Expansion of a ideal gas

In summary, the conversation is discussing a physics problem involving a cylinder filled with gas, a piston, and a spring. The temperature and volume of the gas are changing, and the goal is to determine how high the piston will be raised. The equations used to solve the problem include the ideal gas law, the equation for work done by the gas, and the equation for equilibrium of forces. The conversation also mentions integrating and finding the change in potential energy of the spring.
  • #1
Bob19
71
0
Hi

I got the following physics problem:

A cylinder closed by a piston contains 5,00 L of an ideal gass at a pressure of 1,00 amt at a temperature of 20,0 degrees celcius.

The piston has a cross-centential area of 0,0100 m^2. The mass of the piston is neglible.
The piston is connected to a spring which is relaxed and has a k-value of 2,00 * 10^2 N/m.

If the temperature of the gass is raised from 20 degree's celcius to 250 degree's celcius.

How high the will piston by raised from its current position ?

I would very much apriciate if there is anybody out there who can provide me with a hint on howto calculate this.

Best Regards,

Bob
 
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  • #2
The temparature changes, so you will need to find out the pressure afterwards. What equation should you use?
You may notice however that the new volume is unknown. You need to know the work done by the gas.
What other equation should you use?
 
  • #3
Hi Berislav and thanks for Your answer,

The equation You are referring must be:

[tex]\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_{f}}[/tex]

On can imagine that [tex]P_{f} V_{f}}{[/tex] plays a part in calculating the new position of piston, but how is this connected to the cross-centinial of the piston and the spring constant ?

/Bob

Berislav said:
The temparature changes, so you will need to find out the pressure afterwards. What equation should you use?
You may notice however that the new volume is unknown. You need to know the work done by the gas.
What other equation should you use?
 
  • #4
On can imagine that [snip] plays a part in calculating the new position of piston, but how is this connected to the cross-centinial of the piston and the spring constant ?
Through the equation for work done by the gas.
 
  • #5
Maybe I should elaborate. When you write down the equation for work done by anything you will have an integral. Use the fact that:
[tex]F(x)=p(x)A[/tex]
Integrate and eliminate the unknowns via the first equation. The work done should be equal to the change in potential energy of the spring.
 
  • #6
Hello I have a theory here:

The volume of the gas can increase by pushing the piston partway out of the cylinder. The amount of work done is equal to the product of the force exerted on the piston times the distance the piston is moved.

w = F x d

Secondly:

The pressure (P) the gas exerts on the piston is equal to the force (F) with which it pushes up on the piston divided by the surface area (A) of the piston.

P = F/A

Thirdly:

Thus, the force exerted by the gas is equal to the product of its pressure times the surface area of the piston.

F = P x A

Finally:

Substituting this expression into the equation defining work gives the following result.

w = (P x A) x d


Where as You write the work done is equal to th change in potential energy in the spring.

Am I on the right track?

Best Regards,

Bob

Berislav said:
Maybe I should elaborate. When you write down the equation for work done by anything you will have an integral. Use the fact that:
[tex]F(x)=p(x)A[/tex]
Integrate and eliminate the unknowns via the first equation. The work done should be equal to the change in potential energy of the spring.
 
  • #7
But pressure is not constant. Use a more general formula for work:

[tex]W=\int_{PATH}{F(x)dx}[/tex]
 
Last edited:
  • #8
Hello

By path do You mean the change in position of the piston ?

Best Regards,

Bob

Berislav said:
But pressure is not constant. Use a more general formula for work:

[tex]W=\int_{PATH}{F(x)dx}[/tex]
 
  • #9
You can find out how much the spring has extended without calculating the work done.

Initially, the Pressure is 1.0 atm, the Volume = (Area)(length). (Since the Volume and area are given, the length can be found, after converting into proper units). The Temperature is 293K
Let the spring extend by x.

The new Volume=(Area)(length + x)
The new Temperature is also given. So, the only thing you need to find is the value of new pressure in terms of x. But you also know that the piston is in equilibirium. From that can you find the answer?
 
  • #10
Thank You,

I will try this out :-)

Best Regards,

/Fred

siddharth said:
You can find out how much the spring has extended without calculating the work done.

Initially, the Pressure is 1.0 atm, the Volume = (Area)(length). (Since the Volume and area are given, the length can be found, after converting into proper units). The Temperature is 293K
Let the spring extend by x.

The new Volume=(Area)(length + x)
The new Temperature is also given. So, the only thing you need to find is the value of new pressure in terms of x. But you also know that the piston is in equilibirium. From that can you find the answer?
 
  • #11
The new Temperature is also given. So, the only thing you need to find is the value of new pressure in terms of x. But you also know that the piston is in equilibirium. From that can you find the answer?

Are you certain that's the correct approach? I don't think they're equivalent.
 
  • #12
Hello Berislav,

I have tried that approah but I didn't get a result that I could use.

Looking at Your surgestion.

Which function F(x) do I integrate over to find the work done by the gas ?

