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# The Gas Law

THE GAS LAWS
Specific objectives: by the end of the lesson, the students should be able to
• Identify and use the instrument for measuring pressure,
• Explain Charles and Boyles law of gases,
• Solve simple problems involving gas laws,
• Define general gas law.

THE GAS LAWS (Introduction)
In studying the behavior of gases, there are always three parameters of interest namely, volume, pressure, and temperature of a gas. Pressure is however more relevant because more noticeable effect or changes are observed whenever the pressure of gases are varied.
BOYLES LAW: Boyles law states that the product of the pressure and volume of a fixed mass of gas at constant temperature is constant. That is, PV = K (constant temp).

PV = K (constant temp)
Given that P = force/area
Volume = area × height. Therefore, PV = K = force/ Area ×area × height.
K is the same as the work done.
Experimental Prove:
Using the diagram above, dry air is introduced into the burette and the tap of the burette is closed. Take the atm reading (H) using a barometer. The glass tube on the other side is then lowered and the mercury level(h) is noted and recorded. The pressure is either given as H – h or H + h depending on the two mercury levels.
On plotting the graph of pressure against length = volume of the gas, we have the following graph
See a plot of P Vs 1/V in any physics textbook.
The two graphs show that PV = k i.e. (P1V1 =P2V2) = K, which implies that P  1/V which proves Boyles law.
Charles Law:  It states that at constant pressure, the ratio of the volume of a given mass of gas to its temperature for a fixed mass of gas is a constant. That is:
V/T = K (constant pressure), that is
V1/T1 = V2/T2 ……. = K

In the experiment to prove the law, only the temperature of the water and the volume of the gas varies. By heating the water at higher temperature values of temperature and volume are obtained. The result of the experiment is as shown below:
The above experiment implies that when the temperature is increased at constant pressure, the volume of the gas increases by 1/273 of its original volume. This is Charles’s law.
The Pressure Law: it states that at constant volume, the ratio of the pressure of a gas to its temperature is a constant, i.e. P/T = K
P1/T1 = P2/T2 ……… =K
THE GENERAL GAS LAW
It is the combination of the three stated gas laws i.e. Boyles, Charles and Pressure law. This law states that
PV/T = K
P1V1/T1 = P2V2/T2 = ……… =K.
Gases that obey this equation are referred to as IDEAL GASES. Thus, the equation states that if the amount of gas under consideration is in moles, equation of state then takes the form.
PV/T =nR or PV = nRT
Where n = number of moles
R = molar constant
M = mass of gas
M = molar mass of the gas.

Example 1:
The presence of a fixed mass of gas is 2.0 × 105N/m2 at a known temperature. Assuming that the temperature remains constant, what will be the pressure of the gas if its volume is halved? (SSCE 1996)
SOLUTION
From Boyles Law: P1V1 = P2V2
2 × 105 × V1 = P2 × V1/2
Since V2 = V1/2
P2 = 2 × 2 × 105 Nm-2 = m=4 × 105Nm-2.
Example 2:
A balloon containing 546cm3 of air is heated from OoC to 10oC. If the pressure is kept constant, what will be its volume at 10oC? (SSCE June 1995).
SOLUTION
From Charles Law
V1/T1 = V2/T2
Where V1 = 546cm3, OoC = 273K, T2 = 282K and V2 = ?
546/273 = V2/283
V2 = 546 × 283/ 273 = 566cm3
Example 3:
500cm3 of gas is collection at 10oC and at a pressure of 72.0cm of mercury. What is the vol. of the gas at the same temp. and at a pressure of 76.0cm of mercury ? (SSCE, 1994)
SOLUTION:
From general gas law: (PV/T = K )
= 72 × 500 /273 = 76 × V2 / 273
V2 = 72 × 500 × 273 / 76 × 273
V2 = 473.68 = 474cm3

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