Amount of Substance
- In thermodynamics, the amount of substance is measured in the SI unit ‘mole’
- This has the symbol mol
- The mole is a unit of substance, not a unit of mass
- The mole is defined as:
The SI base unit of an ‘amount of substance’. It is the amount containing as many particles (e.g. atoms or molecules) as there are atoms in $12 \mathrm{~g}$ of carbon-12
- The mole is an important unit in thermodynamics
- If we consider the number of moles of two different gases under the same conditions, their physical properties are the same
The Avogadro Constant
- In AS Physics, the atomic mass unit (u) was introduced as approximately the mass of a proton or neutron $=1.66 \times 10^{-27} \mathrm{~kg}$
- This means that an atom or molecule has a mass approximately equal to the number of protons and neutrons it contains
- A carbon-12 atom has a mass of:
$
12 \mathrm{u}=12 \times 1.66 \times 10^{-27}=1.99 \times 10^{-26} \mathrm{~kg}
$
- The exact number for a mole is defined as the number of molecules in exactly $12 \mathrm{~g}$ of carbon:
$
1 \text { mole }=\frac{0.012}{1.99 \times 10^{-26}}=6.02 \times 10^{23} \text { molecules }
$
- Avogadro’s constant $\left(N_A\right)$ is defined as:
The number of atoms of carbon-12 in $12 \mathrm{~g}$ of carbon-12; equal to $6.02 \times 10^{23} \mathrm{~mol}^1$ - For example, 1 mole of sodium ( $\mathrm{Na}$ ) contain $6.02 \times 10^{23}$ atoms of sodium
- The number of atoms can be determined if the number of moles is known, for example:
- 2.0 mol of nitrogen contains: $2.0 \times \mathrm{N}_{\mathrm{A}}=2.0 \times 6.02 \times 10^{23}=1.20 \times 10^{24}$ atoms
Mole and the Atomic Mass
- One mole of any element is equal to the relative atomic mass of that element in grams
- e.g. helium has an atom mass of 4 – this means 1 mole of helium has a mass of $4 \mathrm{~g}$
- If the substance is a compound, add up the relative atomic masses, for example, water $\left(\mathrm{H}_2 \mathrm{O}\right)$ is made up of
- 2 hydrogen atoms (atomic mass of 1 ) and 1 oxygen atom (atomic mass of 16)
- So, 1 mole of water would have a mass of $(2 \times 1)+16=18 \mathrm{~g}$
Molar Mass
- The molar mass of a substance is the mass, in grams, in one mole
- Its unit is $\mathbf{g ~ m o l}^{-1}$
- The number of moles from this can be calculated using the equation:
$
\text { Number of moles }=\frac{\operatorname{mass}(g)}{\text { molar mass }\left(g \text { mol }^{-1}\right)}
$
Worked example
How many molecules are there in $6 \mathrm{~g}$ of magnesium-24?
Answer/Explanation
Step 1: Calculate the mass of 1 mole of magnesium
One mole of any element is equal to the relative atomic mass of that element in grams
$
1 \text { mole }=24 \mathrm{~g} \text { of magnesium }
$
Step 2:
Calculate the amount of moles in $6 \mathrm{~g}$
$
\frac{6}{24}=0.25 \text { moles }
$
Step 3:
Convert the moles to number of molecules
$
1 \text { mole }=6.02 \times 10^{23} \text { molecules }
$
$
0.25 \text { moles }=0.25 \times 6.02 \times 10^{23}=1.51 \times 10^{23} \text { molecules }
$
Exam Tip
If you want to find out more about the mole, check out the CIE A Level Chemistry revision notes!
