]> Properties of Gases & Gas Laws< matter< chemistry< high school< ICSE CBSE< mentorials.com
Top Box
Section Navigator
Author: S.LAL
Created: 30 Jul, 2010; Last Modified: 04 Nov, 2016

Matter - 05

Properties of Gases

Because the distance between the particles of a gas are orders of magnitude more than the distance between particles in liquids and solids, gases have certain unique properties. These are as follows:

  • Gases are fluids, which means they can flow, like liquids. The particles, being far apart, have great freedom of movement.
  • Gases have very low densities as compared to liquids and solids because of the large particle separations. In fact, most of the volume occupied by a gas is empty space.
  • Gases are highly compressible, since the space between particles is very large compared to the volume taken up by the particles.
  • A gas expands/contracts to completely fill whatever closed container it is put in. This happens because the particles have great freedom of movement and are not constrained by mutually attractive interparticle forces.

The Kinetic-Molecular Theory of Gas

gas-particles.gif
Fig 1: Gas particles. (Source: Bishop, Introduction to Chemistry, p. 484)

The properties of gases as outlined above have their explanations based on the kinetic-molecular theory of gas (KMT), which is essentially the kinetic theory of matter applied to gases (see Fig. ).

According to this theory, the particles of a gas are in constant rapid and random motion. They move in straight lines, till they collide with each other or with the walls of the container, which changes their direction of motion. The particles are very far apart relative to their sizes. This explains the fluidity, low density and compressibility of gases.

This theory states that the pressure exerted by a gas is due to the frequency of collisions of the gas particles with each other and the walls of its container.

Gas Pressure

Our atmosphere, or air, is a mixture of gases, and envelopes the earth. The air particles are pulled towards the earth due to gravity, making the air denser at lower altitudes than at higher altitudes. The collisions of the air particles with anything in contact with the atmosphere, such as the earth's surface, results in atmospheric pressure or air pressure. Since the atmosphere is denser at lower altitudes, the atmospheric pressure at sea level, for example, will be higher than at the top of a mountain.

Measuring pressure

Pressure is defined as the force per unit area.

In SI units, the unit of force is the newton (N), which is the force that gives an object of mass 1kg an acceleration of 1m/s2.

The SI unit of pressure is the pascal (Pa), which is the the force of 1N applied to an area of 1m2. 1Pa is actually a very small unit of pressure.

Pressure is measured using an instrument called a barometer.

Mercury Barometer
mercury-barometer.gif
Fig 2: Measuring pressure with mercury barometer. (Adapted from: Myers, Oldham & Tocci, Holt Chemistry, p. 419)

A mercury barometer consists of a long tube closed at one end, filled with mercury (Hg), and inverted in a vessel containing mercury (see Fig. ).

As depicted in the figure, the air exerts pressure on the surface of mercury in the vessel. When the tube filled with mercury is inverted into the pool of mercury (taking care not to allow air in), the mercury in the tube falls till the pressure exerted downwards by the weight of the mercury column balances the atmospheric pressure at the surface of the pool.

At sea level, it is found that the air pressure can support a mercury column height of 760mm in the barometer. In other words, at sea level, a column of air of height equal to the thickness of the atmosphere (which is thousands of kilometers) exerts a pressure equal to the pressure exerted by a column of mercury 760mm high.

This pressure is read as 760mm Hg or 760torr (after the Italian physicist Torricelli who invented the barometer). 760mm Hg is also called 1atm (atmosphere).

1 atm = 760    mm Hg = 760 torr

.......(1)

STP – standard temperature and pressure

For studying changes in properties of gases with changing temperatures and pressures, a standard for comparison purposes is useful. The scientific community uses a set of standard conditions called standard temperature and pressure, or STP, which stands for a temperature of 273K (0°C) and a pressure of 1atm. STP is also called NTP (normal temperature and pressure).

The Gas Laws

Unlike solids and liquids, gases exhibit remarkably similar behaviour regardless of their chemical structure. In the 17th century, scientists found out that the physical properties of gases can be defined using four variables: pressure (P), volume (V), temperature (T) and amount, or number of moles (n). The gas laws define the relationships between these variables. A gas which exactly follows these relationships is called an ideal gas.

There are four main gas laws – Boyle's law, Charles's law, Gay-Lussac's law and Avogadro's law. The last two laws will be dealt with later on in the subject.