Best Regards,

Bob


Berislav said:
Are you certain that's the correct approach? I don't think they're equivalent.
 
  • #13
Which function F(x) do I integrate over to find the work done by the gas ?
Over force, limits being 0 to final height of the piston (with respect to the initial one).
Force is:
[tex]F(x)=p(x)A[/tex]
[tex]F(x)=\frac{p_i V_i T_f}{T_i V_f} A[/tex]
[tex]F(x)=\frac{p_i A h_i T_f}{T_i (h_i+x)}[/tex]

Equilibrium of forces should be a necessary condition, but IIRC F=kx for a spring isn't applicable if the LHS of the equation isn't a constant.
 
  • #14
Berislav said:
Are you certain that's the correct approach? I don't think they're equivalent.

Initially the pressure is 1 atm (P_0), Volume is 5 L, and the temperature is 293 K.

After expansion of the piston, at equilibirium, the forces acting on the piston upwards are, (P_1)A where P_1 is the new pressure inside the piston, and the forces acting downwards are (P_0)A + kx where P_0 is the atmospheric pressure outside the piston.
So we have
[tex] P_1A = P_0A + kx [/tex]

[tex] P_1 = P_0 + \frac{kx}{A} [/tex]

And the new volume is [tex] A(l + x) [/tex]

So use [tex] \frac{P_0V_0}{T_0} = \frac{P_1V_1}{T_1} [/tex]

[tex] \frac{(1)(.005)}{293} = \frac{[(1)(.01) + 20000x][(.01)(0.5 + x)]}{523} [/tex]

From this, we can solve for x.
 
  • #15
Hi

I have played around with this problem myself and then ended the following solution (please correct me if there are any mistakes)

[tex]V_{f} = V_{i} + Ah[/tex]

Looking at the equation:

[tex]\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_f}[/tex]

Because of the force exerted by the spring connected to the piston:

then

[tex]P_{f} = \frac{k h}{A}[/tex]

The resulting gass equation is then:

[tex]\frac{P_{i} V_{i}}{T_{i}} = \frac{kh}{A T_{f}} (V_{i} + Ah)[/tex]

Finally I insert the the given values for [tex]P_{i}, V_{i}, k, A, T_{f}[/tex]

This yields the equation:

[tex]\frac{101 \cdot 10^{3} \ * 5.00 \cdot 10^{-3}}{293} = \frac{ 2 \cdot 10^{3}h (5.00 \cdot 10^{-3} + 0.01h)}{0.01 * 523} [/tex]

and then solve the above equation for "h" to obtain the new hight!

Is this approach correct ?

Sincerely and Best Regards,

Bob

siddharth said:
Initially the pressure is 1 atm (P_0), Volume is 5 L, and the temperature is 293 K.

After expansion of the piston, at equilibirium, the forces acting on the piston upwards are, (P_1)A where P_1 is the new pressure inside the piston, and the forces acting downwards are (P_0)A + kx where P_0 is the atmospheric pressure outside the piston.
So we have
[tex] P_1A = P_0A + kx [/tex]

[tex] P_1 = P_0 + \frac{kx}{A} [/tex]

And the new volume is [tex] A(l + x) [/tex]

So use [tex] \frac{P_0V_0}{T_0} = \frac{P_1V_1}{T_1} [/tex]

[tex] \frac{(1)(.005)}{293} = \frac{[(1)(.01) + 20000x][(.01)(0.5 + x)]}{523} [/tex]

From this, we can solve for x.
 

1. What is the ideal gas law?

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

2. How does an ideal gas expand?

An ideal gas expands when its volume increases due to a decrease in pressure or an increase in temperature. This is because an ideal gas follows the ideal gas law, which states that as volume increases, pressure decreases and vice versa, assuming constant temperature and number of moles.

3. What is the difference between isothermal and adiabatic expansion of an ideal gas?

Isothermal expansion of an ideal gas occurs at a constant temperature, meaning that the change in internal energy is equal to zero. Adiabatic expansion, on the other hand, occurs without the transfer of heat, so the change in internal energy is also equal to zero. The main difference is that isothermal expansion occurs at a constant temperature, while adiabatic expansion can occur at any temperature.

4. How does the expansion of an ideal gas relate to the laws of thermodynamics?

The expansion of an ideal gas is governed by the first and second laws of thermodynamics. The first law, also known as the law of conservation of energy, states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. The second law states that in a closed system, entropy (disorder) always increases. In the expansion of an ideal gas, the system is doing work on its surroundings, leading to a decrease in internal energy and an increase in entropy.

5. What are some real-life examples of the expansion of an ideal gas?

Some real-life examples of the expansion of an ideal gas include the inflation of a balloon, the expansion of a car tire when heated, and the release of gas from a pressurized aerosol can. These examples demonstrate the relationship between pressure, volume, and temperature in an ideal gas and how changes in these variables can cause expansion or contraction of the gas.

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