Finding the gas based off of mass

[mk9898]

Homework Statement

Die average kinetic energy and impuls squared of a single molecule are:
$$<E_{kin}>= 6.5*10^{-21} J$$ and $$<p^2> = 1.8*10^{-45} kg^2m^2s^{-2}$$

Which gas ist it?

3. The Attempt at a Solution
$\frac{m<v^2>}{2}=\frac{<p^2>}{2m}$ $\Leftrightarrow$ $\frac{<p^2>}{<v^2>} = m^2$

Therefore is $m^2 = 2.77*10^{-25}kg$ and $m = 5.26*10^{-13}kg$

I searched and attempted to find with gas it is based off of the mass and also multiplying it by a mole but to no avail.

 

[Charles Link]

The equation for kinetic energy reads $<E>=\frac{m <v>^2}{2} =\frac{<p^2>}{2m}$ and not $\frac{m}{2 <v>^2}$. Try again=this one should be fairly straightforward.

 

[mk9898]

Yea that's just a typo in the latex of course. You can tell from the next step that the p and v variables are where they need to be. The value for the mass above is correct.

 

[Charles Link]

Yea that's just a typo in the latex of course. The value for the mass above is correct.

There's no $m^2$ in this calculation. You should be able to get $m=\frac{<p^2>}{2 <E_{kin}>}$.

 

[mk9898]

Ah yea of course. The value I get is 1.3846*10^-25. Even then I'm not sure how I can tell which gas that is. If I multiply a mole with it, it's still smaller than hydrogen.

 

[Charles Link]

Ah yea of course. The value I get is 1.3846*10^-25. Even then I'm not sure how I can tell which gas that is. If I multiply a mole with it, it's still smaller than hydrogen.

The proton has mass $m_p=1.67 \cdot 10^{-27} \, kg$. The molecule is much heavier than that.

 

[mk9898]

How can that be the mass of a proton when it states in the exercise, that the molecule has the kinetic energy and the impulse? Meaning, m would be the entire molecule? Or why is it the mass of a proton?

 

[Charles Link]

How can that be the mass of a proton when it states in the exercise, that the molecule has the kinetic energy and the impulse? Meaning, m would be the entire molecule? Or why is it the mass of a proton?

How many protons and/or neutrons (both have approximately the same mass) do you need to get the mass $m$ of the molecule?

 

[mk9898]

Hmm. I have to be honest here...I have no idea. This wasn't covered in the lecture and from the readings I've done, it doesn't touch on this.

 

[Charles Link]

If you take the mass $m$ you computed, and divide by 1.67 E-27, this will tell you the molecular weight of the molecule. I computed this already, but I so far haven't identified what the molecule most likely is.

 

[mk9898]

I would guess it's oxygen but where does the 1.67e-27 come from? My apologies if I'm asking rudimentary questions. I really want to learn this stuff. Thanks a lot for the effort.

 

[Charles Link]

The proton and neutron mass have $m_p \approx m_n \approx 1.67 \cdot 10^{-27}$ kg. If you divide the molecular mass by the mass of the atomic building block, it will tell you how many neutrons and protons you have in the molecule. The problem isn't complete there though. I'm looking over the periodic table and trying to get a match. The closest thing I have paired up is Krypton, which is simply an atom. $SO_2$ and $Cl_2$ are both quite a ways off.

 

[mk9898]

Thanks for the clarification. The number I get is 8.29 which I would assume would give us atomic number 8.

 

[Charles Link]

Thanks for the clarification. The number I get is 8.29 which I would assume would give us atomic number 8.

Try your arithmetic again. I get 83. $\\$ When they tell you it is a molecule, it really should be a molecule, but the only gas I could find with atomic weight near 83 that is at all common is Krypton, at 83.8. They only give 2 sig figs in the statement of the problem. Another possibility though is $HBr$=hydrogen bromide=molecular weight=81. $\\$ As a homework helper, I'm not supposed to supply answers, but, assuming our arithmetic is correct, this question is somewhat sticky for trying to teach concepts. They could have made it something simple like $O_2$ or $SO_2$ or $CO_2$, but that doesn't appear to be the case.

