[beruniy]

A Lithium-6 sample (m = 12,0 mg) which was isolated in ampule V = 200 cm3, was loaded in nuclear reactor and irradiated by neutrons for a long time. Calculate the maximal pressure in reactor, T = 400 K.

The equation is
Li + n = T + α

I calculated the energy of this reaction using formula ΔE = -Δmc2
4,78 MeV for 1 particle, and then calculated it for 12 mg of Lithium.
How can I transfer energy to pressure? I tried to use the kinetic energy but nothing came out of it. Is it possible to use thermodynamic laws?

[Borek]

Isn't it just pV=nRT?

[beruniy]

pV = nRT is used for perfect gas, this equation allows us to calculate pressure without any reaction. I think that nuclear reaction gives a huge energy (9.21×10⁵ kJ), and exactly this energy makes pressure. So I want to know how transfer this energy to pressure. Am I right?

[Borek]

pV = nRT is used for perfect gas

What are products of the reaction?

[beruniy]

Trithium and Helium. They are both perfect gases. But at the moment of releasing the huge energy liberates... Problem says to determine the maximal pressure, that's why I consider to transfer nuclear energy in pressure.

And, in general, why it was necessary to put the statement of a problem this way, if its essence is just the application of Mendeleyev-Clapeyron equation?

[Borek]

What you call Mendeleyev-Clapeyron equation is just another form of ideal gas equation.

As far as I know this name is used only in some post-Soviet countries, don't be surprised if people will not know what you are talking about. I had to google it: http://encyclopedia2.thefreedictionary.com/Clapeyron+Equation

[Arkcon]

pV = nRT is used for perfect gas,

Do you know under what circumstances the Ideal gas law begins to fail for real gasses?  And do those circumstances apply to the situation you have here?

[beruniy]

http://encyclopedia2.thefreedictionary.com/Clapeyron+Equation

Looking through this link I understood that Mendeleev-Clapeyron equation is
pV = nRT or pV = NkT because k = R/N_A and n = N/N_A
We were just talking about it.

Do you know under what circumstances the Ideal gas law begins to fail for real gasses?

Yes, the Ideal gas laws begin to fail near condensation point. Also, polyatomic molecules don't follow the ideal gas laws.

And do those circumstances apply to the situation you have here?

No. We can consider Tritium and Helium as ideal gases.

But the essense of problem is quite another: how is it possible to associate nuclear energy and pressure?

[Borek]

You are stubbornly trying to make this question something that it is not. You are given volume, you are given temperature, you are given information necessary to calculate number of moles. It cries out loud "ideal gas equation!"

Energy by itself is not a source of pressure, for pressure you need a momentum carrier. And the only known momentum carriers here are gas molecules.

[beruniy]

1) Momentum carrier is p = mv and energy is E = mv²/2
Also, pressure P = F/S, F = ma; S is an area and it is possible to associate it with volume. So, it is possible to assosiate pressure with energy!

2) If I use perfect gas equation
pV = nRT

Li + n = T + He
m (Li) = 12 mg = 0.012 g; n (Li) = 0.012 g / 6 g/mole = 0.002 mole
n (T) + n (He) = 0.002 + 0.002 = 0.004 mole.
V = 200 cm³ = 2×10^-4 m³
p = nRT/V =  4×10^-3 × 8.314 × 400 / 2×10^-4 = 66512 (Pa)
that is lower than atmospheric.
Obviously, it doesn't correspond to maximal pressure which is suggested to calculate proceeding from the problem statement.

[DrCMS]

What about the rest of the ampoule?  Was it evacuated or filled with air or inert gas?

You have calculated the pressure if the ampoule only contained lithium and was evacuated prior to sealing.  Why do you think the maximum pressure has to be more than 1 atmosphere

[Borek]

1) Momentum carrier is p = mv and energy is E = mv²/2
Also, pressure P = F/S, F = ma; S is an area and it is possible to associate it with volume. So, it is possible to assosiate pressure with energy!

You have not shown any association, you just listed a bunch of equations. p=mv is not a momentum carrier, it is a linear momentum definition. Momentum carrier is a moving object with known mass.

2) If I use perfect gas equation
pV = nRT

Li + n = T + He
m (Li) = 12 mg = 0.012 g; n (Li) = 0.012 g / 6 g/mole = 0.002 mole
n (T) + n (He) = 0.002 + 0.002 = 0.004 mole.
V = 200 cm³ = 2×10^-4 m³
p = nRT/V =  4×10^-3 × 8.314 × 400 / 2×10^-4 = 66512 (Pa)
that is lower than atmospheric.
Obviously, it doesn't correspond to maximal pressure which is suggested to calculate proceeding from the problem statement.

There is an important error in your calculation (hint: number of moles of gas is not 4 millimoles), but I don't see anything wrong with the result being lower than the atmospheric pressure.

[Borek]

What about the rest of the ampoule?  Was it evacuated or filled with air or inert gas?

I guess "lithium sample was isolated in the ampule" means it was evacuated. But that's just a guess.

[DrCMS]

What about the rest of the ampoule?  Was it evacuated or filled with air or inert gas?

I guess "lithium sample was isolated in the ampule" means it was evacuated. But that's just a guess.

That would be mine as well but with translations it might not be correct to assume that which is why I asked the question. 

However, I missed the mistake in the number of moles

[beruniy]

In origin Lithium was isolated in ampoule so I agree with you that initial pressure was 0.

About number of moles - if from 1 mole Lithium we get 1 mole Trithium and 1 mole Helium - doesn't it mean that generally we have 2 moles of ideal gas? Why not, explain please.