How far does the piston move after energy is added?

In summary, the pressure in the first canister increased by 1.6E5 Pa, and the final volume of the gas is (.25)^3 m^3.
  • #1
sliperyfrog
27
0

Homework Statement


There are two rectangular canisters, both have a .25m cross sectional area that house 4 g of helium at 300 kelvin. The pistons are initially placed so that the enclosed volume is in a cube. The first canister's piston is locked in place, while the second canister is allowed to move freely up or down. 1000 J of energy is added to each canister.

a. Find how far the piston in the second canister moves up or down?
b. Find the final pressure of the gas in the first canister

Homework Equations


[/B]
PV = nRT

ΔQ = ΔW + ΔU

3. The Attempt at a Solution

So for question a I found the initial volume of the canister. Which is (.25)^3 m^3

Then I found the moles of helium which is 4 g / 4 g/mol = 1 mol

Knowing that I found the pressure to be P = ((1 mol)(300 K)(8.314))/(.25 m)^3 = 160,000 Pa

So since the pressure doesn't change in the question a,
ΔW = PΔV
Which means ΔW = 1.6E5 * ( Hf(.25)^2 - (.25)^3) Hf being final height of the piston

ΔQ = 1000J
So 1000 = (1.6E5 * ( Hf(.25)^2 - (.25)^3)) + ΔU

At this point I am stuck because I don't know how to solve for ΔU
 
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  • #2
sliperyfrog said:
So for question a I found the initial volume of the canister. Which is (.25)^3 m^3
Are you sure about that? What is .25 (just reread what the question says about .25, look at the math and see what you did wrong)? Have you tried part b yet?
 
  • #3
Also, is this a perfect isolated system? If yes, part (a) is extremely easy using the conservation of energy and the ideal gas law. The internal energy equation wouldn't even be necessary.

Edit: That is, is energy gained or lost AFTER the 1000J is added.
 
  • #4
The (.25)^3 m^3 is the initial volume I could write it as .015625 m^3 but that is annoying to plug in. After the 1000J is added, I would assume it to be an isolated system. How would I use the the conservation of energy?

No I did not try b yet it is probably done with the same method used to find a. Just instead of constant pressure you would have constant volume instead.
 
  • #5
sliperyfrog said:

Homework Statement


There are two rectangular canisters, both have a .25m cross sectional area
I think the wording here might be causing some confusion. Is this to be interpreted as saying that the cross section is a square with each side of the square being .25 m long?
 
  • #6
TSny said:
I think the wording here might be causing some confusion. Is this to be interpreted as saying that the cross section is a square with each side of the square being .25 m long?
Yes

That was a typo it is supposed to say ".25 m side square cross sections"
 
Last edited:
  • #7
OK.

What is the molar heat capacity at constant pressure, Cp, for helium?
 
  • #8
I am not allowed to use that for the probelm
 
  • #9
OK. Do you know how to relate the internal energy, U, to the temperature for helium?
 
  • #10
No
 
  • #11
Hmm. We're running out of ammunition.

The relation between U and T for monatomic gases is usually covered in introductory courses.
 
  • #12
Okay I am a idiot, I didn't realize helium was monoatomic. So when you mention molar heat capacity I thought it was one of those thing you had to look it up.

So for a mono atomic gas the Cp = 5R/2 = 20.785
So I can use the Q = mol * Cp * ΔT
1000 = (1 mol)(20.785) (ΔT)
ΔT = 48.1 K

So then I can use V1/T1 =V2/T2
(.015625)/(300) = V2/(348.1)
V2 = .0181 m^3
V2 =((.25)^2) *Hf
Hf = .29m
.29 m -.25 m = .04 m

And for b I would just use Cv = 3R/2 instead.
 
  • #13
sliperyfrog said:
Yes

That was a typo it is supposed to say ".25 m side square cross sections"
You said cross sectional area, which is why I said that other thing. You cubed an area and have units of volume.
 
  • #14
sliperyfrog said:
So for a mono atomic gas the Cp = 5R/2 = 20.785
So I can use the Q = mol * Cp * ΔT
1000 = (1 mol)(20.785) (ΔT)
ΔT = 48.1 K

So then I can use V1/T1 =V2/T2
(.015625)/(300) = V2/(348.1)
V2 = .0181 m^3
V2 =((.25)^2) *Hf
Hf = .29m
.29 m -.25 m = .04 m

And for b I would just use Cv = 3R/2 instead.
That looks good.
 

1. How does energy affect the movement of the piston?

Energy is a fundamental property of matter that can be transferred and transformed. In the case of a piston, adding energy in the form of heat or pressure can cause it to move as the molecules within the gas expand and push against the piston walls.

2. What factors determine how far the piston will move after energy is added?

The distance the piston moves after energy is added depends on several factors, including the amount of energy added, the size and shape of the piston, the type and pressure of the gas, and the friction or resistance within the piston's mechanical system.

3. Is there a limit to how far the piston can move after energy is added?

Theoretically, the piston could continue to move infinitely if there is an endless supply of energy being added. However, in practical applications, the movement of the piston will eventually be limited by factors such as the size of the piston chamber and the strength of the piston's materials.

4. How does the movement of the piston affect the surrounding environment?

The movement of the piston can have various effects on the surrounding environment. For example, if the piston is part of an engine, its movement can produce mechanical work that can power a vehicle. In other cases, the movement of the piston may simply produce heat or pressure that can be used for other purposes.

5. Can the movement of the piston be controlled?

Yes, the movement of the piston can be controlled through various methods such as adjusting the amount of energy being added, changing the gas pressure, and using mechanical devices such as valves or gears. This allows for precise control of the piston's movement and makes it a useful tool in many scientific and industrial applications.

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