Discussing gas density is slightly more complex than discussing solid/liquid density. Since gas volume is VERY responsive to temperature and pressure, these two factors must be included in EVERY gas density discussion.
By the way, solid and liquid volumes are responsive to temperature and pressure, but the response is so little that it can usually be ignored in introductory classes.
So, for gases, we speak of "standard gas density." This is the density of the gas (expressed in grams per liter) at STP. If you discuss gas density at any other set of conditions, you drop the word standard and specify the pressure and temperature. Also, when you say "standard gas density," you do not need to add "at STP." STP is part of the definition of the term. It does no harm to say "standard gas density at STP," it's just a bit redundant.
You can calculate the standard gas density fairly easily. Just take the mass of one mole of the gas and divide by molar volume.
For example, using nitrogen, we would have:
28.014 g mol¯1 / 22.414 L mol¯1 = 1.250 g/L
Remember, nitrogen is diatomic, N2, not N.
For water, we have:
18.015 g mol¯1 / 22.414 L mol¯1 = 0.8037 g/L
You could see this used: "ideal standard gas density." The behavior of "real" gases diverges from predictions based on ideal conditions. Small gases like H2 at high temperatures approach ideal behavior almost exactly while larger gas molecules (NH3, for example) at low temperatures diverge the greatest amount. These "real" gas differences are small enough to ignore right now, but in later classes they will become important.
The official IUPAC unit for gas density is kg/m3 (not g/L). However, it turns out that one kg/m3 equals one g/L. Here is a brief video explaining the conversion.
One place teachers like to bring gas density into play is when you calculate a molar mass of a gas using PV = nRT.
Example #1: The density of a gas is measured at 1.853 g / L at 745.5 mmHg and 23.8 °C. What is its molar mass?
1) Plug values into PV = nRT:
(745.5 mmHg/760.0 mmHg/atm) (1.000 L) = (n) (0.08206 L atm / mol K) (296.8 K)
The (745.5/760) term converts mmHg to atm. Celsius has been converted to Kelvin
n = 0.0402753 mol (I'll keep a few guard digits.)
2) Divide grams by moles to get the molar mass:
1.853 g / 0.0402753 mol = 46.01 g/mol
Comment: This solution exploits a rearrangement of the ideal gas law.
1) Write an equality for the number of moles:
mass n = ––––––––– molar mass
2) Substitute into the idea gas law (along with a bit of rearranging):
mass RT P = ––––––––– x –––– molar mass V
3) Write an equality for density:
mass d = ––––– V
4) Substitute and rearrange:
dRT molar mass = ––––– P
5) Plug numbers in:
(1.853 g / 1.000 L) (0.08206 L atm / mol K) (296.8 K) molar mass = –––––––––––––––––––––––––––––––––––––––––––– = 46.01 g/mol (745.5 mmHg / 760.0 mmHg/atm)
Example #2: What is the molar mass of a gas which has a density of 0.00249 g/mL at 20.0 °C and 744.0 mm Hg?
Solution: Convert mmHg to atm (744.0/760.0) and °C to K (20.0 + 273.15). Use 0.001 L (which is 1 mL converted to liters). Plug into PV = nRT and solve for n (the value of which is calculated to be 4.069 x 10¯5 mol).
Then, divide 0.00249 g by the moles just calculated for an answer of 61.2 g/mol.
Please note that I used 273.15 (rather than 273) for the Celsius to Kelvin conversion. Some teachers require the use of 273.15. The ChemTeam always felt that 273 was good enough, but some teachers disagree. Yours may be one of those people (Are teachers people?).
Example #3: Anhydrous aluminum chloride sublimes at high temperatures. What density will the vapor have at 225 degrees Celsius and 0.939 atm of pressure?
1) Use PV = nRT to determine moles of gas present in vapor:
(0.939 atm) (1.00 L) = (n) (0.08206) (498 K)
n = 0.0229776 mol
I assumed 1.00 L because gas densities are measured in g/L.
2) Get grams of AlCl3 in the calculated moles:
0.0229776 mol x 133.34 g/mol = 3.06 g (to three sig figs)
3) Get density:
3.06 g / 1.00 L = 3.06 g/L
I could have assumed any volume I wanted back in the PV = nRT calc. However, I would have then divided the grams by that volume in this last step and wound up with 3.06.
Example #4: Air is a mixture of 21% oxygen gas and 79% nitrogen gas (neglect minor components and water vapor). What is the density of air at 30.0 °C and 1.00 atm?
Comment: For both solutions, we need the "molecular weight" of air:
MW(air) = (%O2 x MWO2) + (%N2 x MWN2)
(0.21 x 32) + (0.79 x 28) = 29 g/mol
1) Use PV = nRT and assume 1.00 L:
(1.00 atm) (1.00 L) = (n) (0.08206) (303 K)
n = 0.0402185 mol (of air at 303 K)
2) Calculate grams of air:
0.0402185 mol times 29 g/mol = 1.17 g
3) Determine density:
1.17 g / 1.00 L = 1.17 g/L
1) Density of air at STP
29 g/mol divided by 22.4 L/ mol = 1.29383 g/L (I'll keep some guard digits
2) As air is heated it gets less dense, so apply a temperature correction which makes the density smaller:
d (g/L) = 1.29383 g/L x (273 K / 303 K) = 1.16 g/L
This is actually a disguised Charles' Law correction:V1 / T1 = V2 / T2
(1.00 L / 273 K) = (x / 303 K)
x = 1.10989 L
1.29383 g / 1.10989 L = 1.16 g/L (to three sig figs)
If the problem had been written with a change in pressure from standard, we would have used the Combined Gas Law rather than Charles' Law.
