15.5: Universal Gas Law
Lesson Objectives
The student will:
 solve problems using the universal gas law, \begin{align*}PV = nRT\end{align*}
PV=nRT .  state Avogadro’s law of equal molecules in equal volumes.
 calculate molar mass (M) from \begin{align*}M = gRT/PV\end{align*} given mass, temperature, pressure, and volume.
Vocabulary
 Avogadro's law
 universal gas law
 universal gas law constant (R)
Introduction
The individual gas laws (Boyle's, Charles's, and GayLussac's) and the combined gas law all require the quantity of gas remain constant. The universal gas law (sometimes called the ideal gas law) allows us to make calculations when there are different quantities of gas.
Avogadro’s Law
Avogadro’s law postulates that equal volumes of gas with the same temperature and pressure contain the same number of molecules. Mathematically, this would be written as: \begin{align*}V = k_4n\end{align*}, where \begin{align*}n\end{align*} represents the number of moles of molecules and \begin{align*}k_3\end{align*} represents a constant. This law was known as Avogadro’s hypothesis for the first century of its existence. Since Avogadro's hypothesis can now be demonstrated mathematically, it was decided that it should be called a law instead of a hypothesis. If we think about the definitions of the volumes, pressures, and temperatures for gases, we can also develop Avogadro’s conclusion.
Suppose we have a group of toy robots that are all identical in strength. They do not have the same size or look, but they all have exactly the same strength at an assigned task. We arrange a tugofwar between groups of robots. We arrange the rope so that we can see one end of the rope, but the other end disappears behind a wall. On the visible end, we place eight robots to pull. On the other end, an unknown number of robots will pull. When we say “go,” both sides pull with maximum strength, but the rope does not move. How many robots are pulling on the hidden end of the rope? Since each robot pulls with the exact same strength, to balance the eight robots we know are on our end, there must be eight robots on the other end. We can think of molecules as robots by recognizing that molecules at the same temperature have exactly the same striking power upon collision. As a result, equal volumes of gas with equal pressures and temperatures must contain an equal number of molecules.
In the early 1800s, the first attempts to assign relative atomic weights to the atoms were accomplished by assigning hydrogen to have an atomic mass of \begin{align*}1.0.\end{align*} and by decomposing compounds to determine the mass ratios in the compounds. Some of the atomic weights found this way were accurate, but many were not. In the 1860s, Stanislao Cannizzaro refined the process of determining relative atomic weights by using Avogadro’s law. If gas \begin{align*}X\end{align*} and gas \begin{align*}Y\end{align*} were heated to the same temperature, placed in equal volume containers under the same pressure (gas would be released from one container until the two containers had the same pressure), Avogadro’s conditions would be present. As a result, Cannizzaro could conclude that the two containers had exactly the same number of molecules. The mass of each gas was then determined with a balance (subtracting the masses of the containers), and the relationship between the masses would be the same as the mass relationship of one molecule of \begin{align*}X\end{align*} to one molecule of \begin{align*}Y\end{align*}. That is, if the total mass of gas \begin{align*}X\end{align*} was \begin{align*}10.\ \mathrm{grams}\end{align*} and the total mass of gas \begin{align*}Y\end{align*} was \begin{align*}40.\ \mathrm{grams}\end{align*}, then Cannizzaro knew that one molecule of \begin{align*}Y\end{align*} must have four times as much mass as one molecule of \begin{align*}X\end{align*}. If an arbitrary value such as \begin{align*}1.0\end{align*} dalton was assigned as the mass of gas \begin{align*}X\end{align*}, then the mass of gas \begin{align*}Y\end{align*} on that same scale would be \begin{align*}4.0\end{align*} daltons.
The Universal Gas Law Constant
We have considered three laws that examine how the volume of a gas depends on pressure, temperature, and number of moles of gas (GayLussac's law requires a constant volume).
 Boyle’s law: \begin{align*}V = \frac {k_1} {P}\end{align*} at constant temperature and constant moles of gas, where \begin{align*}k_1\end{align*} is a constant.
 Charles’s law: \begin{align*}V = k_2T\end{align*} at constant pressure and constant moles of gas, where \begin{align*}k_2\end{align*} is a constant.
 Avogadro’s law: \begin{align*}V = k_4n\end{align*} at constant temperature and pressure, where \begin{align*}n\end{align*} is the number of moles of gas and \begin{align*}k_3\end{align*} is a constant.
These three relationships can be combined to form the expression \begin{align*}PV = nRT\end{align*}, where \begin{align*}R\end{align*} is the combination of the three constants. The equation is called the universal gas law (or the ideal gas law), and \begin{align*}R\end{align*} is called the universal gas law constant. When the pressure is expressed in atmospheres and the volume in liters, \begin{align*}R\end{align*} has the value \begin{align*}0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{1} \cdot \mathrm{K}^{1}\end{align*}. You can convert the value of \begin{align*}R\end{align*} into values for any set of units for pressure and volume. Moles, of course, always have the unit moles, while the temperature must always be Kelvin.
In our analysis of gas behavior, we have used a pair of assumptions that are not always true. We have assumed that the volume of the actual molecules in a gas is insignificant compared to the volume of the empty space between molecules. We have treated the molecules as if they were geometric points that took up no space. For most gases, this assumption will be true most of the time, and the gas laws work well. If, however, a gas is highly compressed (at very high pressure), the molecules will be pushed together very closely, with much of the empty space between molecules removed. Under such circumstances, the volume of the molecules becomes significant, and some calculations with the gas laws will be slightly off.
Another assumption that we have used is that the molecules are not attracted to each other so that every collision is a perfectly elastic collision. In other words, we have assumed that no energy is lost during the collision. This assumption also works well most of the time. Even though the molecules do have some attraction for each other, usually the temperature is high enough that the molecular motion readily overcomes any attraction and the molecules move around as if there were no attraction. If, however, we operate with gases at low temperatures (temperatures near the phase change temperature of the gas), the molecular attractions have enough effect to cause our calculations to be slightly off. If a gas follows the ideal gas laws, we say that the gas behaves ideally. Gases behave ideally at low pressure and high temperature. At low temperatures or high pressures, gas behavior may become nonideal. As a point of interest, if you continue your study of chemistry, you will discover that there is yet another gas law equation for these nonideal situations.
Example:
A sample of nitrogen gas \begin{align*}(\text{N}_2)\end{align*} has a volume of \begin{align*}5.56\ \mathrm{liters}\end{align*} at \begin{align*}0^\circ \text{C}\end{align*} and a pressure of \begin{align*}1.50\ \mathrm{atm}\end{align*}. How many moles of nitrogen are present in this sample?
Solution:
Step 1: Assign known values to the appropriate variable.