Boyle's Law: pressure-volume relationship

boyle's-law-pressure-volume.gif
Fig 3: Pressure-volume inverse relationship.

In 1662, the English scientist Robert Boyle found that, for a gas in a container maintained at a constant temperature, the volume of the gas decreases with rising pressure.

The pressure-volume relationship is an example of an inverse relationship, in which increasing the value of one parameter decreases that of the other, and vice-versa. (see Fig. )

On the basis of the KMT, decreasing the volume of a gas implies that there will be more frequent collisions on the walls of the container by the gas particles since the number of particles is remaining constant. Hence, the pressure, which is a measure of the frequency collisions of the gas particles with the walls of the container, goes up. The opposite is true when the volume of the gas is increased.

boyle's-law-pressure-inverse-volume.gif
Fig 4: Pressure against inverse-volume, at constant temperature.

Boyle, on taking accurate measurements, further found that not only are pressure and volume inversely related, but at constant temperature, they are inversely proportional. A graph of the pressure against the inverse of volume came out to be a straight line (Fig. ).

Boyle's law:
At constant temperature, the pressure of an ideal gas is inversely proportional to its volume.

Thus,

P 1 V P = k V

where: P = pressure, V = volume, k = proportionality constant

Rearranging, we get the mathematical formulation for Boyle's law:

 P  V =  k    (at constant temperature)

.......(2)

Thus, according to Boyle's law, for ideal gases at constant temperature, the product of pressure and volume is constant.

A useful form of the above equation, for the purpose of solving problems, is:

 P 1  V 1 =  P 2  V 2

.......(3)

where: P1 = original pressure, V1 = original volume, P2 = new pressure, V2 = new volume

A sample of helium occupies a volume of 160cm3 at 100kPa and 25°C. What volume will it occupy if the pressure is decreased to 80kPa at constant temperature? (Adapted from: FHSST, Chemistry 10 - 12, Worked Example 26, p. 130)

We have:

V1 = 160cm3 P1 = 100kPa T1 = 25°C
V2 = ? P2 = 80kPa T2 = 25°C

Boyle's law is applicable since this is a constant temperature process. Thus,

 P 1  V 1 =  P 2  V 2

Rearranging to make V2 the subject of the formula:

 V 2 =  P 1  V 1  P 2

Substituting the values:

 V 2 =  P 1  V 1  P 2 = 100 ( kPa ) × 160 2 ( cm 3 ) 80 ( kPa ) = 200 cm 3

Thus, the final volume occupied by helium is 200cm3.

Note 1: In examples like this one, it was not necessary to convert units to SI units during the calculation process. This is so because the change of pressures or volumes to SI units require multiplication with conversion factors, which can as easily be applied to the final result, if the result is required to be in SI units. However, you should make sure that the units should be used consistently throughout the calculation process.

Note 2: The rules for significant figures arithmetic have not been applied.

Charles's Law: volume-temperature relationship

charles's-law-volume-temperature.gif
Fig 5: Volume-temperature direct relationship.

Around 1787, Jacques Charles had found that, at constant pressure, the volume of a gas increases with rising temperature.

This volume-temperature relationship is an example of a direct relationship, in which increasing the value of one parameter increases that of the other.

On the basis of the KMT, increasing the temperature of a gas results in the gas particles becoming more energetic, which increases the frequency and the force of collisions on the walls of the container by the gas particles. If the container walls are flexible, the container will tend to increase in volume.

Charles found that not only are volume and temperature directly related, but at constant pressure, volume and absolute temperature are directly proportional. A graph of the volume against temperature is a straight line (see Fig. ).

From the graph it might appear that as the absolute temperature reduces to 0K (absolute zero), the volume too becomes zero. However, the gas liquefies to a liquid, and finally solidifies to a solid, which has a certain volume, before the temperature reaches absolute zero.

Charles's law:
At constant pressure, the volume of an ideal gas is directly proportional to its absolute temperature.

Thus,

 V  T        V = kT

where: V = volume, T = temperature, k = proportionality constant

Rearranging, we get the mathematical formulation for Charles's law:

 V  T =  k    (at constant pressure)

.......(4)

Thus, according to Charles's law, for ideal gases at constant pressure, the division of volume by temperature is constant.

