r fl 1'- ' f.\ \ \ ' \ \ \ ' f. J \ t ' t I \ \. r f '\ \ \ , I ~;; ' . I : :i ' I I;, ,. I .:~~ ' I ··.' . I ~ \ CHAPTER 8

Gases IN THIS CHAPTER

Pressure and the particle nature of gases Diffusion Kinetic theory of gases Avogadro's law Boyle's law

Charles' law Combined gas law Ideal gas equation Dalton's law of partial pressures Summary

Gases are an important part of the environment; without oxygen and carbon dioxide, we and all the other living organisms of the Earth could not exist. However, gases often go unnoticed, because they are less obvious than solids and liquids. On the other hand, in some aspects of our daily lives gases arc very noticeable- from carbon dioxide adding fizz to soft drinks to LPG (liquefied petroleum gas) for fuelling barbecues and motor vehicles to natural gas for heating ·water, cooking and warming our homes to the propellant in aerosol deodorants.

In this chapter we shall survey some of the common properties of gases.

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8.1 PRESSURE AND THE PARTICLE NATURE OF GASES

We saw in Section 1.2 and Figure 1.1 (c) and (f) on pp. 164- 5 that gases consist of particles that are well separated in space and in continual random motion. Because of the rapid motion, gases always expand to fill the total \'Olume available to them; because the particles are well separated in space, gases are easily compressed into smaller volumes.

We often recognise the presence of a gas from the pressure it exerts.

Pressure Pressure is defined as force per unit area. The systematic unit for pressure is the pascal, Pa, which is a newton (unit of force) per square metre,~ m-2.

This is quite a small pressure, so we frequently use kilopascal, kPa. Standard atmospheric pressure is 1.103 x 105 Pa or 101.3 kPa.

Another common unit for pressure is the atmosphere. It is defined as:

1 atmosphere (attn) = 101.3 kPa

315

I

I 316 CHEMISTRY IN USE

FIGURE 8.1 Compression of a gas increases the number of collisions per unit area and hence pressure

Atmospheric pressure is not always exactly 1.00 atmosphere; it Yaries \\ith weather conditions. While chemists and physicists generally use pascals (or kilopascals) or atmospheres, meteorologists generally use the heccopascal, hPa (= 100 Pa) though sometimes the older units, the bar(= lOs Pa) and millibar(= 100 Pa), are usedt. Standard atmospheric pressure is 1013 hectopascals or 1013 millibars.

Effect of temperature and pressure on volumes of gases If we haYe a sample of gas in a hypodermic syringe that is fined with a well­lubricated and easily moYing plunger and that has the needle outlet sealed off as in Figure 8.1(a), then the pressure exerted by the gas upon the walls and plunger of the syringe is equal to the pressure the atmosphere is exerting upon the barrel and plunger of the syringe. The gas pressure arises because the particles of the gas are continuously hitting the walls of the container and exerting a force (and hence a pressure) upon them.

If we push the plunger in by exerting a force upon it (that is, by increasing the pressure applied to the gas), the gas compresses (that is, its Yolume becomes smaller). The gas volume decreases until the pressure exerted by the gas becomes equal to the applied external pressure. The collisions of gas particles with the walls of the syringe are spread out over a smaller area and hence force per unit area (pressure) increases as shown in Figure 8.1 (b). If\.ve release the plunger so that the external pressure on it falls back to atmospheric, the compressed gas expands and pushes the plunger back until the gas pressure also falls to atmospheric again.

~ :/ ~~1 ::l :[ : needle (a) Pressure is due to collision of particles with walls end sealed off

atmospheric pressure

~~f{j[ : increased pressure

{b) If the gas is compressed (volume decreased), there are more collisions per unit area: pressure increases.

T he volume of gas is very dependent upon the pressure acting upon it. In fact doubling the pressure halves the ·volume.

If the syringe in (a) with atmospheric pressure acting on the plunger is immersed in a beaker of boiling water, the volume increases considerably: 20 mL of gas at 20°C would expand to 25 mL at 1 00°C. By contrast 20 mL water at 20°C heated to 100°C (\vithout being vaporised) would expand to only 20.8 mL. Gases

expand to ajar greater extent than do solids or liquids. The explanation is this: as we heat the gas, we increase the kinetic energy of

the particles (molecules) so they move faster and strike the walls and plunger

t The unusual prefix hecco (= I 00) was used when meteorologists converted to systematic units in order to keep the numerical value of the pressure in new units equal to the value in the old unit, millibars.

-.__,... . -~-

CHAPTER 8 GASES 317 I more frequently and with more force, and so at constant ,·olume would exert greater pressure; because the pressure acting on the plunger is atmospheric, this greater pressure pushes the plunger out until the number of (more forceful) collisions per unit area decreases sufficiently to bring the pressure exerted by the gas back to the external pressure acting on the plunger (atmospheric pressure).

