What are the partial pressures of each gas in the mixture?

[ianc1339]

Homework Statement

A mixture of gases at a pressure of $7.4\times 10^4 Nm^{-2}$ has the volume composition 40% nitrogen, 35% oxygen, 25% carbon dioxide. What will the partial pressures of nitrogen become if the carbon dioxide is removed by the introduction of some sodium hydroxide pellets?

Relevant Equations

pV=nRT
Dalton's law of partial pressures

I've first calculated the partial pressures of each gas:

$N_2: 0.4\times 7.4\times 10^4=3.0\times 10^4 Nm^{-2}\\$
$O_2: 0.35\times 7.4\times 10^4=2.6\times 10^4 Nm^{-2}\\$
$CO_2: 0.25\times 7.4\times 10^4=1.9\times 10^4 Nm^{-2}\\$

From here, I do not know how to continue. Could someone give me a hint on what to do next?

Thanks

 

[chemisttree]

Is volume allowed to change in this experiment? This stuff in a balloon or a rigid-walled vessel?

 

[ianc1339]

Is volume allowed to change in this experiment? This stuff in a balloon or a rigid-walled vessel?

It doesn't say anything about it so I'm assuming it to be constant volume.

 

[etotheipi]

It might help to think that when you have several different types of gases in a container, you can consider the pressure of each independently (this is a sort of definition of a partial pressure: the pressure if only that gas were alone in the container!) and add them all together at the end.

What effect then does removing one component of the mixture do to the partial pressures of the others?

 

[ianc1339]

It might help to think that when you have several different types of gases in a container, you can consider the pressure of each independently (this is a sort of definition of a partial pressure: the pressure if only that gas were alone in the container!) and add them all together at the end.

What effect then does removing one component of the mixture do to the partial pressures of the others?

Oh, so does the partial pressure of the other gases still remain constant even if carbon dioxide was removed?

 

[etotheipi]

Oh, so does the partial pressure of the other gases still remain constant even if carbon dioxide was removed?

Yeah. This might not be too satisfying an answer, so perhaps you can work it through mathematically.

If we now have 0.75 times the original number of total moles, but the same temperature/volume, what is the new total pressure in terms of the original total pressure? What is the new mole fraction of each constituent gas? What is then the new partial pressure of each?

Also have a go algebraically, and see how the cancellation occurs!

 

[FactChecker]

Ok, I'll ask a rookie question.
When you figured the partial pressures of the component gases, you made no use of their molecular weights. Is that correct? I thought that the volumes gave the number of molecules using Avogadro's number and then the partial pressures were calculated from the molecular weights. Am I wrong about that?

 

[etotheipi]

Ok, I'll ask a rookie question.
When you figured the partial pressures of the component gases, you made no use of their molecular weights. Is that correct? I thought that the volumes gave the number of molecules using Avogadro's number and then the partial pressures were calculated from the molecular weights. Am I wrong about that?

I don't think molecular weights play a role here for ideal gases at least; one mole of any ideal gas will occupy the same volume at a given temperature and pressure. I think this is Avogadro's Law.

 

[FactChecker]

I don't think molecular weights play a role here for ideal gases at least; one mole of any ideal gas will occupy the same volume at a given temperature and pressure. I think this is Avogadro's Law.

Ok, I stand corrected. I interpret that to mean that the pressure in Avagadro's Law is partial pressure and is proportional to the volume.

 

[ianc1339]

Yeah. This might not be too satisfying an answer, so perhaps you can work it through mathematically.

If we now have 0.75 times the original number of total moles, but the same temperature/volume, what is the new total pressure in terms of the original total pressure? What is the new mole fraction of each constituent gas? What is then the new partial pressure of each?

Also have a go algebraically, and see how the cancellation occurs!

