Volume-pressure equation for an expanding ideal gas

In summary, the problem is that the equation between pressure and volume does not hold when atmospheric pressure is not present.
  • #1
MatinSAR
522
174
Homework Statement
How do I find the volume-pressure equation for an expanding ideal gas for the figure below?
Relevant Equations
pV=nRT
There is no atmosphere pressure.
1674383660123.png


My work :
pA=k(x-x0) => pA=(k/A)(V-V0)

But this should be false beccause I want to use W=∫PdV to find work done by the gas but my final anwer is wrong ...
Please guide me where my mistake is if you have enough time. Thanks.
 
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  • #2
Please show more details of what you did. Is this expansion by adding heat at constant temperature? Please provide an exact word-for-word statement of the problem.
 
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  • #3
Chestermiller said:
Is this expansion by adding heat at constant temperature?
It just said that we heat the gas slowly.
Chestermiller said:
Please provide an exact word-for-word statement of the problem.
We heat the gas slowly so that it goes from (P1,V1) to (P2,V2). What is the work done on the gas?
We can use Hook's law for the spring. Piston is light(=forget mass of it).

Chestermiller said:
Please show more details of what you did.
I want to find the equation between pressure and volume.
And I don't know why this is wrong :
pA=k(x-x1) => pA=(k/A)(V-V1)
 
  • #4
MatinSAR said:
pA=(k/A)(V-V1)
I think I have founded the problem. It should be (p-p1)A=(k/A)(V-V1).
Is it true?
 
  • #5
MatinSAR said:
I think I have founded the problem. It should be (p-p1)A=(k/A)(V-V1).
Is it true?
yes
 
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  • #6
Chestermiller said:
yes
Thank you for your help and time.
 
  • #7
1674403349213.png


I'm having some doubts on this one?

Ignoring atmospheric pressure on the spring side seems like bit of blunder IMO. The spring would initially have to be compressed at ##x_o##.Bypassing that issue, I think it's understood to be in a state of "quasistatic equilibrium" this implies that the forces are balanced across the piston.

$$ - k ( x - x_o ) + F_g = 0 $$

$$ \implies P_x = \frac{k}{A}( x - x_o )$$

you can see that the absolute pressure in the gas must be 0 at ##x = x_o## in the absence of atmospheric pressure...is that really ok? Don't we have to be at absolute zero in temperature initially for this to be possible? can we have a gas at absolute 0 temp? Some of the question I have.

In terms of volume ## V\llap{-}##, just multiply through by ##A##:

$$ P_x A = \frac{k}{A}( V\llap{-}_x - V\llap{-}_o )$$

$$ \boxed{ P_x = \frac{k}{A^2}( V\llap{-}_x - V\llap{-}_o ) }$$

What am I missing?
 
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  • #8
erobz said:
View attachment 320900

I'm having some doubts on this one?

Ignoring atmospheric pressure on the spring side seems like bit of blunder IMO. The spring would initially have to be compressed at ##x_o##.Bypassing that issue, I think it's understood to be in a state of "quasistatic equilibrium" this implies that the forces are balanced across the piston.

$$ - k ( x - x_o ) + F_g = 0 $$

$$ \implies P_x = \frac{k}{A}( x - x_o )$$
In terms of volume ## V\llap{-}##, just multiply through by ##A##:

$$ P_x A = \frac{k}{A}( V\llap{-}_x - V\llap{-}_o )$$

$$ \boxed{ P_x = \frac{k}{A^2}( V\llap{-}_x - V\llap{-}_o ) }$$

What am I missing?
In my judgment, the confusion with this problem is yours.

The problem statement says that the atmospheric pressure is zero. This is the same as saying that the device is contained within a vacuum chamber. Who says that this is not allowed?

If the atmospheric pressure is zero, this means that the spring is under preload compression so that the forces on both sides of the piston are initially equal. Therefore, $$P_1A=A\frac{nRT_1}{V_1}=\frac{k}{A}(V_1-V_0)$$where ##V_0## would be the volume at which the sprig is unextended. At all other times, the forces on both sides of the piston also match: $$P=\frac{nRT}{V}=\frac{nRT_1}{V_1}+\frac{k}{A^2}(V-V_1)$$

So what's the problem?
 
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  • #9
Chestermiller said:
In my judgment, the confusion with this problem is yours.
Quite likely.
Chestermiller said:
So what's the problem?
One thing, if this is happening in the vacuum of space we have to have a gas at some initial volume ## V\llap{-}_o## ( zero pre-compressed spring state) with absolute zero temperature (with regards to the ideal gas law).
 
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  • #10
erobz said:
Quite likely.

One thing, if this is happening in the vacuum of space we have to have a gas at some initial volume ## V\llap{-}_o## ( zero pre-compressed spring state) with absolute zero temperature (with regards to the ideal gas law).
No. Vo is the chamber volume at which the spring is unextended. This is geometric, and independent of the gas.
 
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  • #11
Chestermiller said:
No. Vo is the chamber volume at which the spring is unextended. This is geometric, and independent of the gas.
Ahh, there is no gas in the chamber when ##V\llap{-} = V\llap{-}_o##. I guess that is the plausible alternative I was missing.

Thank You.
 
Last edited:

1. What is the volume-pressure equation for an expanding ideal gas?

The volume-pressure equation for an expanding ideal gas is known as the ideal gas law, which states that the product of the pressure and the volume of a gas is proportional to the number of moles of the gas and its absolute temperature. It can be written as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the absolute temperature.

2. How does the volume-pressure equation apply to real gases?

The volume-pressure equation is an approximation for real gases at low pressures and high temperatures. At higher pressures and lower temperatures, real gases deviate from ideal behavior due to intermolecular forces and the finite size of gas molecules. In these cases, more complex equations, such as the van der Waals equation, are used to describe the behavior of real gases.

3. What is the significance of the volume-pressure equation for gas laws?

The volume-pressure equation is a fundamental equation in gas laws, which describe the behavior of gases under different conditions. It helps us understand how changes in pressure, volume, temperature, and number of moles affect the properties of gases. The equation is also used in many practical applications, such as in the design and operation of gas cylinders and engines.

4. How does the volume-pressure equation change for a gas at constant temperature?

When the temperature of a gas is held constant, the volume-pressure equation becomes Boyle's law, which states that the pressure and volume of a gas are inversely proportional to each other. This means that as one variable increases, the other decreases, and vice versa. Mathematically, this can be written as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

5. How does the volume-pressure equation relate to the kinetic theory of gases?

The volume-pressure equation is derived from the kinetic theory of gases, which states that gases consist of particles in constant motion and that the pressure of a gas is due to the collisions of these particles with the walls of the container. The equation relates the macroscopic properties of gases, such as pressure and volume, to the microscopic behavior of gas molecules, such as their speed and number. This allows us to make predictions about the behavior of gases based on their underlying molecular properties.

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