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9.2Ideal gases

Pressure

Pressure the normal force on an area per unit area is defined as the normal force applied per unit area. In Figure 9.1a the force is normal to the area A, so the pressure is:

P=F A

Figure 9.1: Pressure is the normal force per unit area.

The force in Figure 9.1b acts at an angle θ, so the pressure on the area A is given by the expression:

P=F cos θ A

The unit of pressure is newton per square metre, N m−2, also known as the pascal, Pa.

Another commonly used non-SI unit is the atmosphere a non-SI unit of pressure, atm, which is equal to 1.013 × 105 Pa.

Figure 9.2 shows a molecule that collides with a container wall. The momentum normal to the wall before the collision is m v cos θ. After the collision momentum normal to the wall is − m v cos θ. So the change in momentum has magnitude 2m v cos θ. The fact that the momentum of the molecule has changed means that a force acted on the molecule (from the wall). By Newton’s third law, therefore, the molecule exerted on the wall an equal and opposite force. Taking into account the forces due to all the molecules colliding with the walls results in a force, and hence pressure, on the walls.

Figure 9.2: A molecule has its momentum changed when it collides with a wall. A force is exerted on the molecule, and so the molecule exerts an equal and opposite force on the wall.

Exam tip

You must be able to give an explanation of pressure in terms of molecules colliding with the container walls.

Worked example 9.4

A solid cylinder of density 8.0 × 103 kg m−3 has height h = 25 cm (see Figure 9.3).

Figure 9.3: For worked example 9.4.

What is the pressure exerted by the cylinder on the horizontal surface on which it rests?

Answer

The surface exerts a normal force on the cylinder equal to the weight of the cylinder, and so the cylinder exerts the same force on the surface.

The weight of the cylinder is:

W=m g=ρ V g=8.0×10 3×A×0.25×9.8 N

where A is the area of the base and ρ is the density.

This weight acts on an area A so the pressure is:

P=8.0×10 3×A×0.25×9.8 A=2.0×10 4 Pa

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CHECK YOURSELF 2

Two solid cylinders made of the same material rest on a horizontal surface (see Figure 9.4). Which cylinder exerts the greater pressure?

Figure 9.4: For check yourself question 2.

aThe taller cylinder

bThe shorter cylinder

cThe heavier cylinder

dThe lighter cylinder

The ideal gas

An ideal gas a theoretical gas in which the particles do not exert forces on each other except during contact is a theoretical model of a gas. It helps us to understand the behaviour of real gases a gas obeying the gas laws approximately for limited ranges of pressures, volumes and temperatures. We assume that an ideal gas obeys the following:

●The molecules are point particles, each with negligible volume.

●The molecules obey the laws of mechanics.

●There are no forces between the molecules except when the molecules collide.

●The duration of a collision is negligible compared to the time between collisions.

●The collisions of the molecules with each other and with the container walls are elastic.

●Molecules have a range of speeds and move randomly.

Exam tip

You must be able to recall and describe a few of these assumptions in an exam.

An ideal gas (unlike real gases) cannot be liquefied or solidified. You should be able to see how some of these assumptions may not be obeyed by a real gas. For example, there will always be forces between molecules of a real gas, not just when the molecules are in contact. In general, we expect that a real gas will behave like an ideal gas when the density is low (so that molecules are not close to each other and hence the forces between them are negligible). We do not expect ideal gas behaviour at high densities (molecules will be too close to each other and will exert forces on each other). Similarly, we do not expect ideal gas behaviour from a real gas at very low temperature because the gas will then become a liquid or even a solid! We also do not want very high temperatures. At very high temperatures collisions between molecules may not be elastic as some energy may be transferred to electrons in atoms forcing them to make transitions to higher energy levels (see Chapter 21).

KEY POINT

A real gas may be approximated by an ideal gas when the density is low and the temperature moderate.

(Low and moderate are vague words because we have to compare them with something else! The interested reader may want to research this further.)

Exam tip

A real gas may be approximated by an ideal gas at low density which is equivalent to low pressure and moderate temperature.

