The units of pressure that are used are pascal (Pa), standard atmosphere (atm), and torr. 1 atm is the average pressure at sea level. It is normally used as a standard unit of pressure. The SI unit though, is the pascal. 101,325 pascals equals 1 atm.
For laboratory work the atmosphere is very large. A more convient unit is the torr. 760 torr equals 1 atm. A torr is the same unit as the mmHg (millimeter of mercury). It is the pressure that is needed to raise a tube of mercury 1 millimeter.
Boyle's law or the pressure-volume law states that the volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are constant.
Another way to describing it is saying that their products are constant.PV = C
When pressure goes up, volume goes down. When
volume goes up, pressure goes down.
From the equation above, this can be derived:
P_{1}V_{1} = P_{2}V_{2} = P_{3}V_{3} etc.
This equation states that the product of the
initial volume and pressure is equal to the product of the volume and pressure
after a change in one of them under constant temperature. For example,
if the initial volume was 500 mL at a pressure of 760 torr, when the volume
is compressed to 450 mL, what is the pressure?
Plug in the values:
P_{1}V_{1} = P_{2}V_{2}
(760 torr)(500 mL) = P_{2}(450
mL)
760 torr x 500 mL/450 mL = P_{2}
844 torr = P_{2}
The pressure is 844 torr after compression.
This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature.
V T
Same as before, a constant can be put in:
V / T = C
As the volume goes up, the temperature also
goes up, and vice-versa.
Also same as before, initial and final volumes
and temperatures under constant pressure can be calculated.
V_{1} / T_{1} = V_{2} / T_{2} = V_{3} / T_{3} etc.
This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
P T
Same as before, a constant can be put in:
P / T = C
As the pressure goes up, the temperature also
goes up, and vice-versa.
Also same as before, initial and final volumes
and temperatures under constant pressure can be calculated.
P_{1} / T_{1} = P_{2} / T_{2} = P_{3} / T_{3} etc.
Avogadro's Law: The Volume Amount Law
Gives the relationship between volume and amount when pressure and temperature are held constant. Remember amount is measured in moles. Also, since volume is one of the variables, that means the container holding the gas is flexible in some way and can expand or contract.
If the amount of gas in a container is increased, the volume increases. If the amount of gas in a container is decreased, the volume decreases.
V n
As before, a constant can be put in:
V / n = C
This means that the volume-amount fraction will always be the same value if the pressure and temperature remain constant.
V_{1} / n_{1} = V_{2} / n_{2} = V_{3} / n_{3} etc.
PV / T = C
As the pressure goes up, the temperature also
goes up, and vice-versa.
Also same as before, initial and final volumes
and temperatures under constant pressure can be calculated.
P_{1}V_{1} / T_{1} = P_{2}V_{2} / T_{2} = P_{3}V_{3} / T_{3} etc.
PV = nRT
Where n is the number of moles of the number of moles and R is a constant called the universal gas constant and is equal to approximately 0.0821 L-atm / mole-K.
EXAMPLE 1:
The balloon used by Charles in his historic
flight in 1783 was filled with about 1300 mole of H_{2}.
If the outside temperature was 21 ^{o}C and the atmospheric pressure
was 750 mm Hg, what was the volume of the balloon?
Quantity | Raw data | Conversion | Data with proper units |
P | 750 mm Hg | x 1 atm / 760 torr = | 0.9868 atm |
V | ? | ? | |
n | 1300 mole H_{2} | 1300 mole H_{2} | |
R | 0.0821 L-atm / mole-K | 0.0821 L-atm / mole-K | |
T | 21 ^{o}C | + 273 = | 294 K |
V = nRT / P ; V = (1300 mole)(0.0821 L-atm/mole-K)(294 K) / (0.9868 atm) = 31798.358 L = 3.2 x 10^{4} L.
PV = gRT / FW
or
FW = gRT / PV
This equation provides a convenient way of determining the formula weight of a gas if mass, temperature, volume and pressure of the gas are known (or can be determined).
EXAMPLE 2:
A 0.1000 g sample of a compound with the empirical
formula CHF_{2} is vaporized into a 256 mL flask at a temperature
of 22.3 ^{o}C. The pressure in the flask is measured to be
70.5 torr. What is the molecular formula of the compound?
Quantity | Raw data | Conversion | Data with proper units |
P | 70.5 torr | x 1 atm / 760 torr = | 0.0928 atm |
V | 256 mL | x 1 L / 1000 mL = | 0.256 L |
g | 0.1000 g sample | 0.1000 g | |
R | 0.0821 L-atm / mole-K | 0.0821 L-atm / mole-K | |
T | 22.3 ^{o}C | + 273 = | 295.3 K |
FW | ? | ? |
FW = gRT / PV ; V = (0.1000 g)(0.0821 L-atm / mole-K)(295.3 K) / (0.0928 atm)(0.256 L) = 102 g / mole
FW of CHF_{2} = 51.0 g / mole ; 102 / 51.0 = 2 ; C_{2}H_{2}F_{4}
If the equation above is rearranged further,
g / V = P x FW / RT = density
you get an expression of the density of the gas as a function of T and FW.