Because 1he volwne of a sample of gas depends upon both the Lemperawre and the applied pressure> it is essential when reporting gas volumes to state the temperature and the pressure of the measure1nent. Hence we talk about 1 0 L of oxygen at 20°C and 1.00 atm pressure, or 50 mL of hydrogen at l00°C and 220 kPa.

Gas volumes are frequently giYen for conditions called standard temperature and pressure, STP.

Standard temperature is 0°C. Standard pressure is exactly 1 atmosphere (atm).

Hence we frequently talk about 2.0 L of gas at STP or 400 mL of nitrogen at STP

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8.2 DIFFUSION Diffusion is the process in which one substance becomes mixed with another substance.

Consider a gas jar of brown nitrogen dioxide placed in contact with another gas jar containing colourless nitrogen as in Figure 8.2. When the coverslip is removed, the two gases quickly diffuse into each other and ''ithin a few seconds the combined volume is seen to contain a gaseous mixture of uniform colour (the same composition throughout). This diffusion of gases occurs much more quickly than does diffusion of liquids or solids.

/coverslip

r-..... [\-. 1 <'

[.vc;:. . <'• .,,,<. <''' .

v '

The much slower diffusion of liquids or solutions can be demonstrated as follmvs. If a test tube is half-filled with a purple solution of potassium permanganate solution, and if water is carefully poured on top of it (pouring it slowly down the side of a slanted test tube), then diffusion will be seen to occur as the purple colour spreads upwards into the clear water, but it will take a few minutes before the colour is uniform (mixing is complete).

Solids diffuse even more slowly. A block of copper can be clamped to a block of aluminium for many years before there is significant diffusion of copper into the aluminium.

Rapid diffusion of gases can be explained in terms of the particle nature of gases. Because the particles of a gas are well separated from one another and because they are in continuous rapid motion, particles of one gas easily penetrate the space occupied by another gas and Yice versa. The particles mm·e such distances between collisions that they rapidly mix together.

FIGURE 8.2 An experiment to demonstrate rapid diffusion of gases

\

EXERCISES 1 A sample of gas is enclosed in a syringe as in Figure 8.1 (a). If the plunger is pushed

in (by exerting an increased pressure on it}, how would you expect the number of molecules per millilitre to change-to increase. decrease or remain the same? If the syringe and the gas it contained were then heated. what would happen to the number of molecules per millilitre if the volume was kept constant? Why would it be necessary to apply a greater pressure to the syringe plunger to keep the volume constant as the gas was heated?

2 When two gases are brought into contact, they mix more quickly than do two liquids similarly brought together. Explain why in terms of the particle nature of gases and liquids.

3 a Convert the following pressures to atmospheres:

50.7 kPa ii 3.68 X 1 03 Pa iii 986 hectopascals

b Convert the following pressures to kPa:

2.34 atm ii 0.52 atm iii 1 072 hectopascals

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8.3 KINETIC THEORY OF GASES The explanations gi,·en for the experimental observations in the preceding two sections are part of what is called the kinetic theory of gases.

The kinetic theory of gases proposes that:

1 Gases consist of molecules (small particles) that are in continual random motion.

2 The actual volume of all the molecules present in a sample of gas is negligible compared to the total volume they occupy.

3 Intermolecular forces are negligible. 4 Pressure is due to collision of the particles with walls of the container. 5 Temperature is a measure of the average kinetic energy (or speed) of the gas

molecules.

The kinetic theory can be used to explain many properties of gases other than those \\'e have just considered.

Refer to Context 4 Air, pp. 130-1 .

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8.4 AVOGADRO'S LAW From what \Ve have just seen about diffusion of gases and the way volumes of gases depend upon pressure and temperature, it appears that the properties of gases depend more upon the number of particles (molecules) present than upon their chemical nature. It is not surprising then that \\'C find that:

Equal \'Olumes of different gases, measured at the same temperature and pressure, contain the same number of molc..:cules.

This is known as Avogadro's law or Avogadro's hypothesis, after the person who first proposed it in 1811.

At constant temperature and pressure 50 mL of oxygen, 0 2, contains the same number of molecules as does 50 mL of nitrogen, N 2, or 50 mL of ammonia, NH3.

Since helium and argon are monatomic gases (the molecules are atoms, Section 1.8 on p. 171), the number of molecules (atoms) in 50 mL of helium or argon is the same as the number of molecules in 50 mL of oxygen.

We saw in Section 3.3 on pp. 220- 2 that a mole 'vvas a fixed number of particles. Hence we can rephrase the above statement as:

Equal volumes of different gases, measured at the same temperature and pressure, contain the same number of 1110/es.

Or turning the statement around:

Ivleasurcd at the same temperature and pressure, 1 mole of any gas has the same volume as 1 mole of any other gas.

We call this the molar volume of a gas. Its value at STP (0°C and 1.00 arm) is 22.4 L.

