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 GayLussac’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:
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, P_{1}V_{ 1 }= P_{2}V_{ 2,} since when you increase the volume of the gas to 20 cm^{3}, the pressure decreases to 65 kPa. And if the volume of the gas is reduced to 15 cm^{3}, 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 P_{1}V_{ 1 }= P_{2}V_{ 2. }When a graph of pressure is plotted against the inverse of volume, a straight line is obtained.
worked Example
A highaltitude 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, P_{1}V_{ 1 }= P_{2}V_{ 2} P_{1 }= 206kPa V_{1}=60L P_{2}= 50kPa V_{2}=? P_{1}V_{ 1 }= P_{2}V_{ 2, }making V_{ 2} the subject of the formula gives; V_{ 2}= P_{1×}V_{ 1 } P_{2} = 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, 273^{◦}c 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 273^{◦}c. 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
V_{1}T_{1} =V_{2}T_{2}
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, V_{1}T_{1} =V_{2}T_{2} T_{1 }= 297K V_{1}=4.00L T_{2 }= 331K V_{2}=? V_{1}T_{1} =V_{2}T_{2} making V_{ 2} the subject of the formula gives; V_{ 2}= T_{2×}V_{ 1 } T_{1}
V_{ 2}= 331K_{×}4.00L 297 = 4.46L 
At a temperature of 57^{◦}c, nitrogen gas occupies a volume of 750cm^{3. }At what temperature will the gas occupy 100cm^{3}? Express your answer in degrees Celsius. Applying Charles’s Law, V_{1}T_{1} =V_{2}T_{2}
T_{1 }=330 K V_{1}=750cm^{3} V_{2 }= 100cm^{3} T_{2}=? V_{1}T_{1} =V_{2}T_{2} making V_{ 2} the subject of the formula gives; T_{ 2}= V_{2×}T_{ 1 }V_{1}
T_{ 2}= 100cm3_{×}330K 750cm3 = 44273 =229^{◦}c

GayLussac’s Law
GayLussac’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 P_{1}/T_{1 }and P_{2}/T_{2 }are equal at constant volume.
P_{1}T_{1} =P_{2}T_{2}
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, P_{1}T_{1} =P_{2}T_{2} T_{1 }=300 K P_{1}=198kPa P_{2 }= 225kPa T_{2}=? P_{1}T_{1} =P_{2}T_{2} making T_{2} the subject of the formula gives; T_{ 2}= P_{2×}T_{ 1 }P_{1} =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:
P_{1×}V_{1 }T_{1}=P_{2×}V_{2 }T_{2.}
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. P_{1×}V_{1 }T_{1}=P_{2×}V_{2 }T_{2} 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. P_{1×}V_{1= }P_{2×}V_{2} This forms Boyle's law. Furthermore, you can derive Charles’s law when pressure remains constant as shown below. P_{1×}V_{1 }T_{1}=P_{2×}V_{2 }T_{2} = V_{1}T_{1} =V_{2}T_{2} GayLussac’s law when volume remains constant gives us; So, P_{1}T_{1} =P_{2}T_{2} 
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.
P_{1×}V_{1 }T_{1×}n_{1}=P_{2×}V_{2 }T_{2×}n_{2}
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_{×22.4L}273K×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