# 12.5: Dalton's Law (Law of Partial Pressures)

- Page ID
- 238708

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Dalton’s Law, or the Law of Partial Pressures, states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the gases in the mixture.

## Dalton's Law of Partial Pressures

Based on the kinetic theory of gases, gas pressure results from collisions between gas particles and the inside walls of their container. If more gas is added to a rigid container, the gas pressure increases. The identities of the two gases do not matter. An ideal gas will expand in a container to fill up the space it is in and does not have any forces of attraction between the molecules. In other words, the different molecules in a mixture of gases are so far apart that they act independently; they do not react with each other. The pressure of an ideal gas is determined by its collisions with the container, not collisions with molecules of other substances since there are no other collisions. Since the gases in a mixture of gases are in one container, the Volume (V) and Temperature (T) for the different gases are the same as well. Each gas exerts its own pressure on the system, which can be added up to find the total pressure of the mixture of gases in a container.

John Dalton, the English chemist who proposed the atomic theory, also studied mixtures of gases. He found that each gas in a mixture exerts a pressure independently of every other gas in the mixture. For example, our atmosphere is composed of about \(78\%\) nitrogen and \(21\%\) oxygen, with smaller amounts of several other gases making up the rest. Since nitrogen makes up \(78\%\) of the gas particles in a given sample of air, it exerts \(78\%\) of the pressure. If the overall atmospheric pressure is \(1.00 \: \text{atm}\), then the pressure of just the nitrogen in the air is \(0.78 \: \text{atm}\). The pressure of the oxygen in the air is \(0.21 \: \text{atm}\).

The **partial pressure** of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture. The partial pressure of nitrogen is represented by \(P_{N_2}\). **Dalton's law of partial pressures** states that the total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases. Dalton's law can be expressed with the following equation:

\[P_\text{total} = P_1 + P_2 + P_3 + \cdots\]

The figure below shows two gases that are in separate, equal-sized containers at the same temperature and pressure. Each exerts a different pressure, \(P_1\) and \(P_2\), reflective of the number of particles in the container. On the right, the two gases are combined into the same container, with no volume change. The total pressure of the gas mixture is equal to the sum of the individual pressures. If \(P_1 = 300 \: \text{mm} \: \ce{Hg}\) and \(P_2 = 500 \: \text{mm} \: \ce{Hg}\), then \(P_\text{total} = 800 \: \text{mm} \: \ce{Hg}\).

## Derivation

We have, from the Ideal Gas Law

\[PV = nRT \tag{2}\]

If we know the molar composition of the gas, we can write

\[n_{total} = n_a + n_b + ... \tag{3}\]

Again, based on the kinetic theory of gases and the ideal gas law, Dalton’s law can also be applied to the number of moles so that the total number of moles equals the sum of the number of moles of the individual gases. Here, the pressure, temperature and volume are held constant in the system. The total volume of a gas can be found the same way, although this is not used as much. This yields the equation,

\[P_{total} V=n_{total} RT \tag{4}\]

We can rearrange this equation to find the total number of moles. Sometimes masses of each sample of gas are given and students are asked to find the total pressure. This can be done by converting grams to moles and using Dalton's law to find the pressure.

## Example

Suppose that we had 0.010 mol of a gas in a 250-ml container at a temperature of 32°C. The pressure would be

\[\begin{align}P & =\frac{RT}{V}\,n =\frac{\text{0}\text{.0820 liter atm mol}^{-\text{1}}\text{ K}^{-\text{1}}\,\times \text{ 305 K}}{\text{0}\text{.250 liter}}\,\times \text{ 0}\text{.010 mol}\\ & =\text{1}\text{.00 atm}\end{align}\]

Now suppose we filled the same container with 0.004 mol H_{2}(*g*) at the same temperature. The pressure would be

\[\begin{align}p_{\text{H}_{\text{2}}} & =\frac{\text{0}\text{.0820 liter atm mol}^{-\text{1}}\text{ K}^{-\text{1}}\,\times \text{ 305 K}}{\text{0}\text{.250 liter}}\,\times \text{ 0}\text{.004 mol}\\ & =\text{0}\text{.40 atm}\end{align}\]

If we put 0.006 mol N_{2} in the container,

\[p_{\text{N}_{\text{2}}}=\frac{\text{0}\text{.0820 liter atm mol}^{-\text{1}}\text{ K}^{-\text{1}}\,\times \text{ 305 K}}{\text{0}\text{.250 liter}}\,\times \text{ 0}\text{.006 mol}=\text{0}\text{.60 atm}\]

Now suppose we put both the 0.004 mol H_{2} and the 0.006 mol N_{2} into the same flask together. What would the pressure be? Since the ideal gas law does not depend on *which *gas we have but only on the amount of *any* gas, the pressure of the (0.004 + 0.006) mol, or 0.010 mol, would be exactly what we got in our first calculation. But this is just the sum of the pressure that H_{2} would exert if it occupied the container alone plus the pressure of N_{2} if it were the only gas present. That is,

\[P_{total} = p_{\text{H}_{2}} + p_{\text{N}_{2}}\]

The figure below demonstrates the concept of partial pressure in more concrete terms, showing the pressure of each gas alone in a container and then showing the gases combined pressure once mixed.

