Vessel A contains O2 gas at 25C and 1.0 atm. Vessel B contains CO2 gas at 10.0C and 0.75 atm. Both vessels have the same volume. Answer each of the following?

Which vessel contains more molecules? Which vessel contains more mass? In which vessel is the root-mean-square speed of the molecules higher?

May 3, 2016

NUMBER OF MOLECULES COMPARISON

If we assume ideality, then we can use the ideal gas law:

$\setminus m a t h b f \left(P V = n R T\right)$

where:

• $P$ is the pressure in $\text{atm}$.
• $V$ is the volume.
• $n$ is the number of $\text{mol}$s of gas.
• $R = 0.082057$ $\text{L"cdot"atm/mol"cdot"K}$ is the universal gas constant.
• $T$ is the temperature in $\text{K}$.

Suppose we compared the two systems.

${P}_{{\text{O"_2)V = n_("O"_2)RT_("O}}_{2}}$

${P}_{{\text{CO"_2)V = n_("CO"_2)RT_("CO}}_{2}}$

To get the number of $\text{mol}$s and consequently the number of molecules:

n_("O"_2) = (P_("O"_2)V)/(RT_("O"_2))

n_("CO"_2) = (P_("CO"_2)V)/(RT_("O"_2))

Now, let's solve to see what it is. Assume that the volume is $\text{1 L}$ for simplicity. It doesn't really matter what the volume is because they're both the same.

$\textcolor{g r e e n}{{n}_{\text{O"_2)) = (("1.0 atm")*("1 L"))/(0.082057 ("L"cdot"atm")/("mol"cdot"K")cdot(25+"273.15 K}}}$

$\approx \textcolor{g r e e n}{{\text{0.0409 mols O}}_{2}}$

$\textcolor{g r e e n}{{n}_{\text{CO"_2)) = (("0.75 atm")*("1 L"))/(0.082057 ("L"cdot"atm")/("mol"cdot"K")cdot(10+"273.15 K}}}$

$\approx \textcolor{g r e e n}{{\text{0.0323 mols CO}}_{2}}$

Since a $\text{mol}$ is just a quantity (like a dozen), it doesn't matter what it is you have a $\text{mol}$ of. You'll still have exactly that many $\text{mol}$s.

So, if you have more $\text{mol}$s, you automatically have more molecules. Thus, for the same volume of gas under the specified conditions, you have more molecules of $\setminus m a t h b f \left({\text{O}}_{2}\right)$. Specifically, you have:

$\left(\text{0.0409" cancel"mols" "O"_2 - "0.0323" cancel"mols" "CO"_2)xx(6.0221413xx10^23 "molecules")/cancel("1 mol any molecule}\right)$

$= \textcolor{g r e e n}{5.176 \times {10}^{21}}$ $\textcolor{g r e e n}{\text{more molecules}}$ of ${\text{O}}_{2}$ than ${\text{CO}}_{2}$.

MASS COMPARISON

Now, when you ask about mass, that's where things can differ between ${\text{O}}_{2}$ and ${\text{CO}}_{2}$, since they do have different molar masses ($31.998$ and $44.009$ $\text{g/mol}$, respectively).

So, convert the $\text{mol}$s you have to grams to see.

$\textcolor{b l u e}{{\text{m"_("O"_2)) = "0.0409" cancel("mols O"_2) xx ("31.998 g O"_2)/cancel("1 mol O}}_{2}}$

$\approx \textcolor{b l u e}{{\text{1.308 g O}}_{2}}$

color(blue)(m_("CO"_2)) = "0.0323" cancel("mols CO"_2) xx ("44.009 g CO"_2)/cancel("1 mol CO"_2)

$\approx \textcolor{b l u e}{{\text{1.421 g CO}}_{2}}$

So there is more ${\text{CO}}_{2}$ in terms of mass in $\text{g}$.

ROOT-MEAN-SQUARE SPEED COMPARISON

And now, root-mean-square speed. That is represented by the following formula:

\mathbf(upsilon_("RMS") = sqrt((3RT)/(M_m))

where:

• $R = \setminus m a t h b f \left(8.314472 \text{J/mol"cdot"K}\right)$ is your universal gas constant in this case, NOT $0.082057 \text{L"cdot"atm/mol"cdot"K}$. That is a common mistake.
• $T$ is the temperature in $\text{K}$ as before.
• ${M}_{m}$ is the molar mass of the gas in $\setminus m a t h b f \left(\text{kg/mol}\right)$, because ${\text{1 J" = "1 kg"cdot"m"^2/"s}}^{2}$ (that way the units cancel out). This is another common mistake. It is NOT in $\text{g/mol}$!

Already, without doing any actual calculations, ${\text{CO}}_{2}$ has the greater molar mass, but the temperatures are similar. So, we expect the root-mean-square speed of ${\text{CO}}_{2}$ to be lower.

(If you are asked this question on a test and are assuming the same temperature conditions, you do not need to do any actual calculations. Just reason it out.)

Let's see what the values are.

$\textcolor{b l u e}{{\upsilon}_{\text{RMS")^("O"_2)) = sqrt((3*8.31447"2 J/mol"cdot"K"*283.15 )/("0.031998 kg/mol}}}$

= sqrt(((3*8.314472cancel"kg"cdot"m"^2)/("s"^2cdotcancel"mol"cdotcancel"K")*283.15 cancel"K")/("0.031998" cancel"kg/mol"))

$\approx$ $\textcolor{b l u e}{\text{469.81 m/s}}$

And for ${\text{CO}}_{2}$, we have:

$\textcolor{b l u e}{{\upsilon}_{\text{RMS")^("CO"_2)) = sqrt((3*8.314472"J/mol"cdot"K"*298.15 )/("0.044009 kg/mol}}}$

= sqrt(((3*8.314472cancel"kg"cdot"m"^2)/("s"^2cdotcancel"mol"cdotcancel"K")*298.15 cancel"K")/("0.044009" cancel"kg/mol"))

$\approx$ $\textcolor{b l u e}{\text{411.08 m/s}}$

No surprise, right? The root-mean-square speed of ${\text{CO}}_{2}$ was lower, because it was heavier (if given the same number of $\text{mol}$s as you have ${\text{O}}_{2}$) AND the temperatures were comparably similar.

That, and the fact that the molar mass was in the denominator tells you that the root-mean-square speed is inversely proportional to the molar mass of the gas.