# Gas Laws

Topics: Temperature, Absolute zero, Thermodynamic temperature Pages: 5 (1333 words) Published: December 10, 2012
Gas Laws
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The 3 Gas Laws
Introduction
The three gas laws include: Gay-Lussac’s law, Boyle’s law and Charles’ law. When combined with Avogadro’s law the three laws can be generalized by the ideal gas law. Gases possess observable properties which include, mass, pressure (P), thermodynamic temperature (T) and volume (V). These properties are related to each other and the state of a gas is determined by their values. The three laws are derived from these properties. Discussion

Boyle’s gas law
Boyle’s law relates the volume and pressure of an ideal gas. It states that when the temperature of a given mass of a confined gas is held constant, its pressure and volume are inversely proportional. In other words, the product of absolute pressure and volume is nearly constant and exactly a constant for an ideal gas (Stoker, 2013). The mathematical equation for Boyle's law is:

PV = k
Where:
P denotes the pressure of the system.
V denotes the volume of the gas.
K is a constant value representative of the pressure and volume of the system. Explanation
Provided a fixed quantity of gas is kept in its original temperature, when the volume V is increased, the pressure P decreases proportionately. The converse is also true. The volumes and pressures before and after change in volume and pressure of the fixed amount of gas are related as follows:P1V1=P2V2 In the expression, P1 and V1 represent the initial pressure and volume, while P2 and V2 is the second pressure and volume respectively.

Diagram showing Boyle’s lawGraphical representation of Boyle's law Example to show application of the formula:
4,000 cubic meters of air are under a pressure of 50 kilopascals. What is the volume if the pressure is increased to 100 kilopascals? Classifying the data: V₁= 4,000 m³, P₁= 50 kPa, P₂= 100 kPa
We must solve for V₂
Solving Boyle's Law for V₂ we get:
V₂= (V1• P₁) ÷ P₂     V₂= (4,000 m³ • 50 kPa) ÷ 100 kPa
V₂= 2000m³
The final volume (V₂) = 2000m³
Charles’ gas law
Charles’s law relates the volume and temperature of an ideal gas that at a constant pressure, its volume is directly proportional to the absolute temperature. An increase or decrease in temperature of a fixed mass of an ideal gas at constant pressure results in an increase or decrease in volume by the same factor. Charles’ law is also called the law of volumes (Stoker, 2013). The relation can be expressed as follows:

V α T
The mathematical equation for Charles’ law is:
V/T = k
Where:
P denotes the pressure of the system.
V denotes the volume of the gas.
K is a constant value representative of the pressure and volume of the system. Explanation
Assuming that pressure remains constant, when the absolute temperature of a gas is doubled, the volume of that gas also doubles. The volume of a gas increases by 1/273 of its volume at 0°C for every degree Celsius of temperature raises. Charles' Law must be used with the absolute temperature scale. At 0 K, there is no kinetic energy (Absolute Zero). Zero Kelvin is equal to -273.15 °C; any Celsius temperature can be converted by to Kelvin by adding 273.15 (273 is often used). The formula that is used to solve for a change in volume or temperature is: V1/T1 = V2/T2

Example to show application of the formula:
At 27°C the gas occupied a volume of 200 ml. What volume will it occupy at 57°C? First, we must convert degrees Celsius to Kelvin. To do this, we add 273 to the Celsius measure. 42°C + 273 = 330 K

27°C + 273 = 300 K
Plugging these values into formula V1/T1 = V2/T2:
In this problem, V1 = 200 ml, T1 = 300 K, and T2 = 330 K. V2 is unknown. Therefore, we can arrange the formula as:
200 ml/300 K =? /330 K
200 ml x 330 K = 300 K x?
This can be divided by 300 K to yield,
200 ml x 330 K / 300 K = V2 = 220 ml

So the final volume (V2) = 220 ml

Graphical representation of Charles’ law...