 \begin{align*} \begin{array}{l} P = 1.50\ \mathrm{atm} \\ V = 5.56\ \mathrm{L} \\ n = \mathrm{unknown} \\ T = 0^\circ \text{C} + 273 = 273\ \mathrm{K} \\ R = 0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{1} \cdot \mathrm{K}^{1} \end{array} \end{align*}

Step 2: Solve the combined gas law for the unknown variable.
\begin{align*}PV = nRT \ \ \ \ \ \text{so} \ \ \ \ \ n = \frac {PV} {RT}\end{align*}
Step 3: Substitute the known values into the formula and solve for the unknown.
\begin{align*}n = \frac {PV} {RT} = \frac {(1.50\ \text{atm}) \cdot (5.56 \ \text{L})} {(0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{1} \cdot \mathrm{K}^{1}) \cdot (273\ \text{K})} = 0.372\ \text{mol}\end{align*}
Example:
\begin{align*}2.00\ \mathrm{moles}\end{align*} of methane gas \begin{align*}(\text{CH}_4)\end{align*} are placed in a rigid \begin{align*}5.00\ \mathrm{liter}\end{align*} container and heated to \begin{align*}100.^\circ \text{C}\end{align*}. What pressure will be exerted by the methane?
Solution:
Step 1: Assign known values to the appropriate variable.


 \begin{align*} \begin{array}{l} P = \mathrm{unknown} \\ V = 5.00 \ \mathrm{L} \\ n = 2.00 \ \mathrm{moles} \\ T = 100.^\circ\text{C} + 273 = 373\ \text{K} \\ R = 0.08206 \ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{1} \cdot \mathrm{K}^{1} \end{array} \end{align*}

Step 2: Solve the combined gas law for the unknown variable.
\begin{align*}PV = nRT \ \ \ \ \ \text{so} \ \ \ \ \ n = \frac {PV} {RT}\end{align*}
Step 3: Substitute the known values into the formula and solve for the unknown.
\begin{align*}P = \frac {nRT} {V} = \frac {(2.00\ \text{mol}) \cdot (0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{1} \cdot \mathrm{K}^{1}) \cdot (373\ \text{K})} {(5.00\ \text{L})} = 12.2\ \text{atm}\end{align*}
Example:
A sample gas containing \begin{align*}0.300\ \mathrm{moles}\end{align*} of helium at a pressure of \begin{align*}900.\ \mathrm{torr}\end{align*} is cooled to \begin{align*}15^\circ \text{C}\end{align*}. What volume will the gas occupy under these conditions?
Note: If we are to use \begin{align*}0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{1} \cdot \mathrm{K}^{1}\end{align*} for the value of \begin{align*}R\end{align*}, then the pressure must be converted from torr to atm.
Solution:
Step 1: Assign known values to the appropriate variable.


