The Ideal Gas Law

Outline

• Ideal Gas Behavior
  • Looking at the pieces: Pressure, Temperature, and Volume
  • The Ideal Gas Law
  • Applications
• Practice with Concepts
• Discussion Questions
• Optional Website Reading

Ideal Gas Behavior

Looking at the pieces: Pressure, Temperature, and Volume

First model: think about gases as collections of very small, very energetic billiard balls. They are constantly in motion. If you heat them up, they move more quickly. They bang against the sides of their container, and if the container has an opening, they fly out. If you mix two different gases, they will operate independently of each other, but combine their total effect on the universe of their container.

The three physical characteristics of a gas are volume, amount (measured usually in number of particles), and temperature. These characteristics are interrelated. If you change one, the others will change to compensate. Keep in mind:

Volume is measured in cubic length units or in volume units. One liter = 10³cm³. The number of molecules involved can be determined from the number of moles involved, which in turn is related to the mass of the gas involved. If you know the mass, you know the number of molecules.

The temperature is usually expressed as an absolute: in Kelvin. (Note that it is 5 Kelvin, but 5 degrees centigrade, 5 degrees fahrenheit). Recall from chapter one that T_K = T_C + 273.15. In other words, the size of the degree in Centigrade (or Celsius) is the same as a unit on the Kelvin scale, but the point of origin of the scale is offset by 273.15 units. Watch the significant figures in the problem to determine whether you need to take the .15 fraction into account.

Pressure is the force exerted by the gas particles on the sides of its container. It is in units of force/area, so pounds/square foot is a legitimate pressure. Normally we use the SI scale: one newton/on square meter is called a Pascal, after the French philosopher. We also use atmospheres, which is roughly 14.7 pounds/square inch (that's the weight of about 8 miles of Earth's atmosphere at sea level). You may be more familiar with barometric pressure, in which 1 atmosphere is 760 mm (about 29.92 inches) from weather reports. The weather reporter announces barometric pressure in inches, usually around 30. Falling barometric pressure indicates a low pressure region of air is moving in (usually a storm), rising barometric pressure usually indicates good weather. Some people can feel the change in pressure and sense that rain is coming.

Given pressure in pascal, you can convert it to atmospheres by the relation

1 atm = 1.01325 * 10⁵ pascal

and to millimeters of mercury by the relationship

1 atm = 760 mm Hg. Of course, you can invert the formulas as needed to get pascal or mm Hg from atmospheres, and even to get mm Hg from pascal:

1 pascal = 1/1.01325 atm = .9869 atm

1 mm Hg = 1/760 atm = .001316 atm

The Ideal Gas Law

The ideal gas law relates pressure, number of gas particles and temperature to each other. It is one of those relationships that took a long time to work out, partly because there are a number of factors involved. Robert Boyle showed that the pressure of a gas is inversely related to volume, at a constant temperature and amount of gas. Jacques Charles was able to show that volume of a gas is directly proportional to temperature, and moreover, that the volume of a gas would be zero at zero Kelvin. Joseph Gay-Lussac showed that if volume was a direct function of the number of molecules, then equal volumes of different gases all contained the same number of molecules. This paved the way for Avogadro to suggest that gases at the same temperature and pressure contained the same number of molecules (the 1811 suggestion was not widely accepted until 1860).

If we break the ideal gas law down into pieces, and assume that our molecules are perfect billiard balls, though, each piece makes sense.

Consider volume and temperature, and a situation where we want to hold pressure constant. We don't want the number of hits of gas molecules against the side of the container to change, and we measure them in hits/second /area, or force/area. If we increase the number of molecules in the container, we have more molecules hitting the sides of the container. In order to make the number of hits/area in a single unit of time remain constant, we have to increase the surface area of the container. Since surface area of a container of fixed shape (but obviously not size) is related to volume, increasing the volume of the container increases the surface area. We can write volume = some constant * number of molecules involved. Increasing the number of molecules directly affects the volume. The constant remains the same (remember we are holding pressure constant) regardless of the changes to volume or number of molecules: V = k * n, where V is volume, k is a constant including pressure in this case, and n is the number of molecules.

By the same reasoning, volume is directly proportional to temperature. If we increase the temperature, the molecules speed up, hitting the sides of the container more often. Increasing the temperature directly increases the volume: V = k * P. This time, k includes a constant absolute temperature.

Now suppose that we want to increase the pressure. To do that we either need to increase the number of molecules (more hits per unit area, because we have more to do the hitting) or increase the temperature (more hits per unit time, because what we have travels faster). Both these options assume we are holding the volume of the container and the other factor constant. That is, we change number of molecules, but not temperature and volume. We can also change pressure by decreasing the size of the container. This means there is less surface area (more hits per unit area) and the molecules don't have to travel as far between collisions with the surface area (more hits per unit time). This is an inverse relationship: as volume decreases, pressure increases: V = k/P, where k includes a constant number of molecules and temperature.

Eventually, all the pieces were put together in one astounding, simple, elegant equation:

$$\mathrm{PV}=\mathrm{nRT}$$

Scientists save the term "elegant" for those really wonderful formulas that describe complex situations with some (relatively) simple set of relationships. The ideal gas law is elegant.