It follows then, that if 1 mole of oxygen gas at STP has a volume of 22.4 L , then 3 moles of chlorine gas at STP has a volume of 67.2 Land 0.5 mole of helium has a volume of 11.2 L at STP.

EXERCISES 4 a How many moles of i carbon dioxide ii helium are present in 375 ml of the gas

at STP?

5

b What volume container would you need to store 0.095 mol of i argon ii ammonia at STP?

In order to verify Avogadro's law a group of students performed the following experiment. They sealed (with silicone adhesive) two glass tubes into a clean empty drink can as in the diagram below. They filled the can with helium gas by connecting a helium cylinder to tube A and flushing out the can (gas leaves via tube B), then sealing off the two tubes with pieces of 'Blue Tack': this leaves the can containing atmospheric pressure of the gas being used. They weighed the can plus gas. They repeated the procedure with several other gases, being careful to use the same pieces of 'Blue Tack' each time and not losing any of it. Results are shown below.

Gas helium

81.41

41

methane

81 .58

A

V sealed with silicone

8

oxygen

81 .81

- empty drink can (approx 330 ml)

carbon dioxide

81.95

sulfur dioxide

82.29

II

They then measured the mass of the can (without any gas in it, not even air) by first measuring the volume of water needed to fill the can and the tubes (using volumetric flasks and burette for accuracy), then weighing the can (with 'Blue Tack' plugs) filled with water.

Volume of water to fill the can = 332 ml

Mass of can plus water = 412.7 g

Temperature of the water = 20.2°C

Density of water at 20.2°C = 0.998 g/mL

a Use these results to determine the mass of the empty can (that is, without any water, air or other gas in it).

b Combine the mass of the empty can with the results in the table to calculate the number of moles of each gas in the can. Comment upon whether or not the results confirm Avogadro's law.

Avogadro's law tells us that at constant temperature and pressure, volume of a gas is proportional to n, the number of moles present, and is independent of the nature of the gas; that is:

V = kn .. . (8.1 )

Let us now try to quantify the dependence of volumes of gases upon pressure and temperature.

Refer to Context 4 Air, pp. 132- 3.

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8.5 BOYLE'S LAW By measuring the pressures and volumes of many samples of different gases at constant temperature, Robert Boyle in 1662 flrst made the following obserYation:

For a given quantity of gas at a constant temperature, the product of its volume, V, and its pressure P, is constant. This is known as Boyle's law.

The following experiment demonstrates the validity of this law. 25 mL of air is dra\vn into a graduated gas-tight syringe and a pressure gauge is attached to it as shO\\n in Figure 8.3. The plunger of the syringe is pushed in and the new pressure and \'olume recorded; this is repeated twice. Then the plunger is released so pressure reduces back to the starting value, and then the plunger is pulled out to increase the volume and two more sets of pressure and volume values recorded. The results are shown in Table 8.1. We sec there that the product PV is constant (\vithin experimental error). This shows the validity of Boyle's la\\'.

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CHAPTER 8 GASES 321 ~

TABLE 8.1 Measurements showing the validity of Boyle's law

Volume, V (ml)

Pressure, P (kPa)

10--3 X PV

25

102

2.55

21

118

2.48

18

134

2.41

16

164

2.62

Average = 2.5 ± 0.1

30

82

2.46

39

66

2.57

No matter what gas was used, providing temperature was constant during the experiment, the product of pressure and YOlume was always constant (as in our example in Table 8.1). Boyle's law, like all scientific laws, is a summary of a large number of experimental observations.

We can write Boyle's law as:

PV = k ... (8.2)

where k is a constant. We say that volume is inversely proportional to pressure: if pressure is doubled, volume is halYed, or if pressure decreases to one-fifth of its initial value, volume increase to five times its initial value.

An alternative statement of Boyle's law is:

For a given quantity of gas at a constant temperature, the volume i~ inversely proportional to the pressure.

A very useful alternative form of Equation 8.2 is:

P2V2 = pl Vl ... (8.3)

(since each of the products, P2 V2 and P 1 Vp is equal to the same constant, k) . The follo\ving example illustrates the use of this equation.

EXAMPLE 1 A sample of gas originally had a volume of 1.5 L at 1.0 atmosphere (atm). What volume would it have at 4.0 atm pressure at the same temperature?

Let us call the initial conditions P1 and V1; P1 = 1.0 atm, V1 = 1.5 L. The final conditions are P2 and V2; P2 = 4.0 atm and V2 is what we want to find. By equation 8.3:

(4.0 atm) V2 = 1.0 x 1.5 atm L

v: _ 1.5 atm L 2 - 4.0 atm

= 0.38 L

In using Equation 8.3, we can use any units we like for pressure and volume, provided we use the same units on both sides of the equation.