The state of a gas a gas with a specific value of pressure, volume, temperature and number of moles is determined when we know the values of the pressure, the volume, the temperature and the number of moles present. The parameters P, V, T and n are related to each other. The equation relating them is called the equation of state or the ideal gas law. Our objective is to discover the equation of state for a gas. To do this a number of simple experiments can be performed as described in the following sections.

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The pressure–volume law (Boyle–Mariotte law)

The equipment shown in Figure 9.5 can be used to investigate the relationship between pressure and volume of a fixed quantity of gas that is kept at constant temperature.

The pump forces oil to move higher, decreasing the volume of the air trapped in the tube above the oil. A pressure gauge reads the pressure of the trapped air, and so the relationship between pressure and volume may be investigated. The changes in pressure and volume must take place slowly so that the temperature stays the same.

Figure 9.5: Apparatus for investigating the pressure–volume law. The pump forces oil to move up the tube, decreasing the volume of air.

The results of a typical experiment are shown in Figure 9.6. We have plotted pressure against the inverse of the volume and obtained a straight line through the origin.

Figure 9.6: Graph of pressure against inverse volume at constant temperature.

This implies that at constant temperature and with a fixed quantity of gas, pressure is inversely proportional to volume, that is:

P∝1 V or P V= constant

Figure 9.7 shows the same data now plotted as pressure against volume.

Figure 9.7: The relationship between pressure and volume at constant temperature for a fixed quantity of a gas. The product PV is the same for all points on the curve.

The curve in the pressure–volume diagram is known as an isothermal curve a curve on a pressure–volume diagram where all points have the same temperature: the temperature at all points on the curve is the same.

Exam tip

In practice we use the relation PV = constant in the equivalent form P 1 V 1 = P 2 V 2 when the initial pressure and volume (P 1, V 1) change to a new pressure and volume (P 2, V 2) at constant temperature.

Worked Example 9.5

The pressure of a fixed quantity of gas is 2.0 atm and its volume is 0.90 dm3. The pressure is increased to 6.0 atm at constant temperature. Determine the new volume.

Answer

Use P 1 V 1 = P 2 V 2.

Substituting the known values we have:

2.0×0.90=6.0×V

V=0.30 dm 3

The volume–temperature law (Charles’ law)

The dependence of volume on temperature of a fixed quantity of gas kept at constant pressure can be investigated with the apparatus shown in Figure 9.8. Air is trapped in a thin capillary tube that is immersed in heated water. The air is trapped by a thin thread of very concentrated sulfuric acid. The thread is exposed to the atmosphere, and so the pressure of the trapped air is constant.

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It is found that the volume increases uniformly with temperature. The striking fact is that when the straight line is extended backwards it always crosses the temperature axis at −273 °C, as in Figure 9.9. This suggests that there exists a minimum possible temperature, namely −273 °C. (With a real gas the experiment cannot be conducted at very low temperatures since the gas would liquefy—hence the dotted line.

Figure 9.8: Apparatus for verifying the volume–temperature law.

Figure 9.9: When the graph of volume against temperature is extended backwards the line intersects the temperature axis at −273 °C.

If this same experiment is repeated with a different quantity of gas, or a gas at a different constant pressure, the result is the same. In each case, the straight-line graph of volume against temperature crosses the temperature axis at −273 °C (Figure 9.10). In Figure 9.11, the same graphs are drawn using the Kelvin temperature scale.

Figure 9.10: When the graph of volume against temperature is extended backwards, all the lines intersect the temperature axis at the same point.

When the temperature is expressed in kelvin we obtain the graphs of Figure 9.11.

Figure 9.11: When temperature is expressed in kelvin, the lines start at zero temperature.

When the temperature is expressed in kelvin, this experiment implies that at constant pressure:

V T=constant

Exam tip

In practice we use the relation V T=constant in the equivalent form as V 1 T 1=V 2 T 2 where the initial volume and temperature of the gas (V 1, T 1) change to a new volume and temperature (V 2, T 2) at constant pressure.

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Worked example 9.6

A gas expands at constant pressure from an original volume of 2.0 m3 at 22 °C to a volume of 4.0 m3. Calculate the new temperature.

Answer

Substituting in V 1 T 1=V 2 T 2 it follows that:

2.0 295=4.0 T⇒T=590 K or 317 °C

Note that we converted the original temperature into kelvin. (It is very easy to forget this conversion and get the incorrect answer of 44 °C.)

