## Collection of Gas Over Water

In many cases, the amount of gas evolved by a reaction is of interest. Since gases have such small densities, it is usually not practical to collect the gas and find its mass. For gases that are not particularly soluble in water, it is possible to collect the evolved gas by displacement of water from a container.

The setup for the collection of a gas over water involves a container in which the reaction takes place and a gas collection container filled with water and inverted in a reservoir of water. The gas evolved from the reaction is collected by attaching one end of a hose to the reaction container and inserting the other up into the inverted gas collection bottle. As the gas is created, it will displace water from the bottle. The volume of gas can be determined by the amount of water that was displaced by the gas.

The volume of gas collected and the gas laws can be used to calculate the number of moles of gas collected.

During the collection, the water level in the container will adjust so that the pressure inside and outside the container are the same. Because of this, if we know the atmoshperic pressure, we also know the pressure of the gas inside the bottle.

The pressure inside the bottle is partially from the gas being collected and partially from the water vapor that has escaped from the surface of the water in the jar. The water inside the jar will reach an equlibruim state where the number of molecules leaving the surface is the same as the number returning. The equilibrium pressure of water is temperature dependent and is called the **vapor pressure of water.**

Dalton's Law of Partial Pressures tells us that the total pressure in the container must be the sum of the pressures of the gas we collected and the water vapor.

**P**_{T}** = P**_{gas}** + P**_{H}_{2}_{O}

This equation can be used to calculate the pressure of the gas collected. Once the pressure of the collected gas is known, the number of moles of gas can be calculated using the ideal gas law:

**PV= nRT**

Where:

- P = Pressure of the gas
- V = Volume of water displaced
- n = number of moles of gas
- R = the ideal gas constant
- T = the temperature of the gas

#### Learning Objective

- Apply Dalton’s Law to determine the partial pressure of a gas collected over water.

#### Key Points

- The total pressure in an inverted tube can be determined by the height of the water displaced in the tube.
- When calculating the amount of gas collected, Dalton’s Law must be used to account for the presence of water vapor in the collecting bottle.

#### Term

- pneumatic troughdevice used to collect a gas over water; the height of water displaced in the tube can be used to determine the total pressure inside the tube

Since gases have such small densities, it can be difficult to measure their mass. A common way to determine the amount of gas present is by collecting it over water and measuring the height of displaced water; this is accomplished by placing a tube into an inverted bottle, the opening of which is immersed in a larger container of water.

### The Pneumatic Trough

This arrangement is called a pneumatic trough, and it was widely used in the early days of chemistry. As the gas enters the bottle, it displaces the water and becomes trapped in the closed, upper part of the bottle. You can use this method to measure a pure gas (i.e. O_{2}) or the amount of gas produced from a reaction. The collected gas is not the only gas in the bottle, however; keep in mind that liquid water itself is always in equilibrium with its vapor phase, so the space at the top of the bottle is actually a mixture of two gases: the gas being collected, and gaseous H_{2}O. The partial pressure of H_{2}O is known as the vapor pressure of water and is dependent on the temperature. To determine the quantity of gas we have collected alone, we must subtract the vapor pressure of water from the total vapor pressure of the mixture.

### Calculating Gas Volume

### Example 1

O_{2} gas is collected in a pneumatic trough with a volume of 0.155 L until the height of the water inside the trough is equal to the height of the water outside the trough. The atmospheric pressure is 754 torr, and the temperature is 295 K. How many moles of oxygen are present in the trough? (At 295 K, the vapor pressure of water is 19.8 torr.)

The total pressure in the tube can be written using Dalton’s Law of Partial Pressures:

[latex]{P}_{total}={P}_{{H}_{2}O}+{P}_{{O}_{2}}[/latex]

Rearranging this in terms of [latex]P_{O_2}[/latex], we have:

[latex]{P}_{{O}_{2}}= {P}_{total} - {P}_{{H}_{2}O}[/latex]

Because the height of the water inside the tube is equal to the height of the water outside the tube, the total pressure inside the tube must be equal to the atmospheric pressure. With substitution, we have:

[latex]P_{O_2}=P_{total}-P_{H_2O}= 754 - 19.8 = 734\;torr = .966\;atm[/latex]

Next, we apply the Ideal Gas Law:

[latex]n=\frac{PV}{RT}[/latex]

[latex]=\frac {(.966 atm)(.155L)}{(.082 L\cdot atm\cdot {mol}^{-1}\cdot {K}^{-1}) (295K)}[/latex]

[latex]=.00619 \; mol\;O_2[/latex]

### Example 2

Oxygen gas generated in an experiment is collected at 25°C in a bottle inverted in a trough of water. The external laboratory pressure is 1.000 atm. When the water level in the originally full bottle has fallen to the level in the trough, the volume of collected gas is 1750 ml. How many moles of oxygen gas have been collected?

If the water levels inside and outside the bottle are the same, then the total pressure inside the bottle equals 1.000 atm; at 25°C, the vapor pressure of water (or the pressure of water vapor in equilibrium with the liquid) is 23.8 mm Hg or 0.0313 atm.

Therefore, the partial pressure of oxygen gas is 1.000 – 0.031, or 0.969 atm.

The mole fraction of oxygen gas in the bottle is 0.969 (not 1.000), and the partial pressure of oxygen also is 0.969 atm. The number of moles is: [latex]n=\frac{PV}{RT }=\frac{(.969 \ atm)( 1730\ cm^{3})}{(82.054\ cm^{3}\ K^{-1}\ mole^{-1}\ )( \ 298 \ K)}[/latex]

n = 0.068 moles O_{2}

## Partial Pressure Calculator

Welcome to the partial pressure calculator - a device that can help you determine and understand partial pressure. Interested in some of the laws of chemistry, such as Dalton's law of partial pressure? Want to know how to calculate partial pressure? Worry no more. We present you with four partial pressure formulas, so keep reading!

Are you're also interested in thermodynamics? If so, check out our combined gas law calculator.

### Dalton's law of partial pressures

Pressure is the force applied orthogonally over a surface. If a mixture of ideal gases (i.e., where the molecules don't interact with each other) is sealed within a container, the gases will diffuse and fill up all of the available space. The partial pressure of one component of this mixture is the pressure that this individual gas exerts.

Dalton's law states that:

the total pressure exerted on a container's walls by a gas mixture is equal to the sum of the partial pressures of each separate gas.

It can also be illustrated with an equation:

,

where , , and so on, up to , represent the partial pressure of each gaseous component.

It can also be presented as follows:

where is the ratio of moles of the selected gas to the moles of the entire gas mixture.

It shows that the partial pressure of one component is proportional to its mole fraction.

The above formula is one of our calculator's four partial pressure formulas. Which one you choose depends on the data you've collected beforehand.

### How to calculate partial pressure? - Ideal gas law

The ideal gas law states that:

where:

- is the pressure of the gas
- is the volume of the gas
- is the number of moles of the gas
- is the gas constant, 8.3145
^{J}/_{mol*K} - is the temperature of the gas

If you want to calculate the partial pressure of one component of a gas mixture, use the following formula (derived from the one above):

where:

- is the partial pressure of the individual gas
- is the amount of moles of the individual gas
- is the temperature of the mixture
- is the volume of the mixture

Now you know how to calculate partial pressure using the ideal gas law. Let's move on to the last two forms of the partial pressure equation - both using Henry's law.

### Another partial pressure equations - Henry's law

Henry's law states that:

the partial pressure of a gas above a liquid is proportional to the amount of gas dissolved in that liquid.

The coefficient of this proportionality is the Henry's law constant. In the table below, you can find its value for some of the most common gases in water at 298 K:

Element | Henry's law constant [^{litre*atm}/_{mol}] | Henry's law constant [atm] |
---|---|---|

O_{2} | 769.23 | 4.259*10^{4} |

H_{2} | 1282.05 | 7.099*10^{4} |

Co_{2} | 29.41 | 0.163*10^{4} |

N_{2} | 1639.34 | 9.077*10^{4} |

He | 2702.7 | 14.97*10^{4} |

Ne | 2222.22 | 12.3*10^{4} |

Ar | 714.28 | 3.955*10^{4} |

CO | 1052.63 | 5.828*10^{4} |

Henry's law is only accurate at *low gas pressures* (pressures < 1000 hPa), *constant temperatures* (usually 293.15 K) and when *the molecules are at equilibrium*.

How to find partial pressure with Henry's law constant? There are two methods:

- Where the concentration of the solute is given:

where

is Henry's law constant in [^{litre*atm}/_{mol}].

- Where the mole fraction of the solute is given:

where

is Henry's law constant in [atm].

Let's use the Henry's law equation in an example. Let's say that you want to calculate the partial pressure of dinitrogen (N_{2}) in a container. Its concentration is 1.5 moles / L. All you need to do is check the Henry's law constant in the table above, and input the numbers into the partial pressure formula:

Simple, right?

### Fun facts about pressure

It is essential to consider pressure if you are an underwater diver. Divers usually breathe a mixture of oxygen and nitrogen. When diving down to about 35 meters, the standard mixture is safe. As the pressure increases during deeper dives, oxygen becomes toxic, potentially causing narcosis. That's why technical divers (those who dive very deep) use different breathing mixtures to casual divers.

In medicine, while performing an arterial-blood gas test, physicians measure the partial pressure of carbon dioxide and oxygen. With these measurements, they can calculate the pH of their patient's blood.

You might find it surprising that air pressure changes with altitude and temperature. It's important for people trekking up high mountains - the lower the pressure, the harder it is to breathe.

## 9.12: Dalton's Law of Partial Pressures

The ideal gas law can also be rearranged to show that the pressure of a gas is proportional to the amount of gas:

\[P=\frac{RT}{V}\,n\label{1}\]

Thus the factor RT/V may be used to interconvert amount of substance and pressure in a container of specified volume and temperature.

Equation \(\ref{1}\) is also useful in dealing with the situation where two or more gases are confined in the same container (i.e., the same volume). Suppose, for example, 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*.

Dalton’s law of partial pressures is most commonly encountered when a gas is collected by displacement of water, as shown in Figure 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

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

### Contributors and Attributions

## Over collecting calculator gas water

Pretend nothing special happened. It would be reasonable, but won't a similar curiosity happen again someday. What if he's in love.

Collecting a Gas Over WaterWhat will it be. I'm very curious, you know, my revenge is terrible. Not only will I not say what it is, but I will not even have sex with you tonight. - He openly laughed at Lisa.

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