 \begin{align*} \begin{array}{l} P = (900. \ \text{torr})\cdot(\frac{1.00 \ \text{atm}} {760. \ \text{torr}}) = 1.18 \ \text{atm} \\ V = \text{unknown} \\ N = 0.300\ \text{mole} \\ T = 15^\circ \text{C} + 273 = 288\ \text{K} \\ R = 0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{1} \cdot \mathrm{K}^{1}\end{array} \end{align*}

Step 2: Solve the combined gas law for the unknown variable.
\begin{align*}PV = nRT \ \ \ \ \ \text{so} \ \ \ \ \ V = \frac {nRT} {P}\end{align*}
Step 3: Substitute the known values into the formula and solve for the unknown.
\begin{align*}V = \frac {nRT} {P} = \frac {(0.300\ \text{mol}) \cdot(0.08206\ \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{1} \cdot \mathrm{K}^{1}) \cdot (373\ \text{K})} {(1.18\ \text{atm})}= 7.08\ \text{liters} \end{align*}
For a video example of solving an ideal gas law problem (4h), see http://www.youtube.com/watch?v=JEsfU7ogbVQ (4:44).
Molar Mass and the Universal Gas Law
The universal gas law can also be used to determine the molar mass of an unknown gas provided the pressure, volume, temperature, and mass are known. The number of moles of a gas, \begin{align*}n\end{align*}, can be expressed as grams/molar mass. If we substitute \begin{align*}g/M\end{align*} for \begin{align*}n\end{align*} in the universal gas law, we get \begin{align*}PV = (\frac {g} {M})RT\end{align*} which can be rearranged to \begin{align*}\ \mathrm{M} = \frac{gRT} {PV}\end{align*}.
Example:
\begin{align*}20.0\end{align*} grams of an unknown gas occupy \begin{align*}2.00 \ \text{L}\end{align*} under standard conditions. What is the molar mass of the gas?
Solution:
\begin{align*}\text{molar mass} = \frac {gRT} {PV} = \frac {(20.0\ \text{g}) \cdot (0.08206\ \text{L} \cdot \text{atm} \cdot \text{mol}^{1} \cdot \ \text{K}^{1}) \cdot (273\ \text{K})} {(1.00\ \text{atm}) \cdot (2.00\ \text{L})} = 224\ \text{g/mol}\end{align*}
Density and the Universal Gas Law
The density of a gas under a particular set of conditions is the mass of the sample of gas divided by the volume occupied at those conditions, \begin{align*}D = m/V\end{align*}. In the universal gas law, both the mass of the sample of gas and the volume it occupies are represented. We can substitute density for \begin{align*}g/V\end{align*} where it appears in the equation and produce an equation that contains density instead of mass and volume. For example, in the equation \begin{align*}M = \frac{gRT} {PV}\end{align*}, mass appears in the numerator and volume appears in the denominator. We can substitute \begin{align*}D\end{align*} for those two variables and obtain the equation \begin{align*}M = DRT/P\end{align*}.
Example:
The density of a gas was determined to be \begin{align*}1.95\ \mathrm{g/mL}\end{align*} at \begin{align*}1.50\ \mathrm{atm}\end{align*} and \begin{align*}27^\circ \text{C}\end{align*}. What is the molar mass of the gas?
Solution
\begin{align*}\text{molar mass} = \frac {DRT} {P} = \frac {(1.95\ \text{g/L}) \cdot (0.08206\ \text{L} \cdot \text{atm} \cdot \text{mol}^{1} \cdot \text{K}^{1}) \cdot (300.\ \text{K})} {(1.50\ \text{atm})} = 32.0\ \text{g/mol}\end{align*}
Lesson Summary
 Avogadro’s law states that equal volumes of gases under the same temperature and pressure contain equal numbers of molecules.
 The universal gas law, also known as the ideal gas law, relates the pressure, volume, temperature, and number of moles of gas: \begin{align*}PV = nRT.\end{align*}
 The universal gas law assumes that the gas molecules are geometric points that take up no space and that they undergo perfectly elastic collision.
 The universal gas law can also be used to determine the molar mass and the density of an unknown gas.
Further Reading / Supplemental Links
This video reviews how to perform ideal gas law (universal gas law) calculations.
Review Questions
 What conditions of temperature and pressure cause gases to deviate from ideal gas behavior?
 What volume will \begin{align*}2.00\ \mathrm{moles}\end{align*} of hydrogen gas occupy at \begin{align*}2.62 \ \mathrm{atm}\end{align*} of pressure and \begin{align*}300.^\circ \text{C}\end{align*}?
 How many moles of gas are required to fill a volume of \begin{align*}8.00\ \mathrm{liters}\end{align*} at \begin{align*}2.00\ \mathrm{atm}\end{align*} and \begin{align*}273\ \mathrm{K}\end{align*}?
 What is the molar mass of a gas if its density is \begin{align*}1.30\ \mathrm{g/L}\end{align*} at STP?
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