# If equal mols of "N"_2 and "Ar" are in the same container with a total pressure of "10 atm", (A) what are their partial pressures? (B) what is the effusion speed of "H"_2 if that of "He" is "600 m/s" at a certain temperature?

Jun 4, 2017

A) $\text{5 atm}$ each.

B) The lighter gas $B$ effuses at a speed ${v}_{B}$ that is $\sqrt{{M}_{A} / {M}_{B}}$ times as fast as ${v}_{A}$, where ${M}_{A}$ is the molar mass of $A$.

A) If the container is filled with an equal number of each particle, then we know that:

${n}_{{N}_{2}} = {n}_{A r}$

When this is the case, assuming ideal gases, their partial pressures should be identical.

Mathematically, this results in the following mol fractions:

${\chi}_{{N}_{2}} = \frac{{n}_{{N}_{2}}}{{n}_{A r} + {n}_{{N}_{2}}}$

${\chi}_{A r} = \frac{{n}_{A r}}{{n}_{A r} + {n}_{{N}_{2}}}$

But since ${n}_{{N}_{2}} = {n}_{A r}$, we expect ${\chi}_{{N}_{2}} = {\chi}_{A r} = 0.5$. Let's just say we had $\text{1 mol}$ of each gas. Then:

${\chi}_{{N}_{2}} = \frac{1}{1 + 1} = 0.5 = {\chi}_{A r}$

Assuming ideal gases, we can show that both gases have the same partial pressure, the pressure exerted by each gas in the mixture:

$\textcolor{b l u e}{{P}_{{N}_{2}}} = {\chi}_{{N}_{2}} {P}_{t o t}$

$= {\chi}_{A r} {P}_{t o t} = \textcolor{b l u e}{{P}_{A r}}$

$= 0.5 \left(\text{10 atm") = color(blue)("5 atm}\right)$

B) I'm not sure what part $B$ has to do with part $A$. I will assume the question means to write $\text{Ar}$ and ${\text{N}}_{2}$ instead of $\text{He}$ and ${\text{H}}_{2}$.

We are given, then, that argon effuses at $\text{600 m/s}$. Effusion can be derived from the expression for some sort of speed for each gas. The root-mean-square speed equation is fine:

${v}_{r m s} = \sqrt{\frac{3 R T}{M}}$

where $R$ and $T$ are known from the ideal gas law, and $M$ is the molar mass in $\text{kg/mol}$.

The rate of effusion, ${z}_{e f f}$ is proportional to the speed $v$, so the proportionality constants cancel out in a ratio:

${z}_{e f f , A r} / \left({z}_{e f f , {N}_{2}}\right) = \boldsymbol{{v}_{A r} / \left({v}_{{N}_{2}}\right)} = \boldsymbol{\sqrt{{M}_{{N}_{2}} / \left({M}_{A r}\right)}}$

The relationship of $\frac{{z}_{e f f , i}}{{z}_{e f f , j}}$ to $\sqrt{{M}_{j} / {M}_{i}}$ is known as Graham's law of effusion, but we will instead be using the relationship with ${v}_{i} / {v}_{j}$, the ratio of the speeds.

We are given that ${v}_{A r} = \text{600 m/s}$, so:

$\textcolor{b l u e}{{v}_{{N}_{2}}} = {v}_{A r} \sqrt{{M}_{A r} / {M}_{{N}_{2}}}$

= "600 m/s" cdot sqrt("0.039948 kg/mol"/"0.028014 kg/mol")

$=$ $\textcolor{b l u e}{\text{716.5 m/s}}$

This should make sense, that the lighter gas, ${\text{N}}_{2}$, effused faster.

If it really is supposed to be $\text{He}$ and ${\text{H}}_{2}$, then you should expect:

$\textcolor{b l u e}{{v}_{{H}_{2}}} = {v}_{H e} \sqrt{{M}_{H e} / {M}_{{H}_{2}}}$

= "600 m/s" cdot sqrt("0.0040026 kg/mol"/"0.0020158 kg/mol")

$=$ $\textcolor{b l u e}{\text{845.5 m/s}}$