FIGURE 8.3 Device for testing Boyle's law

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; 322 CHEMISTRY IN USE

EXERCISES 6 In the table below, P1 and V1 are the initial pressures and volumes, and P2 and v2

the final pressures and volumes, of samples of gases at constant temperatures. Calculate the missing entry in each line.

p1

3.4 atm

87kPa

2.2 atm

130kPa

v1 500ml

4.6 L

1.5 L

2.3 L

p2

0.82 atm

6.5 L

80 kPa

700mL

7 73.3 kPa pressure of gas in a 2.5 L flask is expanded into a total volume of 6.3 L at constant temperature. Calculate the final pressure.

8 In order to measure the volume of a piece of apparatus, a chemist filled a 750 ml flask with 93.3 kPa pressure of gas, then expanded it into the apparatus. The final pressure was 29.9 kPa. Calculate the total volume occupied by the gas, and hence the volume of the apparatus being measured.

Refer to Context 4 Air, p. 134.

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8.6 CHARLES' LAW In 1 787, the French physicist Jacques Charles summarised the results of many experiments in the following law:

Por a fixed quantity of gas at constant pressure, the volume increases linearly \vith temperature.

Clem perature is defined in terms of a mercury-in-glass thermometer on what we now call the Celsius scale.)

We can demonstrate the validity of Charles' law by modifying our previous apparatus as shown in Figure 8.4(a) . The syringe is connected to the pressure gauge by a bent capillary rube (very small \'Olume) and the body of the S)Tingc is immersed in a beaker of water, which can be heated over a Bunsen burner or cooled by adding ice to it. Initially, the gas in the syringe is at atmospheric pressure and room temperature; the volume and temperature are recorded. The water (and so the gas) is heated; the plunger of the syringe is slowly withdrawn to maintain the pressure at its initial value. Three sets of temperature and volume values are recorded. Then the hot \Vater is replaced by cold water and cooled by adding ice, and again constant pressure is maintained (now by pushing the plunger in); sets of temperature and volume values are recorded. Results are shown in Table 8.2.

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~ ~ 1

CHAPTER 8 GASES 323 : I '~

beaker of water to regulate temperature

35

::::r 30 .5. Q)

E :I

~ 25

FIGURE 8.4 20 L-----l...---L------1.--_..l (a) Apparatus for testing

0 25 50 75 1 oo Charles' law Temperature (°C)

TABLE 8.2 Volume as a function of temperature for a sample of gas at constant pressure (using the apparatus in Figure 8.4(a)

Temperature (0 C)

Volume (ml)

4

23

25

25

50

27

72

29

95

31

In Figure 8.4(b), volumes are plotted against temperatures (in degrees Celsius, °C). A straight line with positi,·e intercept results.

When more sophisticated apparatus is used, it is possible to make measurements over a wider range of temperature, and the results shown in Figure 8.5(a) for four different samples of gas are obtained.

-200 0 200

Temperature (0 C) (a)

Q)

E :I

~

400 0 200 400 600

Temperature (K) (b)

(b) A graph of results from the apparatus in (a)

FIGURE 8.5 Volumes of samples of gas at constant pressure versus (a) Temperature in oc, (b) Absolute temperature, K

I 324 CHEMISTRY IN USE

An interesting feature emerges from these plots. If we extrapolate all the lines to lower temperatures, we find that they all intersect the temperature axis at - 273°C (- 273.15°C using more exact methods). This is true for all samples of gas that have been studied.

Consequently, a new scale of temperature is introduced, called the absolute

scale of temperature, or the Kelvin scale of temperature, or the thermodynamic

scale of temperature. It is defined as follows:

which means that

273.15K = 0°C 373.15 K = 100°C

0 K = -273.15°C

The unit of temperature on this new scale is called the kel\'in, symbol K. One adYantage of this Kcl\'in scale of temperature is that all physically accessible temperatures have positive values. 0 K is the absolute zero of temperature: \Ve cannot get any lo\.-ver.

If the data of Figure 8.5(a) are replotted using absolute temperatures instead of Celsius temperatures, then Figure 8.5(b) results. All the lines now pass through the origin of the graph. We can write:

V = kT ... (8.4)

where k is constant (but not the same constant as in Equation 8.2). If y = kx, we say that y is proportional to x. Hence:

At constant pressure, the volume of a fixed quantity of gas is proportional to its absolute (or KeiYin) temperature.

Equation 8.4 can be written as: v = k T

(at constant pressure)

An alternative and more practical form of this equation is:

Vz vl Tz T l

v v since each of the quotients, -2 and - 1 is equal to the same constant k

T2 Tl (of Equation 8.5). The usc of this equation is easily demonstrated.

EXAMPLE 2

.• . (8.5)

.. . (8.6)

A sample of gas at 1.0 atm pressure had a volume of 2.5 L at 1 00°C. What would its volume be at ooc at the same pressure?

