University High School Chemistry: Molar Volume of a Gas

Molar Volume

Lesson Objectives

The student will:

apply the relationship that $1.00\ \mathrm{mole}$ of any gas at standard conditions will occupy $22.4\ \mathrm{L}$ .
convert gas volume at STP to moles and to molecules, and vice versa.
apply Dalton’s law of partial pressures to describe the composition of a mixture of gases.

Vocabulary

$&\mathbf{Dalton's \ law \ of \ partial \ pressures} & &\mathbf{diffusion} & &\mathbf{Graham's \ law} \\ &\mathbf{molar \ volume} & &\mathbf{partial \ pressure}$

Introduction

When molecules are at the same temperature, they have the same kinetic energy regardless of mass. In this sense, all molecules are created equal (as long as they are at the same temperature). They all exert the same force when they strike a wall. When trapped in identical containers, they will exert the exact same pressure. This is the logic of Avogadro's Law. If we alter this situation a little and say we have the same number of molecules at the same temperature exerting the same pressure, then similar logic allows us to conclude that these groups of molecules must be in equal-sized containers.

Molar Volume of Gases at STP

As you know, $1.00\ \mathrm{mole}$ of any substance contains $6.02 \times 10^{23}$ molecules. We also know that Avogadro's law states that equal volumes of gases under the same temperature and pressure will contain equal number of molecules. With a sort of reverse use of Avogadro’s Law, we know then that $6.02 \times 10^{23}$ molecules of any gas will occupy the same volume under the same temperature and pressure. Therefore, we can say that $1.00\ \mathrm{mole}$ of any gas under the same conditions will occupy the same volume. You can find the volume of $1.00\ \mathrm{mole}$ of any gas under any conditions by plugging the values into the universal gas law, $PV = nRT$ , and solving for the volume. We are particularly interested in the molar volume, or the volume occupied by $1.00\ \mathrm{mole}$ , of any gas under standard conditions. At STP, we find that:

$V = \frac {nRT} {P} = \frac {(1.00\ \text{mol}) \cdot (0.08206\ \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}) \cdot (273\ \text{K})} {(1.00\ \text{atm})} = 22.4\ \text{liters}$

This is the volume that $1.00\ \mathrm{mole}$ of any gas will occupy at STP. You will be using this number to convert gas volumes at STP to moles, and vice versa. You will use this value often enough to make it worth memorizing.

Example:

How many moles are present in $100.\ \mathrm{L}$ of hydrogen gas at STP?

Solution:

$(100.\ \text{L}) \cdot \left (\frac {1.00\ \text{mol}} {22.4\ \text{L}}\right ) = 4.46\ \text{liters}$

Example:

What volume will $100.\ \mathrm{grams}$ of methane gas $(\text{CH}_4, \mathrm{molar\ mass} = 16.0 \ \mathrm{g/mol})$ occupy at STP?

Solution:

$(100.\ \text{g}) \cdot \left (\frac {1.00\ \text{mol}} {16.0 \ \text{g}}\right ) \cdot \left (\frac {22.4\ \text{L}} {1.00\ \text{mol}}\right ) = 140. \ \text{liters}$

Dalton’s Law of Partial Pressures

The English chemist, John Dalton, in addition to giving us the atomic theory, also studied the pressures of gases when gases are mixed together but do not react chemically. His conclusion is known as Dalton’s law of partial pressures:

For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone.

This can be expressed mathematically as: $P_{TOTAL} = P_1 + P_2 + P_3 + \ldots$

In the last section, molecules were described as robots. The robots were identical in their ability to exert force. When a group of diverse molecules are at the same temperature, this analogy works well. In the case of molecules, they don’t all look alike, but they have the same average kinetic energy and therefore exert the same force when they collide. If we have $10,000$ gaseous molecules at $100^\circ \text{C}$ in a container that exert a pressure of $0.10\ \mathrm{atm}$ , then adding another $10,000$ gaseous molecules of any substance at the same temperature to the container would increase the pressure to $0.20\ \mathrm{atm}$ . If there are $200\ \mathrm{molecules}$ in a container (and at the same temperature), each single molecule is responsible for $1/200th$ of the total pressure. In terms of the force of collision, it doesn’t make any difference if the molecules have different masses. At the same temperature, the smaller molecules are moving faster than the larger ones, so the striking force is the same. It is this ability to exert force that is measured by temperature. We can demonstrate that different sized objects can have the same kinetic energy by calculating the kinetic energy of a golf ball and the kinetic energy of a bowling ball at appropriate velocities. Suppose a $100.\ \mathrm{gram}$ golf ball is traveling at $60.\ \mathrm{m/s}$ and a $7,200\ \mathrm{gram}$ bowling ball is traveling at $7.1\ \mathrm{m/s}$ .

$& KE_{\text{GOLF}} = \frac{1} {2} mv^2 = \frac{1} {2} (0.100\ \text{kg})(60.\ \text{m/s})^2 = 180\ \text{Joules}\\ & KE_{\text{BOWL}} = \frac{1} {2} mv^2 = \frac{1} {2} (7.2\ \text{kg})(7.1\ \text{m/s})^2 = 180 \ \text{Joules}$

The kinetic energies are the same. If these balls were to strike a pressure plate, they would exert exactly the same force. If the balls were invisible, we would not be able to determine which one had struck the plate.

Now suppose we have three one-liter containers labeled A, B, and C. Container A holds $0.20\ \mathrm{mole}$ of $\text{O}_2$ gas at $27^\circ text{C}$ , container $B$ holds $0.50\ \mathrm{mole}$ of $text{N}_2$ gas at $27^\circ text{C}$ , and container $C$ holds $0.30\ \mathrm{mole}$ of $He$ gas at $27^\circ text{C}$ . The pressure in each of the separate containers can be calculated with $PV = nRT$ .

$& P_{\text{O}_2} = \frac {nRT} {V} = \frac {(0.20 \ \text{mol}) \cdot (0.08206 \ \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}) \cdot (300.\ \text{K})} {(1.00\ \text{L})} = 4.92\ \text{atm} \\ & P_{\text{N}_2} = \frac {nRT} {V} = \frac {(0.50\ \text{mol}) \cdot (0.08206 \ \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}) \cdot (300.\ \text{K})} {(1.00\ \text{L})} = 12.3\ \text{atm} \\ & P_{\text{He}} = \frac {nRT} {V} = \frac {(0.30\ \text{mol}) \cdot (0.08206 \ \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}) \cdot (300. \ \text{K})} {(1.00 \ \text{L})} = 7.39\ \text{atm}$

The sum of these three pressures is $24.6\ \mathrm{atm}.$

If all three gases are placed in one of the containers at $27^\circ \text{C}$ , the pressure in the single container can also be calculated with $PV = nRT$ . (Remember that with all three gases in the same container, the number of moles is $0.20 + 0.50 + 0.30 = 1.00\ \mathrm{mole}$ .

$P_{\text{MIXTURE}} = \frac {nRT} {V} = \frac {(1.00\ \text{mol}) \cdot (0.08206 \ \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}) \cdot (300.\ \text{K})} {(1.00 \ \text{L})} = 24.6\ \text{atm}$

You can quickly verify that the total pressure in the single container is the sum of the individual pressures the gases would exert if they were alone in their own container. The pressure exerted by each of the gases in a mixture is called the partial pressure of that gas. Hence, Dalton’s law is known as the law of partial pressures.

This video presents a discussion of various relationships involved in Dalton's Law of partial pressures (4i): http://www.youtube.com/watch?v=vPWFjmX-1aI (6:22).

University High School Chemistry

Monday, April 29, 2013

Molar Volume of a Gas