The Behaviour Of Gases

An Introduction to the Behaviour of Gases The atmosphere is too vast for us to have any control over atmospheric pressure. Contained gases are a different matter. We can add gas or remove it, shrink or expand the container, or heat or cool the gas. In this chapter we will examine the work of such eminent scientists as Robert Boyle, Jacques Charles, and Joseph Gay-Lussac. These scientists studied the effects of changes in the pressure, volume, and temperature of contained gases. From their results they proposed a set of relationships that together are known as the gas laws. Using the kinetic theory, however, we can often explain how gases will respond to a change of conditions without resorting to formal mathematical expressions. We will therefore begin this chapter with some examples of how simple kinetic theory is used to explain gas behaviour. Our emphasis will be on gas pressure.

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The Effect of Adding or Removing Gas When we pump up a tire, the pressure inside it increases. The pressure exerted by an enclosed gas is caused by collisions of gas particles with the walls of the container. By adding gas we increase the number of gas particles. This increases the number of collisions and therefore the gas pressure. Doubling the number of gas particles doubles the pressure if the temperature of the gas does not change (Figure 1). Tripling the number of gas particles triples the pressure and so forth. With a powerful pump and a strong container we can generate very high gas pressures. Once the pressure exceeds the strength of the container, however, the container will rupture. Over inflated balloons burst for this reason.

100 kPa

200 kPa

+200 kPa

Figure 1: Gas is pumped into a closed rigid container, the pressure increases in proportion to the number of gas particles added. If the number of particles doubles, the pressure doubles. If the pressure exceeds the strength of the container, the container explodes.

Letting the air out of a tire or the gas out of a storage cylinder decreases the pressure in the container. Fewer particles are left. As you probably already guessed, halving the number of particles in a given volume of gas decreases the pressure by one-half. When a sealed container of gas under pressure is opened, the gas always moves from the region of higher pressure to the region of lower pressure. This is because there is more empty space for the gas particles to occupy. Gas particles increase their randomness by moving into this empty space, which is always a favourable process. Expansion continues until the gas pressures inside and outside the container is equal. Once the pressures are equal, the mixing of gases in the container with gases in the surrounding atmosphere is only by diffusion.

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The Effect of Changing the Size of the Container We increase the pressure exerted by a contained gas when we reduce the size of the container. The more a gas is compressed, the greater the pressure it exerts on its container. Reducing the volume of the container by one-half has the same effect on pressure as doubling the quantity of gas while keeping the volume constant (Figure 2). Increasing the volume of the container has just the opposite effect. By doubling the volume we halve the gas pressure. There are now onehalf as many gas particles in a given volume.

1 Litre

0.5 Litre

Figure 2: Doubling the force on a gas reduces the volume of the gas by one-half and doubles the pressure it exerts.

Gases cool when they expand and heat when they are compressed. When the gas is rapidly released from an aerosol can, the can becomes cooler. This is because the expanding gas absorbs some thermal energy from the container and its contents as it escapes. Compressing a gas always increases the temperature.

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The Effect of Heating or Cooling a Gas Raising the temperature of an enclosed gas increases the gas pressure. The kinetic energy of gas particles increases as they absorb thermal energy. Naturally, fast-moving particles bombard the walls of their container harder than slow-moving particles. The average kinetic energy of gas particles doubles with a doubling of the Kelvin temperature. Hence doubling the Kelvin temperature of an enclosed gas doubles the pressure.

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Boyle’s Law for Pressure – Volume Changes When the pressure goes up, the volume goes down. Similarly, when the pressure goes down, the volume goes up. In 1662, the British chemist Robert Boyle proposed a law to describe this behavior of gases. Boyle’s law states that for a given mass of gas at a constant temperature, the volume of the gas varies inversely with the pressure. We can write Boyle’s law as:

P1 X V1 = P2 X V2

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An Example of Boyle’s Law for Pressure – Volume Changes 1. A balloon is filled with 30 L of helium gas at 100 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25 kPa? (Assume that the temperature remains constant).

P1 x V1 = P2 x V2

P1 = 100 kPa

(100 kPa)(30 L) ÷ (25 kPa) = V2

V1 = 30 L

(3000 L) ÷ (25) = V2

P2 = 25 kPa

120 L = V2

V2 = ?  120 L

The balloon will expand to a volume of 120 L at 25 kPa.

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Charles’ Law for Temperature – Volume Changes In 1787 a French physicist, Jacques Charles, investigated the effect of temperature on the volume of a gas at a constant pressure. Charles summarized his observations into a law. Charles’ law states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant. Temperatures in gas law problems are always in Kelvin. We can write Charles’ law as:

V1 ÷ T1 = V2 ÷ T2

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An Example of Charles’ Law for Temperature – Volume Changes 1. A balloon, inflated in an air-conditioned room at 27 ̊C, has a volume of 4.0 L. It is heated to a temperature of 57 ̊C. What is the new volume of the balloon if the pressure remains constant?