We have just worked out an example of **Dalton’s law of partial pressures** (named for John Dalton, its discoverer). This law states that *in a mixture of two or more gases, the total pressure is the sum of the partial pressures of all the components*. The **partial pressure** of a gas is the pressure that gas would exert if it occupied the container by itself. Partial pressure is represented by a lowercase letter *p*.

## Mole Ratio

From the partial pressure of a certain gas and the total pressure of a certain mixture, the mole ratio, called Xi, of a gas can be found. The mole ratio describes what fraction of the mixture is a specific gas. For example, if oxygen exerts 4 atm of pressure in a mixture and the total pressure of the system is 10 atm, the mole ratio would be 4/10 or 0.4. The mole ratio applies to pressure, volume, and moles as seen by the equation below. This also means that 0.4 moles of the mixture is made up of gas i.

\[X_i=\dfrac{P_i}{P_{tot}}=\dfrac{n_i}{n_{tot}}=\dfrac{V_i}{V_{tot}} \tag{5}\]

The mole ratio, (\(X_i\)) is often used to determine the composition of gases in a mixture. The sum of the mole ratios of each gas in a mixture should always equal one since they represent the proportion of each gas in the mixture.

## Collection of a Gas Over Water

The Law of Partial Pressures is commonly applied in looking at the pressure of a closed container of gas and water. The total pressure of this system is the pressure that the gas exerts on the liquid. The gas is made up of whatever sample of gas there is plus the evaporated water. The pressure of the gas on the liquid consists of the pressure of the evaporated water and the pressure of the gas collected. Based on Dalton’s law, the pressure of the gas collected can be calculated by subtracting the pressure of the water vapor from the total pressure. This is commonly encountered when a gas is collected by displacement of water, as shown in Figure 12.5.2.

Because the gas has been bubbled through water, it contains some water molecules and is said to be “wet.” The total pressure of this wet gas is the sum of the partial pressure of the gas itself and the partial pressure of the water vapor it contains. The latter partial pressure is called **the vapor pressure** of water. It depends only on the temperature of the experiment and may be obtained from a handbook or from Table 1.

Temperature(°C) |
Vapor Pressure (mmHg) |
Vapor Pressure (kPa) |
---|---|---|

0 | 4.6 | 0.61 |

5 | 6.5 | 0.87 |

10 | 9.2 | 1.23 |

15 | 12.8 | 1.71 |

20 | 17.5 | 2.33 |

25 | 23.8 | 3.17 |

30 | 31.8 | 4.24 |

50 | 92.5 | 12.33 |

70 | 233.7 | 31.16 |

75 | 289.1 | 38.63 |

80 | 355.1 | 47.34 |

85 | 433.6 | 57.81 |

90 | 525.8 | 70.10 |

95 | 633.9 | 84.51 |

100 | 760.0 | 101.32 |

Example \(\PageIndex{1}\): Volume of Hydrogen

Assume 0.321 g zinc metal is allowed to react with excess hydrochloric acid (an aqueous solution of HCl gas) according to the equation

\[\text{Zn} (s) + 2 \text{HCL} (aq) \rightarrow \text{Zn} \text{Cl}_{2} (aq) + \text{H}_{2} (g)\]

The resulting hydrogen gas is collected over water at 25°C, while the barometric pressure is 745.4 mmHg. What volume of wet hydrogen will be collected?

**Solution** From Table 1 we find that at 25°C the vapor pressure of water is 23.8 mmHg. Accordingly

*p*_{H2} = *p*_{total}– *p*_{H2O} = 754 mmHg – 23.8 mmHg = 721.6 mmHg.

This must be converted to units compatible *R*:

\[p_{\text{H}_{\text{2}}}=\text{721}\text{.6 mmHg }\times \,\frac{\text{1 atm}}{\text{760 mmHg}}=\text{0}\text{.949 atm}\]

The road map for this problem is

\[m_{\text{Zn}}\xrightarrow{M_{\text{Zn}}}n_{\text{Zn}}\xrightarrow{S\left( \text{H}_{\text{2}}\text{/Zn} \right)}n_{\text{H}_{\text{2}}}\xrightarrow{RT/P}V_{\text{H}_{\text{2}}}\]

Thus

\[\begin{align}V_{\text{H}_{\text{2}}} & =\text{0}\text{.321 g Zn }\times \,\frac{\text{1 mol Zn}}{\text{65}\text{.38 g Zn}}\,\times \,\frac{\text{1 mol H}_{\text{2}}}{\text{2 mol Zn}}\,\times \,\frac{\text{0}\text{.0820 liter atm}}{\text{1 K mol H}_{\text{2}}}\,\times \,\frac{\text{293}\text{0.15 K}}{\text{0}\text{.987 atm}}\\ & =\text{0}\text{.126 liter}\end{align}\]

### Real Gases

Real gases are gases that do not behave ideally. That is, they violate one or more of the rules of the kinetic theory of gases. Real gases behave ideally when the gases are at low pressure and high temperature. Therefore at high pressures and low temperatures, Dalton’s law is not applicable since the gases are more likely to react and change the pressure of the system. For example, if there are forces of attraction between the molecules, the molecules would get closer together and the pressure would be adjusted because the molecules are interacting with each other.

