Counting Molecules: pV = NkT | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Counting Molecules: pV = NkT

The last lesson counted the gas in moles. Many questions give the number of molecules instead.

The equation of state can be written for a plain count of molecules. It is the same law, written for one molecule at a time.

From moles to molecules

Start from the molar equation of state.
$pV=nRT$
Replace the amount of gas with the molecule-count relation from lesson 15.01.
$n=\dfrac{N}{N_A}$
$pV=\dfrac{N}{N_A}RT$
The ratio of the molar gas constant to the Avogadro constant is the Boltzmann constant.
$k=\dfrac{R}{N_A}$

pV = NkT

Symbols

$p$ = pressure of the gas (Pa)
$V$ = volume of the gas (m³)
$N$ = number of molecules (no unit)
$k$ = Boltzmann constant, 1.38 × 10⁻²³ (J K⁻¹)
$T$ = thermodynamic temperature (K)

The two constants. links the molar gas constant and the Avogadro constant. Papers ask for this link directly (9702/41/O/N/24 Q3(a)(ii)).
$k=\dfrac{R}{N_A}\qquad N_A=\dfrac{R}{k}$
$R$ is the gas constant for one mole. $k$ is the gas constant for one molecule. The numerical check is shown below.
$k=\dfrac{8.31}{6.02\times10^{23}}$
$k=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$
Which form to use. Use the molar form when the question gives moles or a mass. Use the molecular form when it gives a number of molecules, or asks for one.
Same conditions as before. The gas must be ideal, the amount fixed, and $T$ in kelvin.

Common mistake

Pair the symbols correctly.

n

goes with

R

, and

N

goes with

k

. The two wrong pairings are shown below. They are wrong by a factor of

6.02\times10^{23}

pV\ne NRT\qquad pV\ne nkT

Worked example

Use the molecule-count equation directly

An ideal gas has pressure $2.4\times10^5\ \mathrm{Pa}$ , volume $6.0\times10^{-3}\ \mathrm{m^3}$ and temperature 300 K. Find the number of molecules.

The question asks for a molecule count, so rearrange the molecular form.
$N=\dfrac{pV}{kT}$
Substitute the values in SI units.
$pV=1.44\times10^3\ \mathrm{J}$
$kT=4.14\times10^{-21}\ \mathrm{J}$
$N=\dfrac{1.44\times10^3}{4.14\times10^{-21}}$
$N=3.5\times10^{23}$

Answer

N=3.5\times10^{23}\ \text{molecules}

Look at what the question counts. Use the molar form for moles and the molecular form for molecules. Convert between the two counts with the Avogadro constant.

Your turnquick check

A fixed-volume gas stays at the same temperature while its pressure doubles. What happens to the number of molecules?

Show worked answer

It doubles. With $V$ and $T$ fixed:

pV=NkT

Therefore, $N$ is proportional to $p$ .

Counting Molecules: pV = NkT

The last lesson counted the gas in moles. Many questions give the number of molecules instead.

The equation of state can be written for a plain count of molecules. It is the same law, written for one molecule at a time.

From moles to molecules

Start from the molar equation of state.
$pV=nRT$
Replace the amount of gas with the molecule-count relation from lesson 15.01.
$n=\dfrac{N}{N_A}$
$pV=\dfrac{N}{N_A}RT$
The ratio of the molar gas constant to the Avogadro constant is the Boltzmann constant.
$k=\dfrac{R}{N_A}$

pV = NkT

Symbols

$p$ = pressure of the gas (Pa)
$V$ = volume of the gas (m³)
$N$ = number of molecules (no unit)
$k$ = Boltzmann constant, 1.38 × 10⁻²³ (J K⁻¹)
$T$ = thermodynamic temperature (K)

The two constants. links the molar gas constant and the Avogadro constant. Papers ask for this link directly (9702/41/O/N/24 Q3(a)(ii)).
$k=\dfrac{R}{N_A}\qquad N_A=\dfrac{R}{k}$
$R$ is the gas constant for one mole. $k$ is the gas constant for one molecule. The numerical check is shown below.
$k=\dfrac{8.31}{6.02\times10^{23}}$
$k=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$
Which form to use. Use the molar form when the question gives moles or a mass. Use the molecular form when it gives a number of molecules, or asks for one.
Same conditions as before. The gas must be ideal, the amount fixed, and $T$ in kelvin.

Common mistake

Pair the symbols correctly.

n

goes with

R

, and

N

goes with

k

. The two wrong pairings are shown below. They are wrong by a factor of

6.02\times10^{23}

pV\ne NRT\qquad pV\ne nkT

Worked example

Use the molecule-count equation directly

An ideal gas has pressure $2.4\times10^5\ \mathrm{Pa}$ , volume $6.0\times10^{-3}\ \mathrm{m^3}$ and temperature 300 K. Find the number of molecules.

The question asks for a molecule count, so rearrange the molecular form.
$N=\dfrac{pV}{kT}$
Substitute the values in SI units.
$pV=1.44\times10^3\ \mathrm{J}$
$kT=4.14\times10^{-21}\ \mathrm{J}$
$N=\dfrac{1.44\times10^3}{4.14\times10^{-21}}$
$N=3.5\times10^{23}$

Answer

N=3.5\times10^{23}\ \text{molecules}

Look at what the question counts. Use the molar form for moles and the molecular form for molecules. Convert between the two counts with the Avogadro constant.

Your turnquick check

A fixed-volume gas stays at the same temperature while its pressure doubles. What happens to the number of molecules?

Show worked answer

It doubles. With $V$ and $T$ fixed:

pV=NkT

Therefore, $N$ is proportional to $p$ .