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# 15.6: Molar Volume

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• apply the relationship that 1.00 mole\begin{align*}1.00\ \mathrm{mole}\end{align*} of any gas at standard conditions will occupy 22.4 L\begin{align*}22.4\ \mathrm{L}\end{align*}.
• 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

• Dalton's law of partial pressure
• diffusion
• Graham's law
• molar volume
• 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 mole\begin{align*}1.00\ \mathrm{mole}\end{align*} of any substance contains 6.02×1023\begin{align*}6.02 \times 10^{23}\end{align*} 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 6.02×1023\begin{align*}6.02 \times 10^{23}\end{align*} molecules of any gas will occupy the same volume under the same temperature and pressure. Therefore, we can say that 1.00 mole\begin{align*}1.00\ \mathrm{mole}\end{align*} of any gas under the same conditions will occupy the same volume. You can find the volume of 1.00 mole\begin{align*}1.00\ \mathrm{mole}\end{align*} of any gas under any conditions by plugging the values into the universal gas law, PV=nRT\begin{align*}PV = nRT\end{align*}, and solving for the volume. We are particularly interested in the molar volume, or the volume occupied by 1.00 mole\begin{align*}1.00\ \mathrm{mole}\end{align*}, of any gas under standard conditions. At STP, we find that:

V=nRTP=(1.00 mol)(0.08206 Latmmol1K1)(273 K)(1.00 atm)=22.4 liters\begin{align*}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}\end{align*}

This is the volume that 1.00 mole\begin{align*}1.00\ \mathrm{mole}\end{align*} 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. L\begin{align*}100.\ \mathrm{L}\end{align*} of hydrogen gas at STP?

Solution:

(100. L)(1.00 mol22.4 L)=4.46 liters\begin{align*}(100.\ \text{L}) \cdot \left (\frac {1.00\ \text{mol}} {22.4\ \text{L}}\right ) = 4.46\ \text{liters}\end{align*}

Example:

What volume will 100. grams\begin{align*}100.\ \mathrm{grams}\end{align*} of methane gas (CH4,molar mass=16.0 g/mol)\begin{align*}(\text{CH}_4, \mathrm{molar\ mass} = 16.0 \ \mathrm{g/mol})\end{align*} occupy at STP?

Solution:

(100. g)(1.00 mol16.0 g)(22.4 L1.00 mol)=140. liters\begin{align*}(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}\end{align*}

## 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: PTOTAL=P1+P2+P3+\begin{align*}P_{TOTAL} = P_1 + P_2 + P_3 + \ldots\end{align*}

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\begin{align*}10,000\end{align*} gaseous molecules at 100C\begin{align*}100^\circ \text{C}\end{align*} in a container that exert a pressure of 0.10 atm\begin{align*}0.10\ \mathrm{atm}\end{align*}, then adding another 10,000\begin{align*}10,000\end{align*} gaseous molecules of any substance at the same temperature to the container would increase the pressure to 0.20 atm\begin{align*}0.20\ \mathrm{atm}\end{align*}. If there are 200 molecules\begin{align*}200\ \mathrm{molecules}\end{align*} in a container (and at the same temperature), each single molecule is responsible for 1/200th\begin{align*}1/200\mathrm{th}\end{align*} 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. gram\begin{align*}100.\ \mathrm{gram}\end{align*} golf ball is traveling at 60. m/s\begin{align*}60.\ \mathrm{m/s}\end{align*} and a 7,200 gram\begin{align*}7,200\ \mathrm{gram}\end{align*} bowling ball is traveling at 7.1 m/s\begin{align*}7.1\ \mathrm{m/s}\end{align*}.

