0
$\begingroup$

I have a question on my homework I can't really figure out. It's from Reif, problem 3.6, Fundamentals of Statistic and thermal physics

Question: A glass bulb contains air at room temperature and at a pressure of 1 atmosphere. It is placed in a chamber filled with helium gas at 1 atmosphere and at a room temperature. A few months later, the experimenter happens to read in a journal article that the particular glass of which the bulb is made is quite permeable to helium, although not to any other gases. Assuming that equilibrium has been attained by this time, what gas pressure will the experimenter measure inside the bulb when he goes back to check?

I'm pretty sure I have to assume the volume of the helium is much larger than the volume of the light bulb in order for this to work. I think in equilibrium the number of accessible states is minimized, so that's kind of what my initial thoughts were, but then I didn't know how to progress after that. Thanks for the help!

$\endgroup$
1
  • 1
    $\begingroup$ Believe I found out the answer. Assuming it's an ideal gas, inside the bulb the two gasses will be non-interacting, therefore the pressure inside the bulb resulting from helium at equilibrium will be the same as outside. Assuming there's much more helium than air, the resulting pressure is 1 atm from helium. Adding partial pressures of air and helium, we get 1 atm + 1 atm = 2 atm inside the bulb $\endgroup$ Nov 11 '13 at 21:24
1
$\begingroup$

Have you tried the way via partial pressure? If it is in equilibrium, the partial pressure of Helium inside the bulb should be the same like the outer pressure?

$p_{Helium, room}$ = 1 bar

$p_{bulb, beginning}$ = 1 bar

"Air" consists of roughly 79 vol.-% nitrogen and 21 vol.-% oxygen. We neglect all the other parts.

$p_{bulb, end} = p_{N_2} + p_{O_2} = 1$ bar

$\Rightarrow p_{N_2} = 0.79 \cdot p_{bulb, beginning} = 0.79$ bar,

$p_{O_2} = 0.21 \cdot p_{bulb, beginning} = 0.21$ bar

As $p_{Helium, room} = p_{He, partial}$

Follows:

$p_{bulb, beginning} = p_{N_2} + p_{O_2} + p_{He} = 2$ bar

But I'm unsure if you have to take into account, that Helium is no molecule, so it might have to be factorized by 1/2 to gain the right pressure, which should be 1.5 bar than...

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.