# 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.