Gases behave differently from the other two commonly studied states of
matter, solids and liquids, so we have different methods for treating
and understanding how gases behave under certain conditions. Gases,
unlike solids and liquids, have neither fixed volume nor shape. They
are molded entirely by the container in which they are held. We have
three variables by which we
measure gases: pressure,
volume, and temperature. Pressure is measured as force per area. The
standard SI unit for pressure is the pascal (Pa). However, atmospheres
(atm) and several other units are commonly used. The table below shows
the conversions between these units.
Volume is related between all gases by Avogadro's hypothesis,
which states: Equal volumes of gases at the same temperature and
pressure contain equal numbers of molecules. From this, we derive the molar volume of a gas (volume/moles of gas). This value, at 1 atm, and 0° C is shown below.
Where:
Vm = molar volume, in liters, the volume that one mole of gas occupies under those conditions
V=volume in liters
n=moles of gas
An equation that chemists call the Ideal Gas Law, shown below, relates the volume, temperature, and pressure of a gas, considering the amount of gas present.
PV = nRT
Where:
P=pressure in atm
T=temperature in Kelvins
R is the molar gas constant, where R=0.082058 L atm mol-1 K-1.
The Ideal Gas Law assumes several factors about the molecules of gas. The volume of the molecules is considered negligible compared to the volume of the container in which they are held. We also assume that gas molecules move randomly, and collide in completely elastic collisions. Attractive and repulsive forces between the molecules are therefore considered negligible.
Example Problem: A gas exerts a pressure of 0.892 atm in a 5.00 L container at 15°C. The density of the gas is 1.22 g/L. What is the molecular mass of the gas?
We can also use the Ideal Gas Law to quantitatively determine how changing
the pressure, temperature, volume, and number of moles of substance
affects the system. Because the gas constant, R, is the same for all gases in any situation, if you solve for
R in the Ideal Gas Law and then set two Gas Laws equal to one another, you have the Combined Gas Law:
Where:
values with a subscript of "1" refer to initial conditions
values with a subscript of "2" refer to final conditions
If you know the initial conditions of a system and want to determine the new pressure after you increase the volume while keeping the numbers of moles and the temperature the same, plug in all of the values you know and then simply solve for the unknown value.
Example Problem: A 25.0 mL sample of gas is enclosed in a flask at 22°C. If the flask was placed in an ice bath at 0°C, what would the new gas volume be if the pressure is held constant?
We can apply the Ideal Gas Law to solve several problems. Thus far, we
have considered only gases of one substance, pure gases. We also
understand what happens when several substances are mixed in one
container. According to Dalton's law of partial pressures,
we know that the total pressure exerted on a container by several
different gases, is equal to the sum of the pressures exerted on the
container by each gas.
Pt = P1 + P2 + P3 + ...
Where:
Pt=total pressure
P1=partial pressure of gas "1"
P2=partial pressure of gas "2"
and so on
Using the Ideal Gas Law, and comparing the pressure of one gas to the total pressure, we solve for the mole fraction.
Where:
X1 = mole fraction of gas "1"
And discover that the partial pressure of each the gas in the mixture is equal to the total pressure multiplied by the mole fraction.
Example Problem: A 10.73 g sample of PCl5 is placed in a 4.00
L flask at 200°C.
a) What is the initial pressure of the flask before any reaction takes place?
b) PCl5 dissociates according to the equation: PCl5(g) --> PCl3(g) + Cl2(g). If half of the total number of moles of PCl5(g) dissociates and the observed pressure is 1.25 atm, what is the partial pressure of Cl2(g)?
As we stated earlier, the shape of a gas is determined entirely by the container in which
the gas is held. Sometimes, however, the container may have small holes,
or leaks. Molecules will flow out of these leaks, in a process called effusion.
Because massive molecules travel
slower than lighter molecules, the rate of effusion is specific to each
particular gas.
We use Graham's law to represent the relationship between rates of effusion for two
different molecules. This relationship is equal to the square-root of the inverse of the
molecular masses of the two substances.
Where:
r1=rate of effusion in molecules per unit time of gas "1"
r2=rate of effusion in molecules per unit time of gas "2"
u1=molecular mass of gas "1"
u2=molecular mass of gas "2"
Previously, we considered only ideal gases, those that fit the assumptions of the ideal gas law. Gases, however, are never perfectly in the ideal state. All atoms of every gas have mass and volume. When pressure is low and temperature is low, gases behave similarly to gases in the ideal state. When pressure and temperature increase, gases deviate farther from the ideal state. We have to assume new standards, and consider new variables to account for these changes. A common equation used to better represent a gas that is not near ideal conditions is the van der Waals equation, seen below.
Where the van der Waals constants are:
a accounts for molecular attraction
b accounts for volume of molecules
The table below shows values for a and b of several different compounds and elements.
Practice Ideal Gas Law Problem:
2.00 g of hydrogen gas and 19.2 g of oxygen gas are placed in a 100.0 L container. These gases react to form H2O(g). The temperature is 38°C at the end of the reaction.
a) What is the pressure at the conclusion of the reaction?
b) If the temperature was raised to 77° C, what would the new pressure be in the same container? Ideal gas law solution. Practice Pressure Problem:
1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850°C according to the equation:
4 NH3(g) + 5 O2(g) --> 4 NO(g) + 6 H2O(g)
a) If the total pressure in the container is 5.00 atm, what are the partial pressures for the three gases remaining?
b) Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)?
Pressure solution.
