Ideal gas in a cylinder with a piston

[Elias Waranoi]

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 - F_p 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 F_p is the downward force by the weight of the piston.

F₀ = p₀A = p₀πr²
F_p = 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?

 

[ehild]

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 - F_p 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 F_p is the downward force by the weight of the piston.

F₀ = p₀A = p₀πr²
F_p = 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.

 

[Elias Waranoi]

I don't even know what y<<h is. And simplifying h/(h+y)-1 gives me (h-y)/(h+y)

 

[Chestermiller]

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.

 

[Elias Waranoi]

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?

 

[Chestermiller]

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 $$\frac{y}{h+y}\rightarrow \frac{y}{h}$$