4
$\begingroup$

If the following gasses are to mix, what is the partial pressure of neon?

http://i.stack.imgur.com/8Pn7w.png

The correct answer is 0.1. I cannot figure out how my professor derived this equation. From the ideal gas law, all I can arrive to is this:

$P_x = \frac{P_tV_t}{V_x}$

This gives me an answer of 4.4, which is obviously incorrect.

I've determined the equation he used is as follows:

$P_x = \frac{P_xV_x}{V_t}$

which gives the correct answer of 0.1. How did he arrive to this answer?

$\endgroup$
  • $\begingroup$ you have $P_x$ twice in your second equation. Should one be $P_t$? $\endgroup$ – Ben Norris Dec 12 '12 at 0:20
  • $\begingroup$ @BenNorris: That was done on purpose as that was the only way to get the correct answer. I listed it because it made no sense. $\endgroup$ – Yep Dec 12 '12 at 0:27
  • $\begingroup$ Mathematically, then both instances of $P_x$ would cancel and you would have $\frac{P_x}{P_x}=\frac{V_x}{V_t} \implies 1=\frac{V_x}{V_t}$, which is silly. $\endgroup$ – Ben Norris Dec 12 '12 at 0:32
  • 1
    $\begingroup$ Right. That's why I was confused. $\endgroup$ – Yep Dec 12 '12 at 0:41
3
$\begingroup$

Prepare for an amazing feat of algebra! I sincerely hope there is a shorter way to the endpoint. (there is! See the end of my answer.)

To calculate partial pressure, $P_{\ce{He}}$, we need the total pressure $P_t$ and the fraction of the gas that is He $X_{\ce{He}}$: $$P_{\ce{He}}=X_{\ce{He}} P_t$$

You have $PV=nRT$. $T$ is not given, so assume it is constant. $n$ is unknown, but the system is closed, so $n_x$ for each gas is constant. Thus we have:

$$PV=\text{ constant}$$

Total Pressure

It is tempting to write variations of $$P_1 V_1 = P_2 V_2$$

However, for each gas, both $P$ and $V$ are changing, so we need to consider their products: $$(PV)_1 = (PV)_2$$

Thus, as trb456 suggests, $$P_t V_t = (PV)_t \implies P_t =\frac{\sum{(PV)_i}}{\sum{V_i}}$$

Fraction of He

The fraction of the mixture that is helium is determined by the ratio $\dfrac{n_{\ce{He}} }{n_t}$. Since $n=PV/RT$, and $R$ and $T$ are constant, we can write: $$X_{\ce{He}}=\frac{n_{\ce{He}}}{\sum{n_i}}=\frac{(PV)_{\ce{He}}}{\sum{(PV)_i}}$$

At last!

$$P_{\ce{He}}=X_{\text{He}} P_t = \left( \frac{(PV)_{\ce{He}}}{\sum{(PV)_i}}\right) \left( \frac{\sum{(PV)_i}}{\sum{V_i}}\right)$$ $$P_{\ce{He}_2}=\frac{(PV)_{\ce{He}_1}}{V_t}$$

Or, as it struck me as I finished:

Helium expands to fill the total volume. Now we can use $P_1 V_1 = P_2 V_2$ and completely ignore the other gasses (there is a lot empty space for the He atoms to fit in).

$$P_{\ce{He}_2}=\frac{P_{\ce{He}_1}V_{\ce{He}_1}}{V_t}$$

This equation is very similar to your second equation, except the subscripts are added to denote initial and final pressure of helium.

$\endgroup$
3
$\begingroup$

The total pressure of the system is:

$\frac{1.0\times0.5+1.0\times0.4+2.0\times0.2}{1.0+1.0+2.0}=1.3/4=0.325$

i.e. just the sum of the $PV$s divided by the sum of the $V$s. Then $0.325$ is the sum of the partial pressures by the Law of Partial Pressures. So this at least suggest that the individual partial pressures are near $0.1$.

Can you figure out how to parcel out the individual pressures from here?

$\endgroup$
  • $\begingroup$ Did you derive PV/Vt by using the ideal gas law? I've seen this used before and I think it's the answer to my problem. $\endgroup$ – Yep Dec 11 '12 at 23:56
  • $\begingroup$ Remember that, at constant temperature for the system, $PV$ is proportional to $n$, the moles of gas, by the Ideal Gas Law, yes. So you add up (something proportional to) moles, divide by volume, and you get pressure. Now to allocate partial pressure, use this same idea of moles again! $\endgroup$ – user467 Dec 12 '12 at 1:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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