Advanced Chemistry Made Easy

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/m²
V = volume in m³
n = no. of moles of gas in mol
R = Gas Constant with a value of 8.31 JK^-1mol^-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 ^oC and a pressure of 8 x 10⁴ Pa, 1.00 dm³ 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 10⁴) x (1.00 x 10^-3)

= 261.8

= 262 (to 3 significant figures)

Note:

T/K = T/^oC + 273

1 dm³ = 1.00 x 10^-3 m³

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

PS: if you feel that this blogpost will be useful to your friend, feel free to forward the link to them.

In the previous blog post, i have introduced to you the usefulness of Hess’ Law and Energy Cycle Diagram to determine the enthalpy changes of reactions that cannot be determined directly through experiments.

The two most common applications are:

A) Calculations involving Enthalpy Changes of Formation

B) Calculations involving Enthalpy Changes of Combustion

Let us look at two exam-based questions and see how we apply it to solve the questions.

A) Calculations involving Enthalpy Changes of Formation

Question 1:

Calculate the enthalpy change for the reaction CaF₂ + H₂SO₄ → 2HF + CaSO₄, given that the enthalpies change of formation of CaF₂, H₂SO₄, HF and CaSO₄ are -1220, -814, -271 and -1434 kJmol^-1 respectively.

Solutions:

Energy Cycle Diagram:

By Hess’ Law, ΔH_r^θ = 2(-271) + (1434) – (-1220) – (-814)

= +58 kJmol^-1

B) Calculations involving Enthalpy Changes of Combustion

Question 2:

Calculate the standard enthalpy change of formation of CH₄ given than the standard enthalpies change of combustion of methane, graphite and hydrogen are -890.2, -393.4 and -285.7 kJmol^-1 respectively.

Solutions:

Energy Cycle Diagram:

By Hess’ Law, ΔH_f^θ = -393.4 + 2(-285.7) – (-890.2)

= -74.6 kJmol-1

Hope the above examples are useful in your learning of this topic on Chemical Energetics.

Feel free to forward this post to anyone that you think will be useful to. Cheers!