Does that mean the partial pressures of the gases decrease by 25% ? So do I do:

$N_2: 3.0\times 10^4\times 0.75=2.25\times 10^4 Nm^{-2}\\$
$O_2: 2.6\times 10^4\times 0.75=1.95\times 10^4 Nm^{-2}\\$

I don't have the answers with me so I don't know if my answer is correct

 

[etotheipi]

Does that mean the partial pressures of the gases decrease by 25% ? So do I do:

Not quite; initially $P_1 = \frac{n_1 RT}{V}$, where $n_1$ is the total number of gaseous moles and $P_1$ is the total initial pressure.

After the CO₂ is removed, we now have $n_2 = 0.75 n_1$ total moles. So our new total pressure $P_2 = \frac{n_2 RT}{V} = 0.75 \frac{n_1 RT}{V} = 0.75 P_1$.

But our new mole fraction of, say nitrogen, has increased to $\frac{0.4}{0.75}$! So what must be our new partial pressure of nitrogen?

 

[etotheipi]

Ok, I stand corrected. I interpret that to mean that the pressure in Avagadro's Law is partial pressure and is proportional to the volume.

I was a little sloppy, technically I think Avogadro's law says that for any ideal gas $V \propto n$ with the same constant of proportionality for any choice of gas. The key upshot is that we can ignore the masses of each gas in our mixture and treat them all "as equals".

For a gas at constant T/V, we know that $P \propto n$ or $P = kn$. If we have gases A, B, C... then the total pressure will be $P_A + P_B + P_C + \dots = k(n_A + n_B + n_C + \dots) = kn_T = P_T$.

If we want to reverse the process to figure out a partial pressure, it suffices to compute the mole fraction $x_Q$ of gas $Q$ and multiply it by the total pressure, i.e, $x_Q P_T = \frac{n_Q}{n_T} P_T = n_Q \frac{P_T}{n_T} = kn_Q = P_Q$.

 

[ianc1339]

Not quite; initially $P_1 = \frac{n_1 RT}{V}$, where $n_1$ is the total number of gaseous moles and $P_1$ is the total initial pressure.

After the CO₂ is removed, we now have $n_2 = 0.75 n_1$ total moles. So our new total pressure $P_2 = \frac{n_2 RT}{V} = 0.75 \frac{n_1 RT}{V} = 0.75 P_1$.

But our new mole fraction of, say nitrogen, has increased to $\frac{0.4}{0.75}$! So what must be our new partial pressure of nitrogen?

Oh, so the partial pressure of nitrogen is:

$0.75\times 7.4\times 10^4\times\frac{0.4}{0.75}=2.96\times 10^4 Nm^{-2}$

And since volume is proportional to number of moles, this new mole fraction can be established using the volume percentages?

 

[etotheipi]

Oh, so the partial pressure of nitrogen is:

$0.75\times 7.4\times 10^4\times\frac{0.4}{0.75}=2.96\times 10^4 Nm^{-2}$

That's your answer! Note that it's the same as what you originally had before you removed the CO₂, save for a difference in how you've rounded them.

 

[FactChecker]

I was a little sloppy, technically I think Avogadro's law says that for any ideal gas $V \propto n$ with the same constant of proportionality for any choice of gas. The key upshot is that we can ignore the masses of each gas in our mixture and treat them all "as equals".

Thanks. That makes it clear. I think I was confused by memories of old problems where the mixture was given in weight. Giving the mixture in terms of volumes makes it nice.

 

[Chestermiller]

Volume percentage typically means mole percentage. It has never made much sense to me, but that's the way it goes.

 

[chemisttree]

Volume percentage typically means mole percentage. It has never made much sense to me, but that's the way it goes.

Yes, for an ideal gas it is, provided partial pressures and volume don’t change. Volume is related to moles by a constant.

V = moles (C)
where C = RT/P (T &P are constant)

P in this case is the partial pressure.

This is one of those problems designed to test understanding of partial pressure and its relation to volume percent and mole fraction. Solved by inspection.

This explanation is for the OP. You don’t need me to tell you that Chestermiller!