How far does the piston move after energy is added?

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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
 

Discussion

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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?
 
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.
 
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.
 
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?
 
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:
I am not allowed to use that for the probelm
 
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.
 
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.
 
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.