The two forms are related
Thermal Physics - OCR A-Level Physics
$$pV = nRT$$
$$pV = NkT$$
- The two forms are related: since $N = nN_A$, then $k = R/N_A$.
- Use $pV = nRT$ when given moles or molar mass.
- Use $pV = NkT$ when given number of molecules or working with kinetic theory.
- Standard pressureForce per unit area. Measured in pascals (Pa), where 1 Pa = 1 N m⁻². is $1.01 \times $\(10^{5}\)$ Pa, standard temperature is 273 K.
- Volume must be in \(m^{3}\): 1 litre = $1 \times 10^{-3}$ \(m^{3}\), 1 cm^3 = $1 \times 10^{-6}$ \(m^{3}\).
Worked Example
A cylinder contains 0.040 mol of an ideal gasA theoretical gas composed of many randomly moving point particles that do not interact except during brief elastic collisions. at $27 \degree$C and a pressureForce per unit area. Measured in pascals (Pa), where 1 Pa = 1 N m⁻². of $2.0 \times $\(10^{5}\)$ Pa. Calculate the volume of the gas.
Show Solution
1
Convert temperature
$T = 27 + 273 = 300 K.$
2
Apply pV = nRT: $V = nRT/p$.
3
$V = (0.040 \times 8.31 \times 300) / (2.0 \times 10^5).$
4
$V = 99.72 / (2.0 \times 10^5) = 4.99 \times 10^{-4} m^3.$
5
$V = 5.0 \times 10^{-4} m^3 (or 0.50 litres).$
Answer
$V = 5.0 \times 10^{-4} m^3$
Common Mistake
MEDIUM
Wrong: Leaving volume in litres or cm^3 when substituting into $pV = nRT$.
Right: Volume must be in \(m^{3}\). Multiply litres by \(10^{-3}\) and cm^3 by \(10^{-6}\).
Right: Volume must be in \(m^{3}\). Multiply litres by \(10^{-3}\) and cm^3 by \(10^{-6}\).