1. Dalton’s Law of Partial Pressures
So far, the gas law problems you have been working on have mostly looked like this:
“What pressure would be exerted by 17.2 g of helium gas in a 14.3 L tank at 28°C?
Or “What mass of CO2 would be needed to fill a 450 mL flask at 1.2 atm and -25°C?”
Notice how these have always involved pure substances.
Most gas samples that we encounter in real life, however, are mixtures.

2. Dalton’s Law of Partial Pressures
Air is a good example. (go to Q1)
Most people equate air with pure oxygen, but the fact is that air is only about 21% oxygen.
Air is mostly nitrogen (78%) and a little less than 1% argon. (Q2)
This accounts for % of air.
The remaining 0.038% is mostly CO2 and a few other trace gases as shown in the figure at right.

3. Dalton’s Law of Partial Pressures
But what do these percents represent? (Q3)
Percent by mass? Percent by moles?
Usually, such statistics are referring to “percent by volume.”
But that is a little misleading. (Q4)
If you had a flask filled with a mixture of three different gases, each component wouldn’t have its own confined volume…

4. Dalton’s Law of Partial Pressures
Instead, of course, each component would occupy the entire volume of the flask...
So saying, for example, that a mix-ture is 5% argon by volume doesn’t really make sense.
What makes more sense is to talk about gaseous mixtures in terms of the “partial pressures” of the components. (Q5)

5. Dalton’s Law of Partial Pressures
Take for example the mixture shown at right…
Although they are hard to count while they are moving, there is one yellow particle (we’ll say it represent an argon atom),
four red particles (helium atoms)
and fifteen blue particles (krypton atoms).
= argon (Ar)
= helium (He)
= krypton (Kr)
And they are all moving around randomly hitting each other and the inside walls of the flask…

6. Dalton’s Law of Partial Pressures
Isn’t this a cool pressure gauge I just stuck to the side of the flask?
And let’s say the total pressure exerted by this gaseous mixture is 600 torr.
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We can talk about the argon, helium and krypton as each exerting their own partial pressure.
The partial pressure of the three gases all add up to the total pressure of 600 torr.
= argon (Ar)
= helium (He)
= krypton (Kr)
What fraction of the total pressure do you think each gas is exerting (Q6)

7. Dalton’s Law of Partial Pressures
Hopefully you realized that since there are so many more Kr atoms, Kr will be exerting a much greater fraction of the total pressure than the other two gases.
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In fact, since 15/20 (that’s ¾) of the particles are Kr atoms, then ¾ of the collisions will involve Kr atoms.
= argon (Ar)
= helium (He)
= krypton (Kr)
Thus, Kr will be exerting ¾ of the total pressure. And ¾ of 600 torr is 450 torr.

8. Dalton’s Law of Partial Pressures
See how the pressure drops to 450 torr
See how the pressure drops to 120 torr
Perhaps the simplest way to visualize Kr’s partial pressure is to imagine what the pressure in the flask would be if Kr were the only gas present:
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Likewise, if the helium were the only gas present, the pressure would only be 4/20 (1/5) of what it was originally. (Q7)
= argon (Ar)
= helium (He)
= krypton (Kr)
1/5 of 600 torr = 120 torr.

9. Dalton’s Law of Partial Pressures
And if the one argon atom were the only gas present… (Q8)
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It would exert a pressure of only 30 torr: 1/20 of 600 torr = 30 torr.
So, starting with the Ar at 30 torr… we can add back in the He at 120 torr:
That will make the combined pressure = 150 torr.
= argon (Ar)
= helium (He)
= krypton (Kr)
Then add back in the Kr at 450 torr:
That will restore our original total pressure for the mixture: = 600 torr

10. Dalton’s Law of Partial Pressures
So it should make sense that the greater the fraction of particles a gas has in a mixture, the higher that gas’s partial pressure will be.
But keep in mind that the number of particles in a gas sample is usually an astronomically large number and not very convenient to use.
So rather than deal with particle fractions, it’s usually easier to deal with mole fractions. (Q9)
Mole fractions and particle fractions are really the same thing if you think about it.

11. Dalton’s Law of Partial Pressures
Consider, for example, a mixture containing 2.56 x 1025 molecules of CH4 and 6.77 x 1025 atoms of Ar.
Adding these together gives a total of 9.33 x 1025 particles. (Q10)
The CH4’s particle fraction would be = 0.274
2.56 x 1025
9.33 x 1025
Let’s convert both the CH4 and Ar to moles:
2.56 x 1025 cules of CH4 x = mole CH4
1 mol CH4
6.022 x 1023cules
6.77 x 1025 atoms of Ar x = mole Ar
1 mol Ar
6.022 x 1023atoms
Adding these gives moles total (Q11).
CH4’s mole fraction = = 0.274
42.5
154.9
Same thing!