Note: Temperature in all gas problems is expressed as degrees Kelvin. Therefore we must convert degrees Celsius to degrees Kelvin.

T1 = 27 ̊C + 273 = 300 K T2 = 57 ̊C + 273 = 330 K

V1 ÷ T1 = V2 ÷ T2

V1 = 4.0 L

(4.0 L) ÷ (300 K) = V2 ÷ (330 K)

T1 = 300 K

((330 K) (4.0 L) ÷ (300 K)) = V2

V2 = ?  4.4 L

4.4 L = V2

T2 = 330 K

The balloon will expand from a volume of 4.0 L at 27 ̊C to a volume of 4.4 L at 57 ̊C.

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Gay-Lussac’s Law for Temperature – Pressure Changes On a hot summer day the pressure in a car tire increases. This illustrates a relation discovered in 1802 by Joseph Gay-Lussac, a French chemist. Gay-Lussac’s law states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume is kept constant. We can write Gay-Lusaac’s law as:

P1 ÷ T1 = P2 ÷ T2

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An Example of Gay-Lussac’s Law for Temperature – Pressure Changes 1. The gas in a container has a pressure of 300 kPa at 27 ̊C (300 K). What will the pressure be if the temperature is lowered to -173 ̊C (100 K)?

P1 ÷ T1 = P2 ÷ T2

P1 = 300 kPa

(300 kPa) ÷ (300 K) = P2 ÷ (100 K)

T1 = 300 K

((100 K) (300 kPa) ÷ (300 K)) = P2

P2 = ?  100 kPa

100 kPa = P2

T2 = 100 K

The pressure will be 100 kPa if the temperature is lowered to 100 K.

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The Combined Gas Law The three gas laws found above can be combined into a single expression called the combined gas law. The other laws can be obtained from the combined gas law by holding one quantity (pressure, volume, or temperature) constant. We can write the combined gas law as:

P1 X V1 ÷T1 = P2 X V2 ÷ T2

If you have been wondering how to remember the expressions for the other gas laws, it turns out that there is really no need. The other laws can be obtained from the combined gas law by holding on quantity (pressure, volume, or temperature) constant.

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An Example of the Combined Gas Law 1. A cylinder of compressed oxygen gas has a volume of 30 L and 10,000 kPa pressure at 300 K. The cylinder is cooled until the pressure is 500 kPa. What is the new temperature of the gas in the cylinder?

P1 x V1 ÷ T1 = P2 x V2 ÷ T2

P1 = 10,000 kPa

(10,000 kPa)(30 L) ÷ (300 K) = (500 kPa)(30 L) ÷ T2 T2 = (500 kPa)(300 K) ÷ (10,000 kPa)

V1 = 30 L T1 = 300 K

T2 = 15 K

V2 = 30 L T2 = ?  15 K (-258 ̊C)

K = ̊C + 273 15 = ̊C + 273 15 – 273 = ̊C -258 = ̊C

The new temperature of the gas in the cylinder is 15 K, or, -258 ̊C.

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The Ideal Gas Law Sometimes we wish to calculate the number of moles of a gas in a fixed volume at a known temperature and pressure. Such a calculation is possible if the combined gas law is modified. The modification may be understood by recognizing that the volume occupied by a gas at a specified temperature and pressure is directly proportional to the number of particles in the gas. The number of moles, n, of gas is also directly proportional to the number of particles. Hence, moles must be directly proportional to the volume as well. Therefore, moles may be introduced into the combined gas law by placing n in the denominator on each side of the equation.

P1 X V1 ÷T1 X N1 = P2 X V2 ÷ T2 X N2 OR

PV = NR(L X KPA ÷ K X MOL)T Where “R” is the ideal gas constant, and is equal to 8.314.

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An Example of the Ideal Gas Law 1. A deep underground cavern contains 2.24 x 106 L of methane gas at a pressure of 1500 kPa and a temperature of 42 ̊C. How many grams of methane does this natural gas deposit contain?

PV = nRT

P = 1500 kPa

(1500 kPa)(2.24 x 106 L) = n(8.314 (L x kPa ÷ K x mol))(315 K)

V = 2.24 x 106 L

33.6 x 108 = n(2618.91 mol)

n = ?  1.28 x 106

(33.6 x 108) ÷ (2618.91 mol) = n

R = 8.314

1.28 x 106 = n

T = 315 K

1.28 x 106 mole of CH4

=

16.05 g 1 mol

=

2.05 x 107 g of CH4

There are 2.05 x 107 grams of methane that this natural gas deposit contains.

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