# Gas laws and equations in physics

What are Gas laws ?

To study gas behaviors, we should first of all consider three physical properties of gas, which are pressure volume and temperature. It is based on these three properties that the three gas laws were formulated. Note that among the three states of matter gas is most movable because of high kinetic energy they have.

The three gas laws are
1. Boyle’s law
2. Charles’ law
3. Pressure law or Gay Lussac’s law.

Boyle’s law
This is one of the gas laws, It states that the volume of a fixed mass of gas varies inversely as its pressure provided the temperature remains constant.
Boyle’s law can be mathematically stated as

${ { V=\frac { K }{ P } }\quad or\quad PV=K }$

Or P1V1 =P2V2 , K=constant, p =pressure and V= volume

Applications of boyle’s law
1.It is used in determination of a volume of a given mass of gas at a constant temperature when pressure is known
2. It can also be used to find the pressure of a given mass off gas at constant temperature when the volume is known

One of the gas laws also is Charles’ law
The Charles’ law state that the volume of a fixed mass of gas is directly proportional to its absolute temperature(T), provided the pressure remains constant

Charles law can be represented in mathematical form as

${ V=KT\quad or\quad \frac { { V }_{ 1 } }{ { T }_{ ! } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } }$

V1 and V2 are initial and final volumes while T1 and T2 are initial and final temperatures and K=constant

Applications of Charles’ law
1.With Charles’ law absolute zero temperature, that is the temperature at which volume of gas is theoretically zero can be determined by plotting the volume of the gas against its temperature

1. The cubic expansivity of gas is also a result of Charles’ law. Therefore cubic expansivity is given as

${ \gamma =\frac { Increase\quad in\quad Volume }{ Initial\quad volume\quad at\quad { \theta }^{ o }C\quad \times \quad Change\quad in\quad temperature } }$

Hence Charles’ law can be restated as the volume of fixed mass of gas at a constant pressure increase by 1/273 of its volume at 0oC (Celsius or Kelvin) for every degree rise in temperature

Pressure law or Gay Lussac’s law
This deals with change in pressure and absolute temperature of a gas when the volume is kept constant

It states that the pressure of a fixed mass of a gas is directly proportional to its absolute temperature provided the volume is kept constant
Mathematically it is written as

P=KT, K=  constant, p= pressure and T=temperature

${ \frac { P }{ T } =K }$

${ Or\quad \frac { { P }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 } }{ { T }_{ 2 } } }$

Absolute temperature and pressure In gas laws
In gas laws if P0 = gas pressure at temperature 0oC and PѲ = gas pressure at temperature ѲoC. The pressure law  states mathematical that
P0 = PѲ (1 + γѲ)
Where γ= 1/273k-1 =pressure cubic expansivity at constant volume

General gas laws equation
The general gas laws equation is derived from the combination of these three laws the boyle’s law Charles’ law and  pressure law or Gay Lussac’s law

From boyle’s law PV  =k at constant temperature
From Charles’ law V/T =k at constant pressure
From pressure law P/T =k constant volume
When you combine these three equation of gas laws you abstain general gas laws equation as

${ \frac { PV }{ T } =K\quad Or }$

${ \frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } }$

Standard temperature and pressure
The temperature and pressure of volume of gases can be converted to a standard temperature and pressure. This conversion makes it easy for possible comparison. The standard temperature is taken as 0oC or 273K while the standard pressure is taken as 760mm of mercury.

Van der waals force and equation for real gas

What is Vander waals force? It is the force of attraction between gases molecule. It can also be called weak Vander waal force, because is indeed weak but not negligible.

Vander waals equation

Vander waals equation is given as

${ { (P+\frac { a }{ { V }^{ 2 } } ) }{ (V-b) }{ =RT } }$

Where a =constant, b= volume of gas, (V-b)=volume of the space insider container and R also =constant.

Kinetic molecular theory and gas laws

The kinetic theory state that:

(I). All matters are made of atoms and molecular

(II). The molecules are in a constant state of motion

(II). That the molecule posses kinetic energy because of it continuous motion

(IV). The molecule exert attractive force on one another

(V). The nearer the molecules to one another the greater the attraction force

Using of kinetic theory to explain the gas laws

Explanation of gas laws using kinetic theory
1. Using kinetic theory to explain pressure exerted by a gas, kinetic theory considers gas as made up of large number of molecules which behave like elastic sphere. The molecule moves about in their containers with random velocities, colliding with one another and with the walls of the container. As gas molecules hit the walls of the containing vessel and rebound they experience a change in velocity and therefore change in momentum. The walls of the containers also experience forces due to change in momentum of the molecules of gas, thereby exacting pressure on the walls of the vessel.

