchange with the surroundings.

5. How does the ideal gas law apply to a cylinder with a piston?

The ideal gas law, which states that the product of pressure and volume is proportional to the product of temperature and number of moles of gas, applies to a cylinder with a piston. As the piston moves and changes the volume of the gas, the pressure and temperature also change accordingly.

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