Ideal Gases
An ideal gas is one which obeys the relation:
$
\mathbf{p V} \propto \mathbf{T}
$
- Where:
- $p=$ pressure of the gas $(\mathrm{Pa})$
- $\mathrm{V}=$ volume of the gas $\left(\mathrm{m}^3\right)$
- $\mathrm{T}=$ thermodynamic temperature $(K)$
- The molecules in a gas move around randomly at high speeds, colliding with surfaces and exerting pressure upon them
- Imagine molecules of gas free to move around in a box
- The temperature of a gas is related to the average speed of the molecules:
- The hotter the gas, the faster the molecules move
- Hence the molecules collide with the surface of the walls more frequently
- Since force is the rate of change of momentum:
- Each collision applies a force across the surface area of the walls
- The faster the molecules hit the walls, the greater the force on them
- Since pressure is the force per unit area
- Higher temperature leads to higher pressure
- If the volume $V$ of the box decreases, and the temperature $T$ stays constant:
- There will be a smaller surface area of the walls and hence more collisions
- This also creates more pressure
- Since this equates to a greater force per unit area, pressure in an ideal gas is therefore defined by:
The frequency of collisions of the gas molecules per unit area of a container
Boyle’s Law
- If the temperature $T$ is constant, then Boyle’s Law is given by:
$
\mathrm{p} \propto \frac{1}{V}
$
- This leads to the relationship between the pressure and volume for a fixed mass of gas at constant temperature:
$
P_1 \mathbf{V}_1=\mathbf{P}_2 \mathbf{V}_2
$
- Charles’s Law
- If the pressure $\mathrm{P}$ is constant, then Charles’s law is given by:
$
\mathbf{V} \propto \mathbf{T}
$
- This leads to the relationship between the volume and thermodynamic temperature for a fixed mass of gas at constant pressure:
$
\frac{V_1}{T_1}=\frac{V_2}{T_2}
$
Pressure Law
- If the volume $V$ is constant, the the Pressure law is given by:
$
\mathbf{P} \propto \mathbf{T}
$
- This leads to the relationship between the pressure and thermodynamic temperature for a fixed mass of gas at constant volume:
$
\frac{P_1}{T_1}=\frac{P_2}{T_2}
$
Worked example
An ideal gas is in a container of volume $4.5 \times 10^{-3} \mathrm{~m}^3$. The gas is at a temperature of $30^{\circ} \mathrm{C}$ and a pressure of $6.2 \times 10^5 \mathrm{~Pa}$.
Calculate the pressure of the ideal gas in the same container when it is heated to $40^{\circ} \mathrm{C}$.
Answer/Explanation
Step 1:
Ideal gas relation between pressure, volume and temperature
$
\mathbf{p V} \propto \mathbf{T}
$
Step 2:
Write the equation in full
$\mathbf{p V}=\mathbf{k T} \quad$ where $\mathrm{k}$ is the constant of proportionality
Step 3:
Rearrange for the constant of proportionality
$
\mathrm{k}=\frac{p V}{T}
$
Step 4:
Convert temperature $\mathrm{T}$ into Kelvin
$
\begin{gathered}
\theta^{\circ} \mathrm{C}+273.15=T K \\
30^{\circ} \mathrm{C}+273.15=303.15 \mathrm{~K}
\end{gathered}
$
Step 5: Substitute in known value into constant of proportionality equation
$
k=\frac{6.2 \times 10^5 \times 4.5 \times 10^{-3}}{303.15}=9.203 \ldots
$
Step 6:
Rearrange ideal gas relation equation for pressure
$
\mathrm{p}=\frac{k T}{V}
$
Step 7:
Substitute in new values
$
k=9.203 \ldots
$
$\begin{gathered}V \text { stays the same }=4.5 \times 10^{-3} \mathrm{~m}^3 \\ T=40^{\circ} \mathrm{C}=40+273.15=313.15 \mathrm{~K} \\ p=\frac{(9.203 \ldots) \times 313.15}{4.5 \times 10^{-3}}=640.45 \ldots \times 10^3 \mathrm{~Pa}=640 \mathrm{kPa}\end{gathered}$
Exam Tip
Don’t round too early in your working out! In the worked example, the unrounded value of $k$ is represented by “…” to show its full value is to be carried over to the next step of the calculation. On your calculator, this can be done by using the “ans” button instead of typing in the whole number.