Let the initial conditions be V1 and T1. Hence V1 = 2.5 Land T1 = 100 + 273.2 = 373.2 K. (Caution: In Equation 8.6, temperature must be in kelvin.) Let the final conditions be V2 and T2 . Hence T2 = 0 + 273.2 = 273.2 K and V2 is what we want to find.

By Equation 8.6 2_ 2.5 L

273.2 K 373.2 K

\1. - 2.5 X 273.2 L K 2 - 373.2 K

= 1.8 L

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CHAPTER 8 GASES 325 1 ' '

EXERCISES 9 In the table below, V1 and (temph are the initial volume and temperature and V2

and (temp)2 are the final volume and temperature of samples of gas at constant pressures. Calculate the missing entry in each line of the table.

v1 500ml

~~~-~-----~

1.5 L

c 250mL

d 500 ml

(temp)1

200 K

350K

22.,C

1so•c

v2

2.2 L

1.3 L

(temp)2

400 K

ao•c

10 A sample of gas had a volume of 1. 7 L at 25°C and 1 atm pressure. What would its volume be at 250°C (at the same pressure)? To what temperature would it need to be cooled to have a volume of 900 ml?

11 To test Charles' law, a sample of dry nitrogen gas was trapped in a capillary tube tied to a small ruler as shown in the diagram below. This assembly was then immersed in a series of liquids at different temperatures and the length of the gas sample measured:

methanol cooled to dry ice temperature - 78°C, 22 mm ii ice water, o·c. 31 mm iii room temperature water, 20°C, 33 mm iv boiling water, 1 00°C, 42 mm v heated ethylene glycol, 151 °C, 48 mm

ruler

capi llary tube

Show

a graphically and

open to the - ----, atmosphere

plug of mercury

b numerically

that these results obey Charles' law.

12 The volume of gas in a constant-pressure gas thermometer was 61.0 ml at 280 K. When it was immersed in a liquid of unknown temperature, its volume increased to 73.3 ml. What is the temperature of the liquid?

Refer to Context Air, pp. 135-6.

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1 ~ 326 CHEMISTRY IN USE

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8.7 COMBINED GAS LAW Boyle's law and Charles' Jaw can be combined into the following:

For a fixed quamity of gas,

PV T = k* (a constant)

Another form of Equation 8. 7 is:

P2V2

T2

... (8.7)

.. . (8.8)

This equation is true because each side of it is equal to the same constant k*. However, this equation is only true if temperature is in kelvins (as also was the case with Equation 8.6). An example will illustrate the use of Equation 8.8.

EXAMPLE 3 A certain quantity of gas had a volume of 1.3 L at 1.0 atm pressure and 80°C. What pressure is needed to compress it to 500 mL at 30°C?

Let the initial conditions be P1, V1, T1; hence P1 = 1.0 atm, V1 = 1.3 L, T1 = 80 + 273.2 = 353.2 K. Let P2, V2, T2 be the final conditions; V2 = 500 mL = 0.5 L (units of V2 must be the same as those for V1) and T2 = 30 + 273.2 = 303.2 K. Hence by Equation 8.8:

P2 x 0.50 L 1.0 x 1.3 atm L = 303.2 K 353.2 K

p _ 1.3 x 303.2 atm L K 2 - 353.2 X 0.50 K L

= 2.2 atm

EXERCISES 13 In the table below, P1, V1 and (temp)1 are the initial pressure, volume and

temperature, and P2 , V2 and (temp)2 are the final pressure, volume and temperature of samples of gas. Calculate the missing item in each line.

p 1 v1 (temp)1 p 2 v2 (temp)2

101.3 kPa 1.5 L 300 K 53.3 kPa 450 K

1.3atm 500ml 280K 1.2 L 500K

250ml 25°C 106.6 kPa 0.60 L

1.3L 80°C 3.5 x 1o5 Pa 271°C

20 L 250°C 5.0 L 422°C

23ml 17°C 3.0atm 0.015 L

14 A gas, initially at 292 K, enclosed in a bulb of volume 15.1 mL at a pressure of 90.2 kPa, was first expanded into a reaction vessel so that the volume was 182.2 mL and then heated to 423 K. What is the final pressure in the vessel?

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CHAPTER 8 GASES 327 .: .1.'

15 A steel capsule of volume 25 ml contains carbon dioxide at a pressure of 5000 kPa; the contents are allowed to expand into an empty glass vessel of volume 1 L. In so doing, the temperature of the gas falls from 295 K to 282 K. Calculate the final pressure in the vessel.

16 Some balloons for upper atmosphere research are designed so that their shapes can alter in order to change their volumes, and as a result, the pressure inside the balloon is always equal to the outside atmospheric pressure. Such a balloon had a volume of 10 m3 at 1.0 atm pressure and 20°C. What would its volume be in the upper atmosphere where the pressure was 0.050 atm and the temperature -50°C.

17 A chemist measured the pressure of a sample of nitric oxide in a cylinder of constant volume at several temperatures. The results are given below.

