# 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?

Which vessel contains more molecules?

Which vessel contains more mass?

In which vessel is the root-mean-square speed of the

molecules higher?

##### 1 Answer

**NUMBER OF MOLECULES COMPARISON**

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

#\mathbf(PV = nRT)# where:

#P# is the pressure in#"atm"# .#V# is the volume.#n# is the number of#"mol"# s of gas.#R = 0.082057# #"L"cdot"atm/mol"cdot"K"# is the universal gas constant.#T# is the temperature in#"K"# .

Suppose we compared the two systems.

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

#P_("CO"_2)V = n_("CO"_2)RT_("CO"_2)#

To get the number of

#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

#color(green)(n_("O"_2)) = (("1.0 atm")*("1 L"))/(0.082057 ("L"cdot"atm")/("mol"cdot"K")cdot(25+"273.15 K"))#

#~~ color(green)("0.0409 mols O"_2)#

#color(green)(n_("CO"_2)) = (("0.75 atm")*("1 L"))/(0.082057 ("L"cdot"atm")/("mol"cdot"K")cdot(10+"273.15 K"))#

#~~ color(green)("0.0323 mols CO"_2)#

Since a *what it is* you have a *exactly* that many

So, if you have more ** automatically** have more molecules. Thus, for the same volume of gas under the specified conditions, you have

**more molecules of**

#("0.0409" cancel"mols" "O"_2 - "0.0323" cancel"mols" "CO"_2)xx(6.0221413xx10^23 "molecules")/cancel("1 mol any molecule")#

#= color(green)(5.176xx10^21)# #color(green)("more molecules")# of#"O"_2# than#"CO"_2# .

**MASS COMPARISON**

Now, when you ask about **mass**, that's where things can differ between *molar masses* (

So, convert the

#color(blue)("m"_("O"_2)) = "0.0409" cancel("mols O"_2) xx ("31.998 g O"_2)/cancel("1 mol O"_2)#

#~~ color(blue)("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)#

#~~ color(blue)("1.421 g CO"_2)#

So there is more

**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 = \mathbf(8.314472 "J/mol"cdot"K")# is your universal gas constant in this case, NOT#0.082057 "L"cdot"atm/mol"cdot"K"# . That is a common mistake.#T# is the temperature in#"K"# as before.#M_m# is themolar mass of the gas in#\mathbf("kg/mol")# , because#"1 J" = "1 kg"cdot"m"^2/"s"^2# (that way the units cancel out). This is another common mistake. It is NOT in#"g/mol"# !

Already, without doing any actual calculations, *similar*. So, we expect the root-mean-square speed of **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.

#color(blue)(upsilon_("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"))#

#~~# #color(blue)("469.81 m/s")#

And for

#color(blue)(upsilon_("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"))#

#~~# #color(blue)("411.08 m/s")#

No surprise, right? The root-mean-square speed of **lower**, because it was **heavier** (if given the same number of *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**.