The pressure–temperature law (Gay-Lussac’s law)

What remains now is to investigate the dependence of pressure on temperature of a fixed quantity of gas in a fixed volume. This can be done with the apparatus shown in Figure 9.12. The gas container is surrounded by water whose temperature can be changed. A pressure gauge measures the pressure of the gas. We find that pressure increases uniformly with increasing temperature, as shown by the graph in Figure 9.13.

Figure 9.12: Investigating the pressure–temperature law.

Figure 9.13: The graph of pressure against temperature is a straight line that, when extended backwards, again intersects the temperature axis at −273 °C.

For quantities of gases containing different numbers of moles at different volumes the results are the same, as shown in Figure 9.14. When the temperature is expressed in kelvin, the straight lines all pass through the origin (Figure 9.15).

Figure 9.14: When extended backwards, the graphs of pressure against temperature for three different quantities of gas all intersect the temperature axis at the same point.

Figure 9.15: If temperature is expressed in kelvin, the lines start at zero temperature.

When the temperature is expressed in kelvin, this experiment implies that at constant volume:

P T=constant

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Exam tip

In practice we use the relation P T=constant in the equivalent form P 1 T 1=P 2 T 2 where the initial pressure and temperature of the gas (P 1, T 1) change to a new pressure and temperature ( P 2, T 2) at constant volume. (Remember, T is in kelvin.)

CHECK YOURSELF 3

The graph (Figure 9.16) shows the variation with kelvin temperature of the pressure of equal quantities of two gases at different volume.

Figure 9.16: For check yourself question 3.

Which gas has the smaller volume?

Worked example 9.7

A gas in a container of fixed volume is heated from a temperature of 37 °C and pressure 3.0 × 105 Pa to a temperature of 87 °C. Calculate the new pressure.

Answer

Substituting in P 1 T 1=P 2 T 2 we have:

3.0×10 5 310=P 360

⇒P=3.5×10 5 Pa

(Notice that we had to change the temperature into kelvin.)

The ideal gas law

If we combine the results of the three preceding experiments, we find that:

P V T=constant

What is the value of the constant? To determine that, we repeat all of the preceding experiments, this time using different quantities of the gas. We discover that the constant in the last equation is proportional to the number of moles n of the gas in question:

P V T=n× constant

We can now measure the pressure, temperature, volume and number of moles for a large number of different gases and calculate the value of P V n T.

We find that this constant has the same value for all gases—it is a universal constant. We call this the gas constant the constant, R that appears in the ideal gas law R.
It has the numerical value R = 8.31 J K−1 mol−1.

Thus, finally, the ideal gas law(also known as the equation of state) equation relating pressure, volume and temperature of an ideal gas: PV = RnT (also known as the equation of state) is:

P V=R n T

(Remember that temperature must always be in kelvin.)

In Section 9.3 we will see the importance of a new constant of physics, the Boltzmann constant the ratio of the gas constant to Avogadro’s number:

k B=R N A=8.31 6.02×10 23=1.38×10−23 J K−1

In terms of this constant the equation of state may also written as P V=R N N A T=N k B T.

Exam tip

Since R=P V n T, in practice we use the equation of state in the form P 1 V 1 n 1 T 1=P 2 V 2 n 2 T 2 when a gas changes from values (P 1, V 1, n 1,T 1) to (P 2, V 2, n 2, T 2).

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Worked example 9.8

Estimate how many molecules there are in a gas of temperature 320 K, volume 0.025 m3 and pressure 4.8 × 105 Pa.

Answer

The number is:

N=P V k B T

N=4.8×10 5×0.025 1.38×10−23×320=2.7×10 24.

Worked example 9.9

A container of hydrogen of volume 0.10 m3 and temperature 25 °C contains 3.2 × 1023 molecules.

Calculate the pressure in the container.

Answer

P=N k B T V=3.2×10 23×1.38×10−23×298 0.10=1.3×10 4 Pa

Worked example 9.10

A fixed quantity of gas of volume 3.0 × 10−3 m3, pressure 3.0 × 105 Pa and temperature 300 K expands to a volume of 4.0 × 10−3 m3 and a pressure of 6.0 × 105 Pa. Calculate the new temperature of the gas.