A useful form of the above equation, for the purpose of solving problems, is:

 V 1  T 1 =  V 2  T 2

.......(5)

where: V1 = original volume, T1 = original temperature, V2 = new volume, T2 = new temperature

At a temperature of 25°C, a certain amount of carbon dioxide gas occupies a volume of 6L. What volume will the gas occupy if its temperature is reduced to 0°C at constant pressure? (Adapted from: FHSST, Chemistry 10 - 12, Worked Example 29, p. 135)

We have:

V1 = 6L T1 = 25°C = 298K
V2 = ? T2 = 0°C = 273K

Charles's law is applicable since this is a constant pressure process. Thus,

 V 1  T 1 =  V 2  T 2

Rearranging to make V2 the subject of the formula:

 V 2 =  V 1  T 2  T 1

Substituting the values:

 V 2 =  V 1  T 2  T 1 = 6 ( L ) × 273 ( K ) 298 ( K ) = 5.5 L

Thus, the final volume occupied by carbon dioxide is 5.5L.

Note 1: In this example, the conversion of temperature from celsius to the SI unit of kelvins is required, because the conversion for temperature requires addition of a fixed conversion term, not multiplication by a fixed factor, as in the case of pressure and volume.

Note 2: The rules for significant figures arithmetic have not been applied.

The General Gas Law Equation

Boyle's Law relates the pressure and volume of a gas at constant temperature,  P 1  V 1 =  P 2  V 2 , while Charles's Law relates the volume and temperature at constant pressure,  V 1 /  T 1 =  V 2 /  T 2 . Combination of Boyle's Law and Charles's Law into a single expression gives the general gas law equation:

 P  V  T =  k

.......(6)

A useful form of the above equation is:

 P 1  V 1  T 1 =  P 2  V 2  T 2

.......(7)

where: P1, V1 and T1 are the original pressure, volume, and temperature; P2, V2 and T2 are the new pressure, volume, and temperature

At the beginning of a journey, a truck tyre has a volume of 30dm3 and an internal pressure of 170kPa. The temperature of the tyre is 16°C. By the end of the trip, the volume of the tyre has increased to 32dm3 and the temperature of the air inside the tyre is 40°C. What is the tyre pressure at the end of the journey? (Adapted from: FHSST, Chemistry 10 - 12, Worked Example 30, p. 138)

We have:

P1 = 170kPa V1 = 30dm3 T1 = 16°C = 289K
P2 = ? V2 = 32dm3 T2 = 40°C = 313K

The general gas law equation can be used to solve this problem. Thus,

 P 1  V 1  T 1 =  P 2  V 2  T 2

Rearranging to make P2 the subject of the formula:

 P 2 =  P 1  V 1  T 1  T 2  V 2

Substituting the values:

 P 2 =  P 1  V 1  T 2  T 1  V 2 = 170 × 30 × 313 289 × 32 = 172.6 kPa

Thus, the final pressure of the gas is 172.6kPa.

Note: The rules for significant figures arithmetic have not been applied.

Feedback

List of References

Bishop, M, An Introduction to Chemistry (First Version), viewed 10 February 2007 <http://preparatorychemistry.com/Bishop_Chemistry_First.htm>, 2005.
FHSST, Chemistry Grades 10 - 12, ver 0, viewed 10 March, 2009, <http://www.fhsst.org>, 2008.
Myers, RT, Oldham, KB & Tocci, S, Holt Chemistry, USA:Holt, Rinehart and Winston, 2006.

Bibliography

Bishop, M, An Introduction to Chemistry (First Version), viewed 10 February 2007 <http://preparatorychemistry.com/Bishop_Chemistry_First.htm>, 2005.
FHSST, Chemistry Grades 10 - 12, ver 0, viewed 10 March, 2009, <http://www.fhsst.org>, 2008.
McMurray, J & Fay, RC, Chemistry, 4th edn, USA: Prentice Hall, 2003.
Myers, RT, Oldham, KB & Tocci, S, Holt Chemistry, USA:Holt, Rinehart and Winston, 2006.
Nathan, HD & Henrickson, C, CliffsQuickReview Chemistry, NY, USA: Hungry Minds, 2001.
Uvarov, EB & Chapman, DR, A Dictionary of Science, 5th edn, UK: The Chauser Press, 1979.
Whitten, KW, Davis, RE, Peck, L & Stanley, GG, General Chemistry, 7th edn, Belmont, USA: Thomson Brooks/Cole, 2004.