The Ideal Gas Equation (pV = nRT)
\[ pV = nRT \]
Mandatory SI Units
To use this equation, you must convert all values into strict SI units. Failure to do so guarantees an incorrect answer.
| Symbol | Quantity | SI Unit | Common Conversion |
|---|---|---|---|
| \(p\) | Pressure | Pa (N m⁻²) | kPa × 1000 = Pa |
| \(V\) | Volume | m³ | dm³ ÷ 1000 = m³ ; cm³ ÷ 10⁶ = m³ |
| \(n\) | Moles | mol | - |
| \(R\) | Gas constant | 8.31 J K⁻¹ mol⁻¹ | Given in Data Booklet |
| \(T\) | Temperature | K | °C + 273 = K |
Worked Example: Determining Molar Mass
Calculating Molar Mass via pV = nRT
Problem: A 0.250 g sample of volatile liquid vaporises to 65.0 cm³ at 99.0 °C and 100 kPa. Calculate its molar mass.
- Convert to SI: \(p = 1.00 \times 10^5\text{ Pa}\), \(V = 6.50 \times 10^{-5}\text{ m}^3\), \(T = 99.0 + 273.15 = 372.15\text{ K}\)
- Find moles: \(n = \frac{pV}{RT} = \frac{(1.00 \times 10^5)(6.50 \times 10^{-5})}{(8.31)(372.15)} = 2.102 \times 10^{-3}\text{ mol}\)
- Find molar mass: \(M = \frac{m}{n} = \frac{0.250}{2.102 \times 10^{-3}} = \mathbf{119\text{ g mol}^{-1}}\)
Step 1: Unit conversions
T = 100 + 273 = 373 K
V = 98.0 ÷ 10⁶ = 9.80 × 10⁻⁵ m³
p = 103 × 1000 = 103000 Pa
Step 2: Rearrange for n
\(n = \frac{pV}{RT} = \frac{103000 \times 9.80 \times 10^{-5}}{8.31 \times 373}\)
\(n = 3.255 \times 10^{-3}\) mol
Step 3: Calculate molar mass
\(M = \frac{m}{n} = \frac{0.250}{3.255 \times 10^{-3}}\)
\(M = \) 76.8 g mol⁻¹
SI Unit Conversions in pV = nRT
The vast majority of marks lost in gas calculations stem from incorrect units. Memorise these non-negotiables:
- Volume: \(1\text{ m}^3 = 1000\text{ dm}^3 = 1\,000\,000\text{ cm}^3\) (divide cm³ by 10⁶ to get m³)
- Pressure: \(1\text{ kPa} = 1000\text{ Pa}\)
- Temperature: Always add 273.15 to convert °C to Kelvin
Try the Gas Law Calculator
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The Combined Gas Law
For a sealed system where the amount of gas stays constant:
Combined Gas Law Equation
\[ \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2} \]
Use this when an unchanging mass of gas (fixed \(n\)) transitions between two sets of conditions.
Molar Volume at STP
Standard Temperature & Pressure (STP)
At STP (Standard Temperature and Pressure): \(T = 273\text{ K}\) and \(p = 100\text{ kPa}\). Under these conditions, \(V_m = 22.7\text{ dm}^3\text{ mol}^{-1}\).
Essential Gas Calculation Checklist
- STP vs SATP: IB uses STP (273 K, 100 kPa, 22.7 dm³ mol⁻¹), not SATP (298 K, 24.8 dm³ mol⁻¹).
- R is in the Data Booklet: \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\). Do not memorise approximations.
- Always verify your final sig figs: match the least precise given measurement.
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