Answer

Use P 1 V 1 n 1 T 1=P 2 V 2 n 2 T 2 with n 1 = n 2 to get:

3.0×10 5×3.0×10−3 300=6.0×10 5×4.0×10−3 T

Solving for T gives T = 800 K.

Worked example 9.11

Figure 9.17 shows two isothermal curves for equal quantities of two ideal gases. State and explain which gas is at the higher temperature.

Figure 9.17: Two isothermal curves for equal quantities of two gases.

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Answer

Draw a vertical line that intersects the two isothermals at two points. At these points both gases have the same volume and n is the same.
So, for these points P T is constant. The point on the blue curve has higher pressure, so it must have the higher temperature.

TEST YOUR UNDERSTANDING

10A sealed bottle contains air at 22.0 °C and a pressure of 12.0 × 105 Pa. The temperature is raised to 120.0 °C. Calculate the new pressure.

11A gas has pressure 8.2 × 106 Pa and volume 2.3 × 10−3 m3. The pressure is reduced to 4.5 × 106 Pa at constant temperature. Calculate the new volume of the gas.

12A mass of 12.0 kg of helium is required to fill a bottle of volume 5.00 × 10−3 m3 at a temperature of 20.0 °C. Determine the pressure in helium.

13Determine the mass of carbon dioxide (molar mass 44 g mol−1) required to fill a tank of volume 12.0 × 10−3 m3 at a temperature of 20.0 °C and a pressure of 4.00 atm.

14A flask of volume 300.0 × 10−6 m3 contains air at a pressure of 5.00 × 105 Pa and a temperature of 27.0 °C. The flask loses molecules at a rate of 3.00 × 1019 per second. Estimate how long it takes for the pressure in the flask to fall to half its original value. (Assume that the temperature of the air remains constant during this time.)

15The point marked in Figure 9.18 represents the state of a fixed quantity of ideal gas in a container with a movable piston. The temperature of the gas in the state shown is 600 K. Copy the diagram. Indicate on the diagram the point representing the new state of the gas after the following separate changes.

aThe volume doubles at constant temperature.

bThe volume doubles at constant pressure.

cThe pressure halves at constant volume.

Figure 9.18: For question 15.

16Two ideal gases are kept at the same temperature in two containers separated by a valve, as shown in Figure 9.19. Estimate the pressure when the valve is opened. (The temperature stays the same.)

Figure 9.19: For question 16.

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17Figure 9.20 shows a cylinder in a vacuum, which has a movable, frictionless piston at the top. An ideal gas is kept in the cylinder. The piston is at a distance of 0.500 m from the bottom of the cylinder and the volume of the cylinder is 0.050 m3. The weight on top of the cylinder has a mass of 10.0 kg. The temperature of the gas is 19.0 °C.

Figure 9.20: For question 17.

aCalculate the pressure of the gas.

bDetermine how many molecules there are in the gas.

cThe temperature is increased to 152 °C. Calculate the new volume of the gas.

18The molar mass of a gas is 28 g mol−1. A container holds 2.00 mol of this gas at 0.00 °C and a pressure of 1.00 × 105 Pa. Determine the mass and volume of the gas.

19A balloon has a volume of 404 m3 and is filled with helium of mass 70.0 kg. The temperature inside the balloon is 17.0 °C. Determine the pressure inside the balloon.

20A flask has a volume of 5.0 × 10−4 m3 and contains air at a temperature of 300 K and a pressure of 150 kPa.

aCalculate the number of moles of air in the flask.

bDetermine the number of molecules in the flask.

cEstimate the mass of air in the flask. You may take the molar mass of air to be 29 g mol−1.

21The molar mass of helium is 4.00 g mol−1.

aCalculate the volume of 1.0 mol of helium at standard temperature and pressure (stp), i.e. at T = 273 K, P = 1.0 × 105 Pa.

bDetermine the density of helium at stp.

22 aA gas is heated at constant volume. Give a qualitative reason why the pressure increases.

bA gas is heated at constant pressure. Give a qualitative reason why the volume increases.