EXAMPLE 3:
Compare the density of He and air (average FW = 28 g/mole) at 25.0 ^{o}C and 1.00 atm.
d_{He}
= (4.003 g / mole)(1.00 atm) / (0.0821 L-atm / mole-K)(298 K) = 0.164 g
/ L
d_{air}
= (28.0 g / mole)(1.00 atm) / (0.0821 L-atm / mole-K)(298 K) = 1.14 g /
L
EXAMPLE 4:
Compare the density of air at 25.0 ^{o}C and air at 1807 ^{o}C and 1.00 atm.
d_{He}
= (28.0 g / mole)(1.00 atm) / (0.0821 L-atm / mole-K)(298 K) = 1.14 g /
L
d_{air}
= (28.0 g / mole)(1.00 atm) / (0.0821 L-atm / mole-K)(2080 K) = 0.164 g
/ L
Dalton's Law of Partial Pressures states that the total pressure of a mixture of nonreacting gases is the sum of their individual partial pressures.
P_{total} = P_{a} + P_{b} + P_{c} + ...
or
P_{total} = n_{a}RT / V + n_{b}RT / V + n_{c}RT / V + ...
or
P_{total} = (n_{a}+ n_{b}+ n_{c}+ ... )RT / V
The pressure in a flask containing a mixture of 1 mole of 0.20 mole O_{2} and 0.80 mole N_{2} would be the same as the same flask holding 1 mole of O_{2}.
Partial pressures are useful when gases are collected by bubbling through water (displacement). The gas collected is saturated in water vapor which contibutes to the total number of moles of gas in the container.
EXAMPLE 5:
A sample of H_{2} was prepared in the laboratory by the reaction:
Mg(s) + 2 HCl(aq) MgCl_{2}(aq) + H_{2}(g)
456 mL of gas was collected at 22.0 ^{o}C.
The total pressure in the flask was 742 torr. How many moles of H_{2}
were collected? The vapor pressure of H_{2}O at 22.0 ^{o}C
is 19.8 torr.
Quantity | Raw data | Conversion | Data with proper units |
P_{total} | 742 torr | ||
P_{H2O} | 19.8 torr | ||
P_{H2} | 742 torr - 19.8 torr = 722.2 torr | x 1 atm / 760 torr = | 0.9503 atm |
V | 456 mL | x 1 L / 1000 mL = | 0.456 L |
n | ? | ? | |
R | 0.0821 L-atm / mole-K | 0.0821 L-atm / mole-K | |
T | 22 ^{o}C | + 273 = | 295 K |
n_{H2} = P_{H2}V / RT ; n_{H2} = (0.9503 atm)(0.456 L) / (0.0821 L-atm / mole-K)(295 K) = 0.0179 mole H_{2}.
Non-Ideal Gases
Johannes Diderik van der Waals (1837-1923)
The ideal gas equation (PV=nRT) provides a valuable model of the relations between volume, pressure, temperature and number of particles in a gas. As an ideal model it serves as a reference for the behavior of real gases. The ideal gas equation makes some simplifying assumptions which are obviously not quite true. Real molecules do have volume and do attract each other. All gases depart from ideal behavior under conditions of low temperature (when liquefaction begins) and high pressure (molecules are more crowed so the volume of the molecule becomes important). Refinements to the ideal gas equation can be made to correct for these deviations.
In 1873 J. D. van der Waals proposed his equation, known as the van der Waals equation. As there are attractive forces between molecules, the pressure is lower than the ideal value. To account for this the pressure term is augmented by an attractive force term a/V^{2}. Likewise real molecules have a volume. The volume of the molecules is represented by the term b. The term b is a function of a spherical diameter d known as the van der Waals diameter. The van der Waals equation for n moles of gas is:
The values for a and b, below, are determined
empirically:
Molecule | a (liters^{2}-atm / mole^{2}) | b (liters / mole) |
H_{2} | 0.2444 | 0.02661 |
O_{2} | 1.360 | 0.03183 |
N_{2} | 1.390 | 0.03913 |
CO_{2} | 3.592 | 0.04267 |
Cl_{2} | 6.493 | 0.05622 |
Ar | 1.345 | 0.03219 |
Ne | 0.2107 | 0.01709 |
He | 0.03412 | 0.02370 |