## Ideal gases

Introduction

An ideal gas has a freely moving random particles. They have the least contact between the particles that can hardly produce elastic collisions. The ideal gas concept is useful as the gas obeys the ideal gas law. In normal conditions, the majority of the real gases behave qualitatively like an ideal gas. Some of the common gases such as nitrogen, oxygen, hydrogen, noble gases, and few heavier gases reasonably fall under the ideal gas category.

 Table of Contents 1. Introduction 2. Assumptions of gas laws 3. Variables that describe a gas 4. The Gas Laws     4.1 Boyle’s Law     4.2 Charles’s Law     4.3 Gay-Lussac’s Law     4.4 The Combined Gas Law     4.5 Ideal gas laws

Assumptions of gas laws

The gas laws assume that gases behave in an ideal manner, obeying the assumptions of the kinetic theory. The amount of gas available in the system is an important variable to consider when dealing with ideal gases.  The assumptions that should be put into consideration include:

1. The particles in any gas move rapidly in a constant random motion.
2. A gas is made up of either molecules or atoms. The particles are considered to be small, hard spheres that have insignificant volume and are relatively far away from each other.
3. The collisions are perfectly elastic. This implies that during collisions energy is transferred without loss from one particle to another.
4.

Variables that describe a gas

Generally, four variables are used to describe a gas. They are pressure- (P), Volume ( litres), the temperature measured in kelvins and the number of moles. Understanding gas laws help in predicting the behaviour of gases under specific conditions. The Gas Laws

Boyle’s Law

Robert Boyle was the first person to propose the relationship between pressure and volume in the year 1662.  Boyle’s law states that under constant temperature, the volume of a fixed mass of gas varies and it is inversely proportional with pressure. From the graph, P1V 1 = P2V 2, since when you increase the volume of the gas to 20 cm3, the pressure decreases to 65 kPa. And if the volume of the gas is reduced to 15 cm3, its pressure will increase to 95 kPa.  The product of volume and pressure at any 2 sets of the condition is always a constant at a given temperature. Boyle’s law can be expressed as   P1V 1 = P2V 2. When a graph of pressure is plotted against the inverse of volume, a straight line is obtained. worked Example   A high-altitude balloon contains 60L of helium gas at 206 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 50.0kPa under constant temperature? Solution Applying Boyle’s Law, P1V 1 = P2V 2                         P1 = 206kPa                         V1=60L                         P2= 50kPa                         V2=? P1V 1 = P2V 2, making V 2 the subject of the formula gives;                         V 2= P1×V 1 P2                         = 206 kPa×60L 50L                         = 247.2L  From the above calculation, a decrease in pressure of the gas at constant temperature leads to an increase in volume.

Charles’s Law

Charles's law was put forth by the French scientist Jacques Charles in 1787. He investigated the quantitative effect of temperature on the volume of a gas at constant pressure. According to him, a decrease in the volume of gas leads to a decrease in temperature and vice versa. In practice, the relationship between temperature and volume of a given gas can be measured only over a limited range since gases condense at low temperatures to form liquids. While plotting graphs of the volume of gas against temperature, Charles realized a critical feature. The lines extended/extrapolated to zero volume and intersected the temperature axes at the same point, -273c as shown below. Another scientist, William Thomson realized the significance of the temperature value. It is the temperature at which the average kinetic energy of gas particles are said to be theoretically zero. He, therefore, came up with an absolute scale in which 0K corresponds to -273c.  Charles law states that under constant pressure, the volume of a fixed mass of a gas is directly proportional to its Kelvin temperature. Mathematically, we can express Charles’s law as follows

V1T1 =V2T2

When a graph of volume against temperature is plotted, a straight line is obtained as shown above.

 Worked Example A balloon is inflated in a room at 24◦c has a volume of 4.00L. The balloon is then heated to a temperature of 58◦c. Calculate the new volume of the gas given that temperature is kept constant? Solution Applying Charles’s Law, V1T1 =V2T2                          T1 = 297K                         V1=4.00L                         T2 = 331K        V2=?  V1T1 =V2T2 making V 2 the subject of the formula gives;                         V 2= T2×V 1 T1                           V 2= 331K×4.00L 297               = 4.46L

 At a temperature of 57◦c, nitrogen gas occupies a volume of 750cm3. At what temperature will the gas occupy 100cm3? Express your answer in degrees Celsius. Applying Charles’s Law, V1T1 =V2T2                            T1 =330 K                         V1=750cm3                         V2 = 100cm3                         T2=?                         V1T1 =V2T2 making V 2 the subject of the formula gives;                         T 2= V2×T 1 V1                           T 2= 100cm3×330K 750cm3 = 44-273 =229◦c

Gay-Lussac’s Law

Gay-Lussac’s Law states that, under constant volume, the pressure of a gas is directly proportional to the Kelvin temperature. Since gay Lussac’s law involves direct proportions, the ratios P1/T1 and P2/T2 are equal at constant volume.

P1T1 =P2T2

 Question The pressure in an automobile tire is 198kPa at27◦c. Given that the pressure has risen to 225kPa. Calculate the temperate of the air in the tire assuming that the volume is kept constant. Solution: Applying Gay Lussac’s Law, P1T1 =P2T2                          T1 =300 K                         P1=198kPa                         P2 = 225kPa                         T2=?                         P1T1 =P2T2 making T2 the subject of the formula gives;                         T 2= P2×T 1 P1                   =225kPa×300K 198kPa                   = 340.9090- 273                   = 67.9◦c

The Combined Gas Law

The combined gas law is a single expression that combines the three gas laws. Mathematically, combined gas law is expressed as:

PV1 T1=PV2 T2.

The other laws can be obtained from this law by keeping one quantity constant.

 Quick question: Briefly explain how combining gas law can be reduced to the other three gas laws. P1×V1 T1=P2×V2 T2 The other laws can be obtained from this law by keeping one quantity constant. To illustrate, suppose we hold the temperature constant. Rearranging the combined gas law, we shall obtain the temperature terms on the same side of the equation and then cancelled.                                P1×V1=P2×V2×     P1×V1= P2×V2 This forms Boyle's law. Furthermore, you can derive Charles’s law when pressure remains constant as shown below. P1×V1 T1=P2×V2 T2                                                      V1T1=P2×V2 T2×P1    =   V1T1 =V2T2  Gay-Lussac’s law when volume remains constant gives us;                                                  P1×V1 T1=P2×V2 T2 P1×V1 T1×V2=P2 T2 So,  P1T1 =P2T2

Ideal Gas Law

Other than the three variables; volume, pressure and temperature, there is another crucial variable we should consider when dealing with ideal gases. The number of moles (n) is directly proportional to the number of particles. Hence, moles must be directly proportional to the volume of a fixed mass of gas as well. We can introduce moles in the combined gas law by dividing each side of the equation by n as shown.

PV1 Tn1=PV2 Tn2

From the equation, (P×V)/(T×n) is constant. The constancy holds for what is referred to as ideal gases. A gas behaves ideally if it conforms to the gas laws.  Form the above relationship; it is now possible for us to find the actual value of R, given an important fact about gases: 1mole of every gas occupies 22.4L at STP. When we replace the values of P, V, T, and n into the equation:

(P×V)/(T×n)= R

101.3kPa×22.4L273K×1mol

= 8.31

Thus the ideal gas constant, R, becomes 8.31(L×kPa)/ (K×mol). It follows that the ideal gas law can be expressed as:

R= P×VT×n

Or

P×V=n×R×T   