KEGOLF=12mv2=12(0.100 kg)(60. m/s)2=180 JoulesKEBOWL=12mv2=12(7.2 kg)(7.1 m/s)2=180 Joules\begin{align*}& 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}\end{align*}

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 mole\begin{align*}0.20\ \mathrm{mole}\end{align*} of O2\begin{align*}\text{O}_2\end{align*} gas at 27C\begin{align*}27^\circ \text{C}\end{align*}, container B\begin{align*}B\end{align*} holds 0.50 mole\begin{align*}0.50\ \mathrm{mole}\end{align*} of N2\begin{align*}\text{N}_2\end{align*} gas at 27C\begin{align*}27^\circ \text{C}\end{align*}, and container C\begin{align*}C\end{align*} holds 0.30 mole\begin{align*}0.30\ \mathrm{mole}\end{align*} of He\begin{align*}\text{He}\end{align*} gas at 27C\begin{align*}27^\circ \text{C}\end{align*}. The pressure in each of the separate containers can be calculated with PV=nRT\begin{align*}PV = nRT\end{align*}.

PO2=nRTV=(0.20 mol)(0.08206 Latmmol1K1)(300. K)(1.00 L)=4.92 atm\begin{align*} 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} \end{align*}
PN2=nRTV=(0.50 mol)(0.08206 Latmmol1K1)(300. K)(1.00 L)=12.3 atm\begin{align*} 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} \end{align*}
PHe=nRTV=(0.30 mol)(0.08206 Latmmol1K1)(300. K)(1.00 L)=7.39 atm\begin{align*} 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}\end{align*}

The sum of these three pressures is \begin{align*}24.6\ \mathrm{atm}.\end{align*}

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

\begin{align*}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}\end{align*}

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).

## Graham’s Law of Diffusion

It was mentioned earlier that gases will spread out and occupy any and all space available. The spreading out and mixing of a substance is called diffusion. The small spaces available in liquid structure allow diffusion to occur only slowly. A drop of food coloring in a glass of water (without stirring) might take half an hour or more to spread through the water evenly. With a solid structure we would expect no diffusion or mixing at all, although there are stories of lead and gold bricks stacked together for many years that showed a few molecules have exchanged on the surfaces. Even if the story is true, diffusion in solids is negligible. In comparison, diffusion is very rapid in gases. Not even the presence of other gases offer much obstacle to gases. If someone opens a bottle of perfume or ammonia from across the room, it is only a matter of minutes before you smell it. The molecules evaporate from the liquid in the bottle and spread throughout the room quickly.

It turns out that diffusion is not the same for all gases. If bottles of strong smelling substances are opened at the same time from across the room, the odors will not reach you at the same time. If we pick the strong smelling substances ammonia \begin{align*}(\text{NH}_3)\end{align*} and acetone \begin{align*}(\text{C}_3\text{H}_6\text{O})\end{align*} and opened bottles of these substances, you will always smell the ammonia before the acetone. If we think about this situation a little more, we can probably predict this behavior. Since the two liquids are released in the same room, they are likely to be at the same temperature, so the molecules of both liquids will have the same kinetic energy. Remember that when molecules have the same kinetic energy, the larger molecules move slower and the smaller ones move faster. If we consider the molar mass of these two substances, the molar mass of ammonia is \begin{align*}17\ \mathrm{g/mol}\end{align*}, and the molar mass of acetone is \begin{align*}58\ \mathrm{g/mol}\end{align*}. We would have realized, then, that the ammonia molecules are traveling quite a bit faster than the acetone molecules in order for them to have the same kinetic energy. As a result, we would have expected the ammonia to diffuse through the room faster.

A demonstration commonly used to show the different rates of diffusion for gases is to dip one cotton ball in an ammonia solution and another cotton ball in a dilute \begin{align*}\text{HCl}\end{align*} solution and stuff both cotton balls at opposite ends of a glass tube. This setup is shown in the diagram below. When \begin{align*}\text{NH}_3\end{align*} and \begin{align*}\text{HCl}\end{align*} react, they form \begin{align*}\text{NH}_4\text{Cl}\end{align*}, a white powdery substance. Molecules of \begin{align*}\text{NH}_3\end{align*} and \begin{align*}\text{HCl}\end{align*} will escape the cotton balls at opposite ends of the tube and diffuse through the tube toward each other. Since the molar mass of \begin{align*}\text{HCl}\end{align*} is slightly more than double that of \begin{align*}\text{NH}_3\end{align*}, the \begin{align*}\text{NH}_3\end{align*} will travel further down the tube than the \begin{align*}\text{HCl}\end{align*} by the time they meet.