2. Using kinetic theory to explain Boyle’s law
If the Volume of the vessel containing gas is reduced at constant temperature the gas molecules will have less spaces to occupy. Therefore they take less time to hit the vessel. In this case more bombardments are made on the wall per unit time and therefore the pressure exerted by the gas increases.

On the other hand when the volume of gas is increased the molecules will take more time to hit the wall, and this lead to the decrease in pressure of gas. In conclusion the pressure of a gas is therefore inversely proportional to the volume at a constant temperature( Boyle’s law)

3 Using kinetic theory to explain Charles’s law
If you Increase the temperature of gas molecule at constant pressure the monocle we gain kinetic energy and thereby increasing the velocities of the molecules. With the increase in velocities the molecules will hit the wall more frequently and this brings about the increase pressure but to keep the pressure constant the volume must increase. therefore an increase in temperature at constant pressure will lead to an increase volume (Charles’s law)

4 using kinetic theory to explain pressure law(Gay Lussac’s Law)
If the temperature of a given mass of gas is increased at constant volume the average kinetic energy of the molecules increases and this results in the increase in speed of molecules which make them to hit the wall of containing vessel more frequently. Consequently the pressure of this given mass of gas increases with increase in temperature at constant volume. (Gay Lussac’s Law)

Questions and answers on gas laws

1. Calculate the volume of fraction change of a fixed mass of gas whose pressure is tripled at constant temperature

${ \frac { { P }_{ 1 } }{ { V }_{ 1 } } =\frac { { P }_{ 2 } }{ { V }_{ 2 } } }$

But P2 = 3P1, therefore

${ \frac { { P }_{ 1 } }{ { V }_{ 1 } } =\frac { { 3P }_{ 2 } }{ { V }_{ 2 } } \Longrightarrow 3{ V }_{ 2 }={ V }_{ 1 } }$

${ \therefore \frac { { V }_{ 1 } }{ { V }_{ 2 } } =3\quad and\quad { v }_{ 2 }=\frac { { V }_{ ! } }{ 3 } }$

So the final volume becomes one-third of the initial volume

1. The pressure of fixed mass of gas is given as 850cmHg when its volume is 20cm3. Calculate the volume of the gas when its pressure is 600cmHg

P1=pressure= 850cmHg, P2=600cmHg, V1=20cm3

850 x 20 = 600 x V2

${ { V }_{ 2 }=\frac { 855\times 20 }{ 600 } =28.33c{ m }^{ 2 } }$

1. A fixed mass of gas with volume 600cm3 at 0oC is heated at constant pressure. Calculate the volume of the gas at 130oC

${ \frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } }$

T1=0 + 273 =273K, T2= 130 + 273 =403K, and V1 = 600cm3

${ \frac { 600 }{ 273 } =\frac { { V }_{ 2 } }{ 403 } }$

${ { V }_{ 2 }=\frac { 600\quad \times 403 }{ 273 } =885.71c{ m }^{ 3 } }$

1. The pressure of a gas at constant volume is 100cmHg at 27oC. Calculate the pressure at 87oC

Answer:  using pressure law ( Gay lussac’s law)

${ \frac { { P }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 } }{ { T }_{ 2 } } }$

T1= 27 + 273 =300K

T2= 87 + 273 =360K

P1=100cmHg, so

${ \frac { 100 }{ 300 } =\frac { { P }_{ 2 } }{ 360 } }$

${ { P }_{ 2 }=\frac { 100\times 360 }{ 300 } =120cmHg }$

1. Some quantity of dry air has a volume of 500cm3 at 40oC and under a pressure of 76cmHg. At a temperature of 120oC its volume becomes 600cm3, calculate the pressure of the gas at this temperature.

Answer: Using gas laws general equation

${ \frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } }$

P1=76cmHg, p2=?, V1=500cm3, V2= 600cm3, T1 = 40 + 273=313K, and T2= 120 + 273 =393K

${ \therefore { P }_{ 2 }=\frac { { P }_{ 1 }{ V }_{ 1 }{ T }_{ 2 } }{ { V }_{ 2 }{ T }_{ 1 } } }$

${ { P }_{ 2 }=\frac { 76\times 500\times 393 }{ 600\times 313 } =79.5cmHg }$

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