*a Determine graphically the relation which exists between the pressure and temperature.

b Is this what you expect from the combined Boyle's and Charles' law? Explain.

Temperature (0 C}

Pressure (atm)

0

1.05

25

1.15

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8.8 IDEAL GAS EQUATION When dealing with gases in many real-life situations such as scuba diving and in using anaesthetics, all of the conditions of pressure, temperature, volume and quantity of gas can be varied. To handle such situations we need to combine Boyle's and Charles' laws with Avogadro's law (in the form of Equation 8.1) to get:

PV - = constant nT

... (8.9)

Part of Avogadro's law is that at constant P, T and n, the volume is constant for all gases; this means that the constant in Equation 8.9 is the same for all gases: 112,

He, 0 2, N 2 , Ar, C02 and so on. We use the symbol R for this constant and call it the universal gas constant.

Hence

or

PV = R nT

PV = nRT

This is called the ideal gas equation or the general gas equation.

•.. (8.10)

•.• (8.11)

This equation is called the ideal gas equation because real gases deviate slightly from it, particularly at high pressures and low temperatures.

The value of R can be calculated from the molar volume of a gas at STI~ namely 22.4 L. Hence putting P = 1.00 atm, V = 22.4 L, n = 1.00 mol and T = 273.2 K into Equation 8.10 gives:

R = 1 x 22.4 atm L 1 x 273.15 mol K

that is R = 0.0820 L atm K- 1 mol-1

(or 0.08206 using a more precise value for V).

This value of R uses a non-systematic unit for pressure, namely the atmosphere. To obtain R in 'proper' units, we put P = 1.013 x 105 N m-2 (that is, pascals,

Ill

~ .

I 328 CHEMISTRY IN USE

using the conversion factor at the bottom of p. 315) and put volume in cubic metres, that is, V = 22.4 x 10-3m3 (because 1 m3 = 103 L).

Hence: R = 1.013 X 105

X 22.4 X 10-3 1 X 273.15

= 8.31 N m K-1 mol-l

Using more precise values for P and V, the value is 8.314. Now a newton metre, N m, is a joule,J (the unit of work or energy), so that:

R = 8.314 J K-1 m ol-l

The following examples illustrate the use of the ideal gas equation.

EXAMPLE 4 What volume will 2.50 moles of carbon dioxide occupy at 400 kPa pressure and 1 00°C?

As we are given P, T and n, and we have to find V, Equation 8.11 is required.

Hence

P = 400 kPa

= 4.00 x 105 Pa

T = 100 + 273.2

= 373.2 K

n = 2.50 mol

R = 8.314 J K-1 mol-1

V = nRT p

(and J = N m)

2.50 x 8.314 x 373.2 mol N m K-1 mor-1 K = 4.00 X 105

= 1 .94 X 1 o-2 m3

Or to put in more familiar units:

V = 19.4 L

The only difficulty in using Equation 8.11 is getting the units right. A foolproof approach is to appreciate that a symbol such as P, V or T represents a physical quantity and that the value of a physical quantity consists of a number and a unit: Pis not just 1 os but 105 Pa; Vis not 2.3 but 2.3 L. Hence when we substitute a value for a symbol, we should put in boch che number and the unit. If this is done for all physical quantities in an equation, then the units should cancel or combine to give a suitable unit for the physical quantity being calculated. This practice has been followed in Examples 1 to 4 above. i'Ou are strongly urged to adopt it too. If an inconsistent set of units is inad\·ertently fed into an equation, then the units will not cancel or combine to give sensible units for the answer, and the error is quickly recognised.

The use of systematic units is strongly recommended:

Pin pascals, Pa ( = N m-2)

V in cubic metres, m3

Tin kelvins, K W'e then useR = 8.314 J K-1 mol- l

= 8.314 N m K- 1 mol- 1

\ t ~· I

CHAPTER 8 GASES 329 : :~J;;,

It is perhaps worth noting that Pin kilopascals and V in litres is equivalent to P in pascals and V in cubic metres, because:

1 kPa X 1 L = 1 X 103 Pa X 1 X w-3 m3 = 1 Pa m 3

Hence an alternative to putting Pin pascals and V in cubic metres is to put Pin kilopascals and V in litres.

The key to handling units in the ideal gas equation is to recognise that:

Pa x m3 = J or kPa x L = J

Although in systematic units R is always given as:

R = 8.314 J K-1 mol-1

you may like to 'translate' that into:

R = 8.314 Pa m3 K.-1 mol-1

or to R = 8.314kPaLK- l mol- l

If R is available in litre atmosphere units (that is R = 0.0820 L atm K-1 mol- 1)

and if pressure is given in atmospheres, it is quicker to work the problem in atmospheres rather than convert to pascals.

EXAMPLE 5 Calculate the molar mass of a gas for which 8.79 g is needed to fill a 2.55 L flask to a pressure of 1.30 atm at 21.2°C.

In order to calculate the molar mass, we need to know how many moles there are in 8. 79 g, that is how many moles are in the flask. Hence we use the ideal gas equation to calculate n.

P = 1.30 atm, V = 2.55 L, T = 21.2 + 273.2 = 294.4 K

From Equation 8.11:

n = PV RT

= 1.30 x 2.55 atm L 0.0820 x 294.4 L atm K-1 mol-1 K

= 0.137 mol

That is, 8. 79 g is 0.137 mol.

Hence 1 mol = 8. 79 0.137

= 64.1 g

The molar mass is therefore 64.1 g/mol.

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I ' 330 CHEMISTRY IN USE

EXERCISES 18 Pressure, volume, temperature and number of moles of different samples of gases

are listed in the table. Calculate the missing entry in each line.

p v Temperature n

5.0 l 300 K

450K

c 0.01 0 m3 292 K 0.055 mol

500ml

2.0 l 26'C

35°C

19 What pressure in kilopascals will 2.0 g argon exert when placed in a 500 ml cylinder at 18°C?

20 What mass of nitric oxide, NO, is present in a 2.5 L flask in which the pressure is 6.67 x 104 Paat310K?

21 What volume is needed to store 0.800 g sulfur dioxide at a pressure of 1 01.3 kPa at 1JOC?

22 To what pressure must a piece of equipment be evacuated in order that there be only 1 os molecules per ml at 25°C?

23 5.0 g of an unknown gas produced a pressure of 19.4 kPa in a 1 0.0 L flask at 25°C. Calculate the molar mass of the gas.

24 A chemist measured the pressure generated at 80°C in a cylinder of fixed volume by different masses of different gases. The results are given below. Show graphically that these results are in accord with the ideal gas equation.

·--·-·-

so2

3.2

54

NO

2.40

89

HCI

3.65

109

C02

5.72

145

NH3

3.06

197

25 A gas exerts a pressure of 0.33 atm in a 1. 7 L flask at 28°C. How many moles are there in the flask?

26 What pressure (in atm) will be exerted by 87.6 g ammonia in a 5.2 L steel cylinder at 24.2°C?

27 A steel cylinder of volume 50 L is filled with argon to a pressure of 1 05 atm at 19°C. What mass of gas is in the cylinder?

28 Some chemists, trying out a new technique for synthesis, prepared a pure gaseous compound containing sulfur and fluorine only. They collected a sample of their compound in a small flask of volume 25.7 ml. At 27.2°C the sample exerted a pressure of 0.50 atm and had a mass of 0.0772 g. Which of the compounds SF2,

SF 4 or SF 6 had they prepared? Explain your answer fully.

Most our discussion of gases so far has concerned single pure gases. I ,et us now look at mixtures of gases.

Refer to Context 4 Air, pp. 137-8.

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~ I ~· X ,

CHAPTER 8 GASES 331 ,~ 1.,~ ;

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8.9 DALTON'S LAW OF PARTIAL PRESSURES In the early 1800s, John Dalton made the observation which is no" called Dalton's law of partial pressures:

If two or more gases arc present in the same container, the total pressure exerted is the sum of the pressures each gas \\·ould exert if it alone occupied the container .

where P101

is the total pressure and PA, P 8 , P c, ... are what are called partial pressures of gases A, B, C ...

... {8.12)

The partial pressure of a gas in a mixture of gases is the pressure which that gas

would exen if it alone were present in the container. An alternative statement of the law of partial pressures is:

The total pressure in a gas mixture is the sum of the partial pressures of the individual gases.

Tf 0.10 mole of nitrogen and 0.050 mole of chlorine are present in a 2.0 L flask at 20°C, the partial pressure of nitrogen, PN, is calculated by putting 11 = 0.10 mole, T = 293.2 K, V = 2.0 x 10-3m3 into Equation 8.11:

...:.0.:....:.1:..::0_x____::.8...::.3...::1.:....:4_x--'2:;:..;9.:....:3.:....:·.;;;.2 mol Pa m 3 K- 1 mol-1 K PK =

2.0 X 10- 3

= 1.22 x 105 Pa

The partial pressure of chlorine, P CJ> is similarly calculated:

0.050 x 8.314 x 293.2 mol Pa m 3 K-1 mol- 1 K p CI = ..;;_;;...:...;;_____::...:.c::._;__ __ _

2.0 X 10- 3

= 6.09 x 104 Pa

The total pressure, P10

t, is the sum of these partial pressures:

prot = PC! + p't\

= (0.61 + 1.22) x 105 Pa

= L83 x 105 Pa

As this illustration shows, the law of partial pressures means that in a mixture, the

gas equation can be applied to each gas separately to calculate the partial pressure of gas, i, in terms of n,, the monber of moles of gas, i, present; alternatively the gas law can be used to calculate the total pressure of all the gases in the containe1; pnn·ided tlzat n, in that case, is the total number of moles present.

That is, in a mixture of gases:

... {8.13)

where P; is the partial pressure of gas, i, and n; is the number of moles of gas, i,

and . .. (8.14)

\Vhere P101 is the total pressure and nrot is the total number of moles of gas. We can use the ideal gas equation in whichever form is more convenient for

the problem at hand. In the illustration just discussed we used it in the form of Equation 8.13. Let us now illustrate the use ofEquation 8.14.

I I

332 CHEMISTRY IN USE

EXAMPLE 6 If 4.0 g carbon dioxide and 5.0 g carbon monoxide are placed in a 5.0 L flask at 15°C, what is the total pressure of the gas mixture?

To use the gas law, we need the total number of moles of gas. The molecular weights of C02 and CO are 44 and 28 respectively. Hence:

total number of moles = 4·0 + 5·0 44 28

= 0.27 mole

Putting n = 0.27 mole, T = 288.2 K, V = 5.0 x 1 o-3 m3 into Equation 8.14:

p = nRT v

0.27 x 8.314 x 288.2 mol Pa m3 K-1 mol-1 K

EXERCISES

= 5.o x 10-3

= 1.3 x 10s Pa

= 130 kPa

29 Two small flasks with closed stopcocks are attached by short tubes to a large flask. The large flask and connecting tubing is well evacuated. One small flask, of volume 25 ml, contains oxygen at a pressure of 101.3 kPa, while the other, of volume 75 ml, contains helium at a pressure of 70.9 kPa. The stopcocks of the small flasks are opened so that the two gases mix and expand into the evacuated volume. The total volume of the whole apparatus is 1 0.0 L. Calculate the partial pressures of oxygen and helium and the total pressure.

75ml__L

total volume 10.0 L

30 A 5.0 L cylinder containing hydrogen at a pressure of 500 kPa is connected to a 3.0 L cylinder containing argon at a pressure of 300 kPa, and sufficient time is allowed for complete mixing. Calculate the partial pressures of hydrogen and argon and hence the total pressure in the joined cylinders.

31 0.5 g carbon dioxide and 0.7 g neon were placed in a 2.0 L flask. Calculate the partial pressures at 25°C and the total pressure.

Refer to Context 4 Air, pp. 14D-1 .

II

~ ' ,:).· CHAPTER 8 GASES 333 . , ,

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8.10 SUMMARY This chapter has shown how to calculate one of the properties of a gas (pressure, volume, temperature, number of moles) from known values of other properties and how to calculate the change in one property when other properties are changed. In this regard the key equations are 8.8 (of which 8.3 and 8.6 are special cases) and 8.11. The key to success in using these equations is using correct or consistent units.

\Vhen mixtures are im·oh·ed, we use Dalton's law of partial pressures, Equation 8.12, and its consequences, Equations 8.13 and 8.14.

IMPORTANT NEW TERMS You should know the meaning of the following terms:

absolute scale of temperature (Kelvin scale) (p. 324)

atmosphere (unit of pressure) (p. 315) Avogadro's law (hypothesis) (p. 319) Boyle's law (p. 320) Charles' law (p. 322) combined gas law (p. 326) Dalton's law of partial pressures (p. 331) diffusion (p. 317) ideal gas equation (general gas equation)

(p. 327) Kelvin scale of temperature (absolute

scale) (p. 324)

TEST YOURSELF

kinetic theory of gases (p. 318) molar volume of a gas (p. 319) partial pressure (p. 331) pascal (Pa) and kilopascal (kPa) (p. 315) pressure (p. 315) standard temperature and pressure

(STP) (p. 317) thermodynamic scale of temperature

(p. 324) universal gas constant, R (p. 327)

1 Explain the meaning of each of the items in the 'Important new terms' section above.

2 State the basic postulates of the kinetic theory of gases.

3 What is the systematic unit for pressure?

4 How is the absolute or Kelvin scale of temperature defined? What is its relationship to the Celsius scale?

5 What is meant by STP?

6 Describe an experiment for demonstrating the validity of Boyle's law.

7 Describe an experiment that could be performed to test Charles' law.

8 Explain the following:

a The pressure in bicycle tyres increases on a hot day. b A weather balloon increases in volume as it rises.

9 Two flasks of equal volume are filled with nitrogen and helium respectively to the same pressure at the same temperature. Compare the two samples in terms of:

a the number of molecules present b the mass of gas present.

10 What experimental fact can be used to calculate a value of the universal gas constant, R?

11 By volume the atmosphere is 20% oxygen, 79% nitrogen and 1% argon. On a day when atmospheric pressure is 1 00 kPa, what are the partial pressures of these gases?