**JAMB Tutorial on Chemistry Part 1 ( Gas Laws- Charles Law)**

One of the Most important topic for **JAMB** students and **Post Jamb** students doing Chemistry is Gas Law.

Gases are substances moving randomly in an enclosed system, they have no Fixed mass as they take the shape of the container they are in.

So on that note, we will go to the Laws governing calculations involving Gaseous Substances.

**What is Charles Law’s?**

**What is Charles Law’s?**

**It states that the volume of a given mass of gas is directly proportional to its temperature in kelvin provided that pressure remains constant.**

To explain this briefly, it means if the volume is increasing, the temperature will also increase and if the volume of a given gas is decreasing, the temperature also will decrease at the same Rate provided that Pressure is constant.

- Mathematically it means

(pressure is constant)**Volume ~ Temperature**

~ means variation sign)

V= KT ( by introducing a constant K)

K = V/T

K= V1 = V2

——- ——-

T1 T2

- The volume of a gas is theoretically Zero at -273 Celsius.

**In any question, alwalys remain to convert from Celsius to kelvin.**

**Here is how to do that:**

Kelvin = ( Celsius +273)

Let’s move on to example

1. The volume of a gas in a cylinder is 40cm3 and 0’C. The temperature at which the volume expands to is 54.70cm3 under same pressure is ?

Solution

**Start by stating parameters**

**V1 = 40cm3**

**T1 = 0’C …… make sure you convert to kelvin**

**(0+273) = 273K**

**V2 = 54.70cm3**

**T2 = ?**

**V1 = V2**

**—— ——**

**T1 T2**

**40=54.7**

**—- ——**

**273 T2**

**T2 = 273 X 54.7**

**————-**

**40**

**T2 = 373.3K**

2. If 60cm3 of a gas is heated from 27’C to 50’C. What is the new volume of the gas at constant pressure?

**V1 = 60cm3**

**T2 = 27’C**

*Is that all?*

**Don’t forget to convert to convert to Kelvin**

K = 273+27

K = 300’

V2 =?

T2 = 50’C == 323K

V1 = V2

—- ——

T1 T2

V2 = 60 X 323

————-

300

V2 =64.6cm3

*Recommended:**JAMB TUTORIAL: PHYICS( GUIDELINES TO MOTION)*

That’s all for Part 1, thanks for taking your time to go through this and we hope it is helpful, kindly share.