Calculating Piston Height in Equilibrium for an Ideal Gas System

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In summary, the conversation discusses a problem involving a piston pressing down on an enclosed gas and the goal of finding the height of the piston when the system is in equilibrium. The conversation also touches on the use of the equation PV = nRT and the calculation of the necessary pressure. However, there is some confusion about the given height of the cylinder and the resulting answer seems to be unrealistic. The conversation ends with a request for more data to solve the problem accurately.
  • #1
winhog
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I'm having some trouble with a question involving a piston pressing down on an enclosed gas. I'm given the height of the cylinder, the number of moles of gas, the mass of the piston, and told that initially the gas is at STP. I need to find the height of the piston when the system is in equilibrium.

I thought I knew how to do it, using PV = nRT. I calculated the area of the piston at (about) 1 square meter, so I decided the pressure pushing up should be about the piston's mass * acceleration due to gravity, correct?

Since the mass of the piston is only 1.4 kg, I get about 14 Pa as the necessary pressure. But this is only an extremely small fraction of the original pressure of 1 atm, so the height difference seems negligible to me. Am I missing something, or am I going to be giving my answer in millimeters?
 
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  • #2
winhog said:
I'm having some trouble with a question involving a piston pressing down on an enclosed gas. I'm given the height of the cylinder, the number of moles of gas, the mass of the piston, and told that initially the gas is at STP. I need to find the height of the piston when the system is in equilibrium.

I thought I knew how to do it, using PV = nRT. I calculated the area of the piston at (about) 1 square meter, so I decided the pressure pushing up should be about the piston's mass * acceleration due to gravity, correct?

Since the mass of the piston is only 1.4 kg, I get about 14 Pa as the necessary pressure. But this is only an extremely small fraction of the original pressure of 1 atm, so the height difference seems negligible to me. Am I missing something, or am I going to be giving my answer in millimeters?
The system will be in equilibrium when the pressure of the gas multiplied by the area of the piston is equal to the weight of the piston. So:

PA = mg = nRTA/V = nRT/h where h = height of piston from end of cylinder
h=nRT/mg

AM
 
  • #3
Thanks very much!

Though, it seems odd that the initial height of the cylinder they gave me was useless...wasn't it? :confused:
 
  • #4
Sorry, I'm still having trouble. Using your formula, with m = 1.4 kg, T = 273 K, n = 0.1 mol

h = (.1)(8.314)(273) / ((1.4)(9.8)) = 16.5 meters.

The units all work out and everything, but the given height of the cylinder was only 2.4 meters. I know the piston would not shoot up to 16.5 meters, or go down to the floor...what am I doing wrong? Thanks in advance for any help.
 
  • #5
winhog said:
Sorry, I'm still having trouble. Using your formula, with m = 1.4 kg, T = 273 K, n = 0.1 mol

h = (.1)(8.314)(273) / ((1.4)(9.8)) = 16.5 meters.

The units all work out and everything, but the given height of the cylinder was only 2.4 meters. I know the piston would not shoot up to 16.5 meters, or go down to the floor...what am I doing wrong? Thanks in advance for any help.
In my expression I omitted the atmospheric pressure Pa. It should be:

PA = mg + PaA = nRTA/V = nRT/h where h = height of piston from end of cylinder
h=nRT/(mg + PaA)

Try that.

AM
 
  • #6
Hmmm, I'm back to my initial problem then, where my answer comes out to something like 0.2 millimeters. I guess it makes some sense since .1 moles of gas can't hold much up unless very compressed, but I've never been asked a problem with such a ridiculous answer from this book before.
 
  • #7
winhog said:
Hmmm, I'm back to my initial problem then, where my answer comes out to something like 0.2 millimeters. I guess it makes some sense since .1 moles of gas can't hold much up unless very compressed, but I've never been asked a problem with such a ridiculous answer from this book before.
You will have to give me all the data. I am missing area A, height of cylinder, h.

AM
 
  • #8
They give height = 2.4 meters, n = 0.1 mol, mass = 1.4 kg, and that the gas is initially at STP. I calculated the area at 0.93 m^2.
 
  • #9
winhog said:
They give height = 2.4 meters, n = 0.1 mol, mass = 1.4 kg, and that the gas is initially at STP. I calculated the area at 0.93 m^2.
That seems like a huge piston. What is the diameter of the piston?
 

1. What is the ideal gas law?

The ideal gas law is a fundamental equation in thermodynamics that describes the relationship between the pressure, volume, and temperature of an ideal gas. It is written as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

2. What is the difference between an ideal gas and a real gas?

An ideal gas is a theoretical concept that follows the ideal gas law exactly under all conditions. Real gases, on the other hand, deviate from the ideal gas law at high pressures and low temperatures due to intermolecular forces and the finite size of gas molecules.

3. How does a piston affect the behavior of an ideal gas?

In an ideal gas/piston problem, the piston acts as a boundary and can change the volume of the gas. When the piston moves, it either compresses or expands the gas, causing a change in pressure and temperature. The ideal gas law can be used to calculate these changes.

4. How do you solve ideal gas/piston problems?

To solve an ideal gas/piston problem, you first need to identify the known variables (pressure, volume, temperature, and number of moles) and the unknown variable. Then, use the ideal gas law to create an equation with these variables. Finally, solve for the unknown variable using algebraic manipulation.

5. Can the ideal gas law be applied to all gases?

The ideal gas law is a good approximation for most gases at low pressures and high temperatures. However, it becomes less accurate for real gases at high pressures and low temperatures. This is because real gases behave differently depending on their chemical composition and intermolecular forces.

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