# Expansion of a ideal gas

#### Bob19

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

Related Introductory Physics Homework Help News on Phys.org

#### Berislav

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?

#### Bob19

The equation You are refering must be:

$$\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_{f}}$$

On can imagine that $$P_{f} V_{f}}{$$ 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?

#### Berislav

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.

#### Berislav

Maybe I should elaborate. When you write down the equation for work done by anything you will have an integral. Use the fact that:
$$F(x)=p(x)A$$
Integrate and eliminate the unknowns via the first equation. The work done should be equal to the change in potential energy of the spring.

#### Bob19

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:
$$F(x)=p(x)A$$
Integrate and eliminate the unknowns via the first equation. The work done should be equal to the change in potential energy of the spring.

#### Berislav

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

$$W=\int_{PATH}{F(x)dx}$$

Last edited:

#### Bob19

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:

$$W=\int_{PATH}{F(x)dx}$$

#### siddharth

Homework Helper
Gold Member
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?

#### Bob19

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?

#### Berislav

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.

#### Bob19

Hello Berislav,

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

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.

#### Berislav

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:
$$F(x)=p(x)A$$
$$F(x)=\frac{p_i V_i T_f}{T_i V_f} A$$
$$F(x)=\frac{p_i A h_i T_f}{T_i (h_i+x)}$$

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.

#### siddharth

Homework Helper
Gold Member
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
$$P_1A = P_0A + kx$$

$$P_1 = P_0 + \frac{kx}{A}$$

And the new volume is $$A(l + x)$$

So use $$\frac{P_0V_0}{T_0} = \frac{P_1V_1}{T_1}$$

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

From this, we can solve for x.

#### Bob19

Hi

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

$$V_{f} = V_{i} + Ah$$

Looking at the equation:

$$\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_f}$$

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

then

$$P_{f} = \frac{k h}{A}$$

The resulting gass equation is then:

$$\frac{P_{i} V_{i}}{T_{i}} = \frac{kh}{A T_{f}} (V_{i} + Ah)$$

Finally I insert the the given values for $$P_{i}, V_{i}, k, A, T_{f}$$

This yields the equation:

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

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

Is this aproach 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
$$P_1A = P_0A + kx$$

$$P_1 = P_0 + \frac{kx}{A}$$

And the new volume is $$A(l + x)$$

So use $$\frac{P_0V_0}{T_0} = \frac{P_1V_1}{T_1}$$

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

From this, we can solve for x.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving