# Gas laws

#### Gas pressure

Pressure is defined as the force per unit area acting on the surface.

The unit of pressure is Newton per metre squared [N/m2]. (or Pascal[[Pa]: 1Pa=1 N/m2)

The gas molecules are in random and continuous motion. They exert a force on the wall of container when they collide to it. Since the force is exerted over an area, pressure is produced.

The pressure of a gas of constant volume increases when;

- there are more molecules in the gas

- the molecules move faster

- the molecules have a greater mass.

#### Boyle’s law

Law For a fixed mass of a dry gas at constant temperature, The product of its volume and pressure is constant.

Formula for Boyle’s law

$\mathrm{PV}=constant$

P: Pressure [N/m2]

V: Volume [m3]

If the initial pressure and volume are P1 and V1, and the final ones are P2 and V2,

$\mathrm{P1 V2}=P2 V2$

#### Example

A gas occupies a volume of 2m3 at 25oC and pressure of 200N/m2. What would be the volume of the gas if the pressure is reduced to 100N/m2 at the same temperature?

Solution

$\mathrm{P1 V2}=P2 V2$

Given that

P1 = 200N/m2

V1 = 2m3

P2 = 100N/m2

V2 = ?

$\mathrm{200}×2=\mathrm{100}×\mathrm{V2}$

$\mathrm{400}=\mathrm{100V2}$

$\frac{\mathrm{100V2}}{100}=\frac{\mathrm{400}}{100}$

Answer: $V={\mathrm{4m}}^{3}$

#### Kelvin temperature scale

SI unit of temperature is Kelvin [K].

- The size of the degree in Kelvin is the same as in Celsius.

- According to the calculations (Charle’s law), a gas would contract as it cools until at -273 oC. Then, the gas has no volume at -273 oC.

- -273 oC is called absolute zero (0K).

- 273 must be added to convert Celsius into Kelvin.

Formula to find temperature in Kelvin

$\mathrm{Tk}=Tc + 273$

Tk: Temperature in Kelvin scale [K]

Tc: Temperature in Celsius scale [oC]

#### Example

Convert (a) 0 oC and (b)100 oC into K.

Solutions

(a) TK = 0 oC + 273 = 273K

(b) TK = 100 oC + 273 = 373K

#### Charles’ law

Law The volume of a fixed mass of a gas at constant pressure is directly proportional to its Kelvin temperature.

Formula for Charles’ law

$\frac{V}{T}=constant$

V: Volume [m3]

T: temperature [K]

If the initial volume and temperature are V1 and T1, and the final ones are V2 and T2,

$\frac{\mathrm{V1}}{T1}=\frac{\mathrm{V2}}{T2}$

#### Example

The sun heats 15m3 of dry air at 27 oC until its volume increases to 16 m3 under the atmospheric pressure. Calculate the temperature of the air.

Solution

$\frac{\mathrm{V1}}{T1}=\frac{\mathrm{V2}}{T2}$

V1 = 15m3

T1 = 27oC = 300k (= 27 + 273)

V2 = 16m3

T2 = ?

$\frac{\mathrm{15}}{300}=\frac{\mathrm{16}}{T2}$

Cross multiply

$\mathrm{15T2}=\mathrm{16}×300$

$\mathrm{15T2}=\mathrm{4800}$

$\frac{\mathrm{15T2}}{15}=\frac{\mathrm{4800}}{15}$

$\mathrm{T2}=\mathrm{320K}$

Answer: TC = TK – 273 = 320 –273 = 47 oC

#### Combination of Boyle’s and Charles’ laws

Formula for Boyle’s and Charles’ laws

$\frac{\mathrm{P1 V1}}{T1}=\frac{\mathrm{P2 V2}}{T2}$

This equation is called the general gas equation.

#### Example

15m3 of gas is at a pressure of 70N/m2 and a temperature of 27 oC. Find its volume when it is at a temperature of 127oC and a pressure of 35N/m2.

Solution

$\frac{\mathrm{P1 V1}}{T1}=\frac{\mathrm{P2 V2}}{T2}$

Given that

P1 = 70N/m2

V1 = 15m3

T1 = 27 oC = 300K (= 27 + 273)

P2 = 35N/m2

V2 = ?

T2 =127 oC = 400K (= 127 + 273)

$\frac{\mathrm{70 \left(15\right)}}{300}=\frac{\mathrm{35\left(V2\right)}}{400}$

$\frac{\mathrm{1050}}{300}=\frac{\mathrm{35V2}}{400}$

Cross multiply

$\mathrm{1050\left(400\right)}=\mathrm{\left(300\right)35V2}$

$\mathrm{10500V2}=\mathrm{420000}$

$\frac{\mathrm{10500V2}}{10500}=\frac{\mathrm{420000}}{10500}$

$\mathrm{V2}={\mathrm{40m}}^{3}$