Given this relationship, we can come up with a value for R, the equation constant. First of all, let's consider what happens at STP, that is, Standard Temperature (= 0.0 degrees Centigrade) and Pressure (1.00 atmosphere). A mole of any gas occupies about 22.4 liters (determined by many experiments). Putting all this into the equation for ideal gases and solving for R gives

$$R=\frac{\mathrm{PV}}{\mathrm{nT}}=1.00\mathrm{atm}\ast \frac{24.4L}{1.00\mathrm{mole}}\ast 273K=0.0821\frac{L\ast \mathrm{atm}}{\mathrm{Mole}\ast K}$$

We can use the constant now in the ideal gas law, regardless of the type of gas, pressure, temperature, or volume ranges involved.

$$\mathrm{PV}=(0.0821\frac{L\ast \mathrm{atm}}{\mathrm{Mole}\ast K})\mathrm{nT}$$

Applications

Now we can do some calculations. We can set up ratios between two different states of a gas. Suppose we change volumes, while holding temperature and number of molecules constant. In the each case, P₁V₁ and P₂V₂ = nRT. Since both P₁V₁ and P₂V₂ are equal to nRT, they are equal to each other $$P_1{V}_1=\mathrm{nRT}=P_2{V}_2$$ Solving for pressure on one side and volume on the other gives $$\frac{P_1}{P_2}=\frac{V_2}{V_1}$$

Knowing any three of the values will allow us to calculate the fourth. For example, what happens to the pressure in a can of gas when 75% of the gas escapes? Assume a starting pressure of 255 lbs/square inch.

First, solve for the relationship: $$P_{1}V=n_1\mathrm{RT}$$ $$P_{2}V=n_2\mathrm{RT}$$ Factoring for the constant parts gives $$\frac{P_1}{n_1}=\frac{\mathrm{RT}}{V}=\frac{P_2}{n_2}$$ We want the final pressure P₂, so we solve for that: $$P_2=\frac{n_2}{n_1}{P}_1$$

Note that we don't need the absolute number of molecules, only the ratio: n₂ = .25n₁. SO finally: $$P_2=\frac{(0.25)n_1}{n_1}(255\frac{\mathrm{lbs}}{\mathrm{sq}\mathrm{in}})=65\frac{\mathrm{lbs}}{\mathrm{sq}\mathrm{in}}$$

Because we lost so much gas out of the can, only 1/4 of the original pressure is exerted by the remaining gas.

We can get very fancy, and put everything about the gas law together with the other things we know about molecule. Given a sample of gas, we discover that it has a density of .00249 g/mL at 20 degrees C and 744 mm Hg. What is the molar mass (which we could use to identify the gas)?

We have to get the equation into the units of R: L*atm/mole*K. First convert the density: $$0.00249\frac{g}{\mathrm{mL}}\ast 1000\frac{\mathrm{mL}}{L}=2.49\frac{g}{L}$$

Then convert mm Hg into atmospheres: $$744\mathrm{mmHg}\ast \frac{1\mathrm{atm}}{760\mathrm{mmHg}}=0.979\mathrm{atm}$$

Now we can substitute into the ideal gas law, noting that molar mass (g/mole) is a function of the mass available divided by the number of molecules (expressed as moles):

$$\mathrm{PV}=\mathrm{mass}\ast \frac{\mathrm{RT}}{\mathrm{molar}\mathrm{mass}}\Rightarrow \mathrm{molar}\mathrm{mass}=\mathrm{mass}\frac{\mathrm{RT}}{\mathrm{PV}}$$

Substituting in values for the equation variables, $$\mathrm{molar}\mathrm{mass}=2.49g\ast [0.0821\frac{L\ast \mathrm{atm}}{\mathrm{mole}\ast K}]\ast \frac{(20+273.15)K}{1L\ast 0.979\mathrm{atm}}=61.38\frac{g}{\mathrm{mole}}$$

Adding this information to other information (like a reaction which allows us to determine the simplest formula of the gas) would allow us to determine the molecular formula for the gas.

Practice with the Concepts

Unit conversions

How would you set up a conversion for pascal to mm Hg?

Using PV=nRT

We can substitute values into the actual ideal gas law as well. Determine the pressure exerted by 15.0 moles of oxygen gas in a 50.0L container at 25 degrees C (about room temperature).

Determining molar mass from gas forms

A flask weights 52.693 g when empty, and 53.117 grams when filled with acetone vapor at 100 degrees C and 752 mm Hg. Assuming the volume of the flask is 226.2 mL (ascertained by filling it with water and weighing the total), can we find the molar mass of acetone?

First, solve the ideal gas law for molar mass: _M_ = mass*RT/PV. Then substitute:

(53.117-52.693g) * 0.0821(L*atm/mole*K) * (100 + 273)K / (752/760 atm)*(0.2262 L) = 58.0 g/mole. Done.

Discussion Questions

• Why is it necessary to use standard temperature and pressure in creating tables of gas information?
• How did Avogadro's theory make determining molecular formulae possible?

Optional Readings

The ChemTeam explains the mathematical background to Avogadro's law.

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