Example #5: Two equal-volume balloons contain the same number of atoms. One contains helium and one contains argon. Comment on the relative densities of the balloons.
Here is a wrong answer: To determine density, you have to divide mass by volume. They are equal volume containers, and they contain the same number of atoms, then the densities of the balloons are equal."
Response to wrong answer: The answer above would be right, if the two sets of atoms had identical weights. However, an atom of argon weighs more than an atom of helium.
Therefore, at a condition of equal numbers of atoms, the argon balloon would be the denser one.
Note the implied Avogadro's Hypothesis in this question: equal volumes of gas contain equal numbers of atoms. Therefore, they MUST be at equal pressure and equal temperature.
Example #6: Calculate the density of radon at 292 K and 1.10 atm of pressure.
1) Begin with the ideal gas law
PV = nRT
2) Let n (number of moles) = m (mass) / M (molar mass). Substituting that gives:
PV = (m/M) RT
3) Rearranging this gives:
m/V = PM / RT
4) Since m/V = density, we have:
d = [(1.10 atm) (222 g/mol)] / [(0.08206 L atm / mol K) (292 K]
d = 10.2 g/L
Note: since radon has no stable isotopes, the mass number of its longest-lived isotope is used for the molar mass.
Example #7: Dry ice is solid CO2, with density of 1.56 g/mL. A small block of dry ice, measuring 1.25 cm x 1.90 cm x 2.80 cm, is sealed inside a balloon and allowed to sublime at room temperature, 22.0 degrees C, and 749 torr. What is the final volume of the balloon?
1) Determine volume of the dry ice:
1.25 cm x 1.90 cm x 2.80 cm = 6.65 cm3
2) Determine mass of dry ice:
(6.65 cm3) (1.56 g/mL) = 10.374 g
Since 1 cm3 = 1 mL, the units cancel.
3) Determine moles of dry ice:
10.374 g / 44.009 g/mol = 0.2357245 mol
4) Determine volume of balloon:
PV = nRT
(749 torr / 760 torr/atm) (V) = (0.2357245 mol) (0.08206 L atm / mol K) (295K)
V = 5.79 L
Example #8: What is the density of a 1.50 g sample of chlorine gas, confined in a flask, that exerts a pressure of 735.0 mmHg at 25.0 C?
1) Determine moles of chlorine gas, Cl2:
1.50 g / 70.906 g/mol = 0.021155 mol
2) Determine the volume the gas occupies:
PV = nRT
(735.0 mmHg / 760.0 mmHg/atm) (V) = (0.021155 mol) (0.08206 L atm mol¯1 K¯1) (298 K)
V = 0.53492 L
3) Determine the density:
1.50 g / 0.53492 L = 2.80 g/L (to three sig figs)
Example #9: What is the density of carbon tetrachloride vapor at 714 torr and 125 ¯C?
1) We need to first determine the moles of CCl4:
PV = nRT
(714 torr / 760. torr/atm) (1.00 L) = (n) (0.08206 L atm / mol K) (398 K)
n = 0.028765 mol (keep a few extra digits)
Note the use of 1.00 L
2) Determine how many grams 0.028765 mol of CCl4 is:
(0.028765 mol) (153.823 g/mol) = 4.42 g (to three sig figs)
3) Determine the density:
4.42 g / 1.00 L = 4.42 g/L
I deliberately used 1.00 L up above because I knew the density unit would be grams per one liter.
Example #10: Air has a density of 1.29 g/L at STP. Calculate its density on Pikes Peak (outside Colorado Springs, Colorado) where the pressure is 450. torr and the temperature is −19.0 °C.
1) Let assume the presence of 1.00 L of air and see what happens to the volume at the top of Pikes Peak. To do this, we use the Combined Gas Law:
P1V1 P2V2 ––––– = ––––– T1 T2
(760. torr) (1.00 atm) (450. torr) (x) ––––––––––––––––– = –––––––––––– 273 K 254 K
x = 1.57135 L
2) Calculate the new density:
1.29 g / 1.57135 L = 0.821 g/L (to three sig figs)
By the way, if you are ever in Colorado Springs, I recommend you make time for a visit to Cheyenne Mountain Zoo. You will not regret it.
Bonus Example: What is the density of nitrogen gas at 39.0 °C and 735.0 mmHg pressure?
1) Let's assume 1.00 mole of N2 is present. What volume does it occupy?
PV = nRT
(735.0 mmHg / 760.0 mmHg/atm) (V) = (1.00 mol) (0.08206 L atm / mol K) (312 K)
V = 26.47356 L
2) Determine the density:
28.014 g / 26.47356 L = 1.06 g/L (to three sig figs)
1) Given the following equations:
(a) PV = (moles)RT
(b) moles = mass / MM <--- where MM means molar mass
(c) d = mass / volume
2) Substitute (b) into (a):
PV = (mass / MM)RT
3) Multiply both sides by MM, divide both side by V:
PMM = (mass / V)RT
Note that mass/V is density, which is equation (c) in the list just above.
4) Divide by RT:
mass / V = PMM/RT
5) Let's solve the problem (use d for mass/V):
d = [(0.967105 atm) (28.014 g/mol)] / [(0.08206 L atm / mol K) (312 K)]
d = 1.06 g/L
1) You could use the combined gas law to get the new volume:
P1 = 760.0 mmHg P2 = 735.0 mmHg V1 = 22.414 L V2 = x T1 = 273 K T2 = 312 K
Note the presence of the molar volume. We are assuming 1.00 mole of the gas is present. At STP, this 1.00 mole would occupy molar volume, 22.414 L
V2 = 26.4873 L
28.014 g / 26.4873 L = 1.06 g/L
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