# Ideal_Gases

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12/4/2011
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```							                                        Ideal Gas Problems

Gases at low pressures obey the ideal gas law,

pV = nRT                                    (1)

where R is a constant (known as the gas constant) which has the value

R = 0.08206 L atm K-1 mol-1                                 (2)

Appropriate units to use for p, V, n, and T in the ideal gas equation are those used for R
above. Thus the pressure (p) should be in atm, the volume (V) in L, the temperature (T) in
degrees K, and the amount of gas (n) should be in moles. Useful conversion factors are

Pressure:      1 atm = 760 Torr = 760 mmHg = 101.3 kPa = 1.013 bar
Temperature:      K = 273 + oC
Volume:        1 L = 1000 mL = 1000 cm3

pV
Since       R , and R is a constant, it follows that
nT
p1V1 p2V 2
                                     (3)
n1 T1 n2 T2

where the subscript “1” represents one set of conditions, and the subscript “2” represents
another set of conditions. More specialized equations may be derived from Eq(3) when one
or more of the variables is held constant. For example, you can easily derive the familiar
equations given below in this manner (convince yourself that this works!):

Boyle’s law:     p1V1  p2V2 (obtained when n1 = n2 and T1 = T2)

V1          V2
Charles’s law:           T1         T2 (obtained when n1 = n2 and p1 = p2)

V1          V2
Avogadro’s Principle:            n1         n2 (obtained when T1 = T2 and p1 = p2)

STP
Often you will see gas volumes reported at STP (standard temperature and pressure).
STP is defined as T = 273 K (0oC) and p = 1 atm. Substitution of these values into Eq(1)
shows that the volume of 1 mol of any gas is approximately 22.4 L at STP. (You should
verify this for yourself using Eq(1)!).

Gas Density (d) and Molar Mass (M)
Rearranging the ideal gas equation and using the definitions of density d and molar mass
M gives

page 1
n(mol)    p                 g  n(mol)     g  pM
         and    d                M                           (4)
V(L)    RT                 L V(L)       mol  RT

Note: M (in italics) is molar mass in g/mol, while M (no italics) is molarity in g/L.

page 2
___________________________________________________________________________
1. What is the volume occupied by 35.4 g of nitrogen gas at 35oC and 735 Torr?

___________________________________________________________________________
2. A scuba diver inhales a lung-full (350 mL) of air at a depth of 33 ft where the pressure is
approximately 2.0 atm and the water temperature is 18oC. If the diver holds her breath (not a
good idea!!), what volume will the same amount of air occupy at sea level where the pressure
is approximately 1.0 atm and the air temperature is 35oC?

Answer: 741 mL (Note: You can either use Eq(3) or you can use Eq(1) twice to solve this problem.)
___________________________________________________________________________
3. (a) Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide gas. If 35.3
g of calcium carbonate reacts with 100 mL of 6.00 M HCl, how many liters of carbon dioxide
gas will be produced at 745 mmHg and 23.0oC?
Hints:
• Begin by writing the balanced equation for the reaction.
• This is a limiting reagent problem (why?), so you will next need to determine whether calcium carbonate or
hydrochloric acid is the limiting reagent.
• Once you have determined the identity of the limiting reagent, you can calculate the moles of carbon
dioxide produced.
• The last step is to find the volume of carbon dioxide using the ideal gas law.

(b) What volume of carbon dioxide gas would be obtained at STP?

page 3
Answers: (a) 7.43 L of carbon dioxide; (b) 6.72 L of carbon dioxide

page 4

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