12. Dalton’s Law of Partial Pressures
If you were at the grocery store and got it in your mind that you wanted to determine the fraction of eggs that were “Large” (don’t ask me why!), you would not need to open every carton and count every egg. (Q12)
Instead, you could just count the dozens. Say you counted 23 dozen large out of a total of 78 doz. The dozen fraction that are “Large” would be = 0.295 23 78
It should be easy to see that the egg fraction that are “Large” would be this same number!

13. Dalton’s Law of Partial Pressures
Now, back to moles.
Recall that number of moles is a variable designated by the letter “n” (as in PV = nRT!).
The mole fraction for gas A in a mixture is usually designated by XA.
The equation for mole fraction is part over whole – just like percentage.
XA = nA
ntotal
For example, a mixture contains 3.42 moles of CO2 and 1.56 moles of Ne (Q13).
3.42
4.98
The mole fraction for CO2 would be: XCO2 = = (Q14)
The mole fraction for Ne would be: XNe = = 0.313
1.56
4.98
Notice how mole fractions are written as decimals and they always add to 1.00, just like percentages always add up to 100%.

14. Dalton’s Law of Partial Pressures
And if that mixture of 3.42 moles of CO2 and 1.56 moles of Ne were at a total pressure of 824 torr, then what would the two partial pressures be? (Q15)
The partial pressure of the CO2 (which by the way is designated as PCO2) would simply be the total pressure (PT) times CO2’s mole fraction (XCO2) =
PCO2 = PT·XCO2 = 824 torr x = 566 torr
There are two ways now that we can determine the partial pressure of the Ne (Q16).
One is simply to use the equation: PNe = PT·XNe = 824 torr x = 258 torr
The other way is simply to subtract PCO2 from PT :
824 torr – 566 torr = 258 torr.

15. Dalton’s Law of Partial Pressures
So, here are the three new equations we use when it comes to partial pressures gas problems:
* A’s mole fraction = moles A/total moles:
* A’s partial pressure = total pressure x A’s mole fraction: PA = PT·XA
* Partial pressures add up to the total pressure: PT = PA + PB + …
XA = nA
ntotal
So, let’s try a few problems. Work these out on the note sheet (links to Jing videos are given if you need them)
(Q 17) A mixture contains He and O2 at a total pressure of 3.47 atm. The partial pressure of the He is 2.10 atm. What is the partial pressure of the O2?
ANS: 1.37 atm

16. Dalton’s Law of Partial Pressures
(Q 18) moles of neon and 6.24 moles of chlorine are mixed together at a total pressure of 68.2 psi. What are the mole fractions and partial pressures of the two gases?
ANS: XNe = XCl2 = PNe = 16.0 psi PCl2 = 52.2 psi
(Q19) x 1021 CH4 molecules and 1.68 x 1022 Xe atoms are mixed in a tank at a total pressure of 316 kPa. What are the mole fractions and partial pressures of the two gases?
ANS: XCH4 = XXe = PCH4 = 76.2 kPa PXe = kPa

17. Dalton’s Law of Partial Pressures
(Q 20) g of He, 25.6 g of Ar and 46.7 g of Kr are mixed together in a L tank at a total pressure of 9.78 atm. What are the partial pressures of the three gases and what is the temp (°C) of the mixture?
ANS: PHe = 6.88 atm PAr = 1.55 atm PKr = 1.35 atm T = 316°C
(Q21) A canister contains moles of oxygen and an unknown amount of fluorine all at a total pressure of 875 torr and a temp of 28.3°C. Oxygen’s partial pressure is 687 torr. How many moles of fluorine are in the canister, and what is the volume of the canister?
ANS: nF2 = mol V = 20.4L

18. Dalton’s Law of Partial Pressures
(Q 22) What mass of nitrogen gas must be added to a 32.0 L tank containing 30.0 g of Ne at 37.0°C to bring the total pressure up to 225 kPa?
ANS: nN2 = 1.30 mol
(Q23) A tank contains 3.86 g of CO and 5.77 g of Ne. The partial pressure of the CO is 437 torr. What is Ne’s mole fraction, Ne’s partial pressure, and what is the total pressure in the tank?
ANS: XNe = PNe = 906 torr PT = 1343 torr

19. Dalton’s Law of Partial Pressures
Q(24) Conceptual: 2.5 moles of gas X and 1.5 moles of gas Y are mixed together in the same tank. Mark the following as Always true, Sometimes true or Never True.
X has a higher partial pressure than Y.
X has a greater volume than Y.
There’s a greater mass of X present than Y.
X is at a higher temperature (C) than Y.
X is at a higher temperature (K) than Y.
X particles have a greater average kinetic energy than Y particles
X particles are hitting the inside walls of the tank harder than Y particles.
X particles are hitting the inside walls of the tank more often than Y particles.
X particles are colliding with Y particles more often than Y particles are colliding with X.
X particles are more concentrated in the tank than Y particles. A N S