Ideal Gas Law

The ideal gas law is on of the easier concepts of chemistry to understand and which can be applied directly to student’s lives, and this is why many students get interested in ideal gas law rather than stoichiometry of solid elements.

The law is introduced in detail in grade 11 chemistry, but references to it are made in grade 5 and in some countries it can be introduced as early as elementary school chemistry. The law is somewhat hard to understand, but applying the law is very easy.

The ideal gas law includes:

1.) Pressure (P)
2.) Temperature (T)
3.) Volume (V)
4.) Moles (n)
5.) Constant Number (R)
6.) And another set of temperature, and pressure etc. For example P1 and P2 means two different pressures.

The constant number is 0.0821 which has been figured out after thousands of experiments by many famous chemists.

Famous chemists to introduce this law are:

– Boyle
– Charles

*All gases that follow this law are known as ideal gases.

The general law states the following:

PV=nRT

1.) If there is a constant amount of gas at constant pressure, the product of the pressure and volume of the gas will also be constant

2.) Pressure and volume can be found out using the constant and temperature.

* Note that temperatures that are much higher and at low pressures, the ideal gas law are an “ideal” theory which can explain the behavior of gases.

The ideal gas law is widely accepted, but there are gases that deviate from the law and are called deviation gases or non ideal gases. On top of that the law is not very accurate and there have to changes or corrections that have to be made.

The temperature used to work out a problem is Kelvin and not Celsius, but for the final answer, the temperature must be in Celsius. 273k is 0 Celsius, which means depending on the number of Celsius the number has to be increased or decreased accordingly to solve a problem.

Here are some practice ideal gas problems

Question#1 If 4.55 moles of Helium gas has a pressure of 5.0 atm at 243 K, what volume must the gas occupy?

To solve this question you will need to apply the constant and use atm as pressure. Remember that pressure can also be in kPa also known as kilopascals. The constant is 0.0821

Answer#1: 4.55×0.0821x243k = 90.77

You must use the constant for every ideal gas problem unless stated otherwise. In the above example you can see how the constant is applied in a problem. The number 0.0821 can also be written as 8.21 R in some situations.