## Summary

- The total pressure in a system is equal to the sums of the partial pressures of the gases present.

## Contributors and Attributions

Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.

## Problems

- A sample of gas A evaporates over water in a closed system. What is the pressure of gas A if the total pressure is 780 torr and water vapor pressure is 1 atm?
- There is a mixture of 4 moles of hydrogen gas, 8 moles of oxygen, 12 moles of helium, and 6 moles of nitrogen in a closed container. What is the total number of moles in this system?
- If there is a mixture of hydrogen and oxygen gas in a container with 10 moles total and the mole ratio of hydrogen is .67 mol
_{Hydrogen}to 1 mol_{total}, how many moles of each gas are there? - 24.0 L of nitrogen gas at 2 atm and 12.0 L of oxygen gas at 2 atm are added to a 10 L container at 273 K. Find the partial pressure of nitrogen and oxygen and then find the total pressure.
- Flourine gas is in a 5.0 L con tainer that is 25 C and 2 atm. A certain amount of hydrogen with a partial pressure of .5 is added to the container. What is the mole ratio of hydrogen?

## Solutions

1:

- Convert pressure to same units so 780 torr=1.03 atm
- Subtract water vapor pressure from total pressure to get partial pressure of gas A: P
_{A}=1.03 atm- 1 atm=**0.03 atm**

*2. *The law of partial pressures also applies to the total number of moles if the other values are constant, so

4 mol Hydrogen+8 mol Oxygen+12 mol Helium+6 mol Nitrogen=**30 moles total**

**3. **

- 10 mol
_{total}x .67 mol_{Hydrogen}/ 1 mol_{total}= 6.7 moles H_{2}gas - 10 mol
_{total}-6.7 mol_{Hydrogen}=3.3 moles O gas

4.

- Find the number of moles of oxygen and nitrogen using PV=nRT which is n=PV/RT
- oxygen: ((1 atm)(12L))/(0.08206 atm L mol
^{-1}K^{-1})(273 K)=0.536 moles oxygen - nitrogen: ((1 atm)(24.0L))/(.08206 atm L mol
^{-1}K^{-1})(273 K)=1.07 moles of Nitrogen - add to get n
_{tot}: .536 mol_{Oxygen}+1.07 mol_{Nitrogen}=1.61 moles total

- oxygen: ((1 atm)(12L))/(0.08206 atm L mol
- Use PV=nRT or P=(nRT)/V to find the total pressure
- Ptot=((1.61 mol
_{total})(0.08206 atm L mol^{-1}K^{-1})(273 K))/(10.0 L)=3.61 atm

- Ptot=((1.61 mol
- P
_{A}/P_{tot}=n_{A}/N_{tot }can be rearranged to P_{A=(}P_{tot)(}n_{A}/N_{tot) }to find the partial pressures- P
_{oxygen}=(3.61 atm_{total})(.536 mol_{Oxygen}/1.61 mol_{total})=1.20 atm_{Oxygen} - P
_{nitrogen}=3.61 atm_{total}-1.20 atm_{Oxygen}=2.41 atm_{Nitrogen}

- P

5.

- total pressure is 2 atm
_{Flourine}+.5 atm_{Hydrogen}=2.5 atm_{total} - remember that p
_{H}/p_{tot}=n_{H}/n_{tot}=v_{h}/v_{tot}so the mole ratio of hydrogen to the mixture is is .5/2.5=**.2 mol**_{Hydrogen}to 1 mol_{total}

## References

- Dutton, F.B. "Dalton's Law of Partial Pressures." Journal of Chemical Education Aug 1961. http://pubs.acs.org/doi/pdfplus/10.1021/ed038pA545.1
- Blauch, David. "Gas Laws." Davidson Chemistry. 2009. http://www.chm.davidson.edu/vce/gaslaws/daltonslaw.html
- Petrucci, Ralph H. "General Chemistry: Principles & Modern Application, 9th Edition." New Jersey: Pearson Prentice Hall. 2007.
- Toupadakis, Andreas. "Chemistry Reader 2A." Kendall Hunt Publishing. 2009.
- http://pubs.acs.org/doi/pdfplus/10.1021/ed038pA545.1
- http://pubs.acs.org/doi/pdf/10.1021/ed084p469