 

[mk9898]

Hm. 1.38e-25/1.67e-27 is roughly 8.29. What did you use to get 83?

 

[Charles Link]

Hm. 1.38e-25/1.67e-27 is roughly 8.29. What did you use to get 83?

(.83)100=83

 

[mk9898]

But from the division of the two variables it gives us 8.29. I don't understand where .83*100 comes from.

 

[Charles Link]

But from the division of the two variables it gives us 8.29. I don't understand where .83*100 comes from.

$\frac{10^{-25}}{10^{-27}}=10^2=100$, and 1.38/1.67=.83.

 

[mk9898]

Yes but 1.38 is smaller than 1.67. Therefore it's only one decimal place higher.

 

[Charles Link]

Yes but 1.38 is smaller than 1.67. Therefore it's only one decimal place higher.

Yes. It's less than 1.0. 1.38/1.67=0.83 .

 

[Charles Link]

.83=83/100

 

[mk9898]

Ah crap sorry. Haha oh boy. I had my calculator on scientific and thought that the 8.3e1 just meant 8.3. Silly. Sorry about that. I understand the problem now. I could have just taken a second and thought about it but just let the calculator do the simple work.

 

[Charles Link]

Ah crap sorry. Haha oh boy. I had my calculator on scientific and thought that the 8.3e1 just meant 8.3. Silly. Sorry about that. I understand the problem now. I could have just taken a second and thought about it but just let me calculator do the simple work.

Very good. Now that we agree the "molecular weight" is 83 or something close to that, this one needs some guesswork, because none of the common molecules have M.W.$\approx$ 83. The krypton atom could also be considered a molecule, so perhaps that is the correct answer. And yes, krypton is a gas.

 

[mk9898]

I think the Krypton answer is a good one and I think they wanted the answer to deviate a bit from an actual element. They have an additional short question asking, "What additional information could help better determine the gas?" My answer would be knowing the temperature and the degrees of freedom so then we could use the E = fkT/2 equation. And that would perhaps help in determining the element.

 

[mk9898]

Try your arithmetic again. I get 83. $\\$ When they tell you it is a molecule, it really should be a molecule, but the only gas I could find with atomic weight near 83 that is at all common is Krypton, at 83.8. They only give 2 sig figs in the statement of the problem. Another possibility though is $HBr$=hydrogen bromide=molecular weight=81. $\\$ As a homework helper, I'm not supposed to supply answers, but, assuming our arithmetic is correct, this question is somewhat sticky for trying to teach concepts. They could have made it something simple like $O_2$ or $SO_2$ or $CO_2$, but that doesn't appear to be the case.

Hm. Yes I understand your point and I'm going to have to look into this. Why isn't it the case that $O_2$ or $SO_2$ are valid options?

 

[Charles Link]

Hm. Yes I understand your point and I'm going to have to look into this. Why isn't it the case that $O_2$ or $SO_2$ are valid options?

$O_2$ has M.W.=32 . $SO_2$ has M.W.=64 .

 

[mk9898]

Ok thanks I thought that's what you meant.

 

[mk9898]

Or I could just begin to add various elements together and see if it is "the" molecule. But then that doesn't say anything of the validity of such a molecule existing.

 

[Charles Link]

Or I could just begin to add various elements together and see if it is "the" molecule. But then that doesn't say anything of the validity of such a molecule existing.

I've got a reasonably good chemistry background, and looked it over, (the periodic table, as well as googling for molecular weights of common substances), pretty carefully. You need to make sure what you assemble is a gas. HBr =hydrogen bromide with M.W.=81 is also a possibility. I couldn't come up with any molecule that has M.W.= 83. It's possible I missed something...

 

[mk9898]

Thanks for your expertise. Just to remove any doubt I have two questions: 1) how is the division from the mass/proton mass = atomic weight? That would just be a scalar. 2) How does one find the total amount of the protons of an element?