As a result, the white cloud of \begin{align*}\text{NH}_4\text{Cl}\end{align*} always forms closer to the \begin{align*}\text{HCl}\end{align*} end. If this experiment is done carefully and the distances are measured accurately, a reasonable ratio for the molar masses of these compounds can be determined.

Thomas Graham (1805-1869), an English chemist, studied the rates of diffusion of different gases and was able to describe the relationships quantitatively in what is called Graham’s law:

Under the same conditions of temperature and pressure, gases diffuse at a rate inversely proportional to the square root of the molecular masses.

Mathematically, this expression would be written as:

\begin{align*}\frac{V_1} {V_2} = \sqrt{\frac{M_2} {M_1}} \ \ \ \ \ (M \ \text{is the molecular mass of the particle})\end{align*}

The video demonstrates diffusion in gases and liquids. It clearly shows the differences in rate of diffusion due to molar mass as well as higher temperatures (4b): http://www.youtube.com/watch?v=H7QsDs8ZRMI (4:57).

## Lesson Summary

• At STP, one mole of any gas occupies \begin{align*}22.4\ \mathrm{liters.}\end{align*}
• When gases are mixed and do not react chemically, the total pressure of the mixture of gases will be equal to the sum of the partial pressures of the individual gases (Dalton's law of partial pressures).
• Gases diffuse at rates that are inversely proportional to the square roots of their molecular masses (Graham's law).

This video shows an experiment that demonstrates the molar volume of a gas at STP is 22.4 liters.

## Review Questions

1. A \begin{align*}1.00\ \mathrm{L}\end{align*} container of helium gas at \begin{align*}1.00\ \mathrm{atm}\end{align*} pressure and a \begin{align*}1.00\ \mathrm{L}\end{align*} container of hydrogen gas at \begin{align*}2.00\ \mathrm{atm}\end{align*} are both transferred into a \begin{align*}1.00\ \mathrm{L}\end{align*} container containing nitrogen gas at \begin{align*}3.00\ \mathrm{atm}\end{align*}. What is the final pressure in the final container holding all three gases (assuming no temperature change)?
2. For the situation described in problem #1, what will be the partial pressure of the helium in the final container?
3. At STP, how many molecules are in \begin{align*}89.6\ \mathrm{liters}\end{align*} of gas?
4. If \begin{align*}1.00\ \mathrm{liter}\end{align*} of gas A at STP and \begin{align*}1.00\ \mathrm{liter}\end{align*} of gas \begin{align*}B\end{align*} at STP are both placed into a \begin{align*}2.00\ \mathrm{liter}\end{align*} evacuated container at STP, what will the pressure be in the \begin{align*} 2.00\ \mathrm{liter}\end{align*} container?
5. Consider the gases \begin{align*}\text{CO}\end{align*} and \begin{align*}\text{N}_2\end{align*}. Which of the following will be nearly identical for the two gases at \begin{align*} 25^\circ \text{C}\end{align*} and \begin{align*}1.0\ \mathrm{atm}\end{align*}? i. average molecular speed ii. rate of effusion through a pinhole iii. density
1. i only
2. iii only
3. i and ii only
4. ii and iii only
5. i, ii, and iii
6. The density of an unknown gas at \begin{align*}2.0\ \mathrm{atm}\end{align*} and \begin{align*}25^\circ \text{C}\end{align*} is determined to be \begin{align*}3.11\ \mathrm{g/L}\end{align*}. Which of the following gases is the unknown most likely to be?
1. \begin{align*}\text{CH}_4\end{align*}
2. \begin{align*}\text{F}_2\end{align*}
3. \begin{align*}\text{N}_2\text{O}_4\end{align*}
4. \begin{align*}\text{O}_2\end{align*}
5. \begin{align*}\text{CF}_2\text{Cl}_2\end{align*}

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