Compressibility and Ideal Gas Approxim
| Units of Pressure | |
|---|---|
| 1 pascal (Pa) | 1 N*m-2 = 1 kg*m-1*s-2 |
| 1 atmosphere (atm) | 1.01325*105 Pa |
| 1 atmosphere (atm) | 760 torr |
| 1 bar | 105 Pa |
| Vm = |  n |
= 22.4 L at 0°C and 1 atm |
Vm = molar volume, in liters, the volume that one mole of gas occupies under those conditions
V=volume in liters
n=moles of gas
An equation that chemists call the Ideal Gas Law, shown below, relates the volume, temperature, and pressure of a gas, considering the amount of gas present.
P=pressure in atm
T=temperature in Kelvins
R is the molar gas constant, where R=0.082058 L atm mol-1 K-1.
The Ideal Gas Law assumes several factors about the molecules of gas. The volume of the molecules is considered negligible compared to the volume of the container in which they are held. We also assume that gas molecules move randomly, and collide in completely elastic collisions. Attractive and repulsive forces between the molecules are therefore considered negligible.
Example Problem: A gas exerts a pressure of 0.892 atm in a 5.00 L container at 15°C. The density of the gas is 1.22 g/L. What is the molecular mass of the gas?
| Answer: | |||||||||||
| PV = nRT | |||||||||||
| T = 273 + 15 = 228 | |||||||||||
| (0.892)(5.00) = n(.0821)(288) | |||||||||||
| n = 0.189 mol | |||||||||||
|
|||||||||||
| x = Molecular Weight = 32.3 g/mol | |||||||||||
n1T1 |
= | n2T2 |
values with a subscript of "2" refer to final conditions
If you know the initial conditions of a system and want to determine the new pressure after you increase the volume while keeping the numbers of moles and the temperature the same, plug in all of the values you know and then simply solve for the unknown value.
Example Problem: A 25.0 mL sample of gas is enclosed in a flask at 22°C. If the flask was placed in an ice bath at 0°C, what would the new gas volume be if the pressure is held constant?
| Answer: | ||||||||||
| Because the pressure and the number of moles are held constant, we do not need to represent them in the equation because their values will cancel. So the combined gas law equation becomes: |
||||||||||
|
||||||||||
|
||||||||||
| V2 = 23.1 mL | ||||||||||
Pt=total pressure
P1=partial pressure of gas "1"
P2=partial pressure of gas "2"
and so on
Using the Ideal Gas Law, and comparing the pressure of one gas to the total pressure, we solve for the mole fraction.
Pt |
= | nt RT/V |
= | nt |
= X1 |
Where:
X1 = mole fraction of gas "1"
And discover that the partial pressure of each the gas in the mixture is equal to the total pressure multiplied by the mole fraction.
| P1 = | nt |
Pt = X1Pt |
a) What is the initial pressure of the flask before any reaction takes place?
b) PCl5 dissociates according to the equation: PCl5(g) --> PCl3(g) + Cl2(g). If half of the total number of moles of PCl5(g) dissociates and the observed pressure is 1.25 atm, what is the partial pressure of Cl2(g)?
| Answer: | ||||||||||
|
||||||||||
| PV = nRT | ||||||||||
| T = 273 + 200 = 473 | ||||||||||
| P(4.00) = (.05146)(.0821)(473) | ||||||||||
| P = 0.4996 atm | ||||||||||
| b) | PCl5 | → | PCl3 | + | Cl2 | |||||
| Start: | .05146 mol | 0 mol | 0 mol | |||||||
| Change: | -.02573 mol | +.02573 mol | +.02573 mol | |||||||
| Final: | .02573 mol | .02573 mol | .02573 mol | |||||||
|
||||||||||
|
||||||||||
| PCl2 = .4167 atm | ||||||||||
|
= |   |
|
 |
Where:
r1=rate of effusion in molecules per unit time of gas "1"
r2=rate of effusion in molecules per unit time of gas "2"
u1=molecular mass of gas "1"
u2=molecular mass of gas "2"
Previously, we considered only ideal gases, those that fit the assumptions of the ideal gas law. Gases, however, are never perfectly in the ideal state. All atoms of every gas have mass and volume. When pressure is low and temperature is low, gases behave similarly to gases in the ideal state. When pressure and temperature increase, gases deviate farther from the ideal state. We have to assume new standards, and consider new variables to account for these changes. A common equation used to better represent a gas that is not near ideal conditions is the van der Waals equation, seen below.
|
P + |
|
|
|
|
- b | |
= RT |
Where the van der Waals constants are:
a accounts for molecular attraction
b accounts for volume of molecules
The table below shows values for a and b of several different compounds and elements.
| Species | a (dm6 bar mol-2) | b (dm3 mol-1) |
|---|---|---|
| Helium | 0.034598 | 0.023733 |
| Hydrogen | 0.24646 | 0.026665 |
| Nitrogen | 1.3661 | 0.038577 |
| Oxygen | 1.3820 | 0.031860 |
| Benzene | 18.876 | 0.11974 |
Practice Ideal Gas Law Problem:
2.00 g of hydrogen gas and 19.2 g of oxygen gas are placed in a 100.0 L container. These gases react to form H2O(g). The temperature is 38°C at the end of the reaction.
a) What is the pressure at the conclusion of the reaction?
b) If the temperature was raised to 77° C, what would the new pressure be in the same container? Ideal gas law solution. Practice Pressure Problem:
1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850°C according to the equation:
a) If the total pressure in the container is 5.00 atm, what are the partial pressures for the three gases remaining?
b) Using Graham's Law, what is the ratio of the effusion rates of NH3(g) to O2(g)?
Pressure solution.
Compressibility and Ideal Gas Approxim
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