Hi A-Level Chemistry students.

Today, we will discuss on the topic of **IDEAL GAS LAW**, which might be known as **THE GASEOUS STATE** in your Chemistry syllabus.

Now, in Singapore *GCE A-Level H2 Chemistry syllabus*, candidates are required to know the following:

1) Basic assumptions of the kinetic theory as applied to an ideal gas

2) Conditions necessary for a gas to approach ideal behavior

3) Limitations of Ideality at very high pressures and very low temperatures

4) General gas equation pV=nRT and the determination of M_{r}

In the prevIous blog posts, we have discussed on point 1, 2 and 3. You can do a search for those posts by keying in “*Ideal Gas Law*” on the top RHS field.

Today we will go through the general gas equation which is:

**pV=nRT**

which can be expressed as

**pV=(m/M _{r})RT **since n = m / M

_{r}

It is important to note the units to be used in each term in the equation. Examiners’ reports have always identified *wrong use of units* to be one of the *most common error in A-Level exams*.

p = pressure in Pa or N/m^{2}

V = volume in m^{3}

n = no. of moles of gas in mol

R = Gas Constant with a value of 8.31 JK^{-1}mol^{-1}

T = temperature in K

m = mass of gas in g

Let’s take a look at an exam-based question to see how we can apply the equation in examination.

Quick Check 1:

At 27

^{o}C and a pressure of 8 x 10^{4}Pa, 1.00 dm^{3}of vapour was produced when 8.40 g of a compound was completely vapourised. Determine the relative molecular mass of the compound.

Suggested Solution:

Using the ideal gas equation

pV=nRT

we will have the following

pV=(m/M_{r})RT

and after rearrangement we will have the following

M_{r} = (mRT)/pV

= 8.40 x 8.31 x (27 + 273) / (8 x 10^{4}) x (1.00 x 10^{-3})

= 261.8

= 262 (to 3 significant figures)

**Note:**

T/K = T/^{o}C + 273

1 dm^{3} = 1.00 x 10^{-3 }m^{3}

Hope you have learned something today. In the next post we will look at more examples of solving questions using ideal gas equation.

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