At GCSE, you learned that one mole of any gas occupies the same volume at room temperature and pressure. At A-Level, we expand this concept to look at how gases behave under any combination of temperature and pressure, using the ideal gas equation.
🔑 Key Principle
An ideal gas is a theoretical gas composed of randomly moving point particles that interact only through elastic collisions. In reality, real gases behave very close to ideally at low pressures and high temperatures, allowing us to use this relationship with high accuracy.
The Ideal Gas Equation Formula
The equation relates five variables:
Where:
- \( p \) = pressure in Pascals (\( \text{Pa} \)) or Newtons per square meter (\( \text{N m}^{-2} \))
- \( V \) = volume in cubic meters (\( \text{m}^3 \))
- \( n \) = amount of substance in moles (\( \text{mol} \))
- \( R \) = gas constant (\( 8.31 \text{ J K}^{-1} \text{ mol}^{-1} \), provided on the data sheet)
- \( T \) = temperature in Kelvin (\( \text{K} \))
The most common source of lost marks in AQA exams is failing to convert units into the correct SI formats. Always double check your units before plugging them into the equation!
Unit Conversion System
You must be comfortable making the following conversions before performing any calculations:
| Variable | Given Unit | Target Unit | Conversion Step |
|---|---|---|---|
| Pressure (\( p \)) | Kilopascals (\( \text{kPa} \)) | Pascals (\( \text{Pa} \)) | Multiply by \( 10^3 \) (\( \times 1000 \)) |
| Pressure (\( p \)) | Megapascals (\( \text{MPa} \)) | Pascals (\( \text{Pa} \)) | Multiply by \( 10^6 \) (\( \times 1,000,000 \)) |
| Volume (\( V \)) | Cubic decimeters (\( \text{dm}^3 \)) | Cubic meters (\( \text{m}^3 \)) | Divide by \( 10^3 \) (\( \times 10^{-3} \)) |
| Volume (\( V \)) | Cubic centimeters (\( \text{cm}^3 \)) | Cubic meters (\( \text{m}^3 \)) | Divide by \( 10^6 \) (\( \times 10^{-6} \)) |
| Temperature (\( T \)) | Celsius (\( ^\circ\text{C} \)) | Kelvin (\( \text{K} \)) | Add 273 |
Worked Examples
Step 1: Identify variables and convert units.
- \( p = 120 \text{ kPa} = 120,000 \text{ Pa} \)
- \( n = 2.50 \text{ mol} \)
- \( T = 25.0 \text{ }^\circ\text{C} = 25.0 + 273 = 298 \text{ K} \)
- \( R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1} \)
Step 2: Rearrange the equation to make \( V \) the subject.
\[ V = \frac{nRT}{p} \]
Step 3: Substitute the values into the equation.
\[ V = \frac{2.50 \times 8.31 \times 298}{120,000} = \frac{6190.95}{120,000} = 0.05159 \text{ m}^3 \]
Step 4: Convert the target volume from \( \text{m}^3 \) into \( \text{dm}^3 \).
\[ V = 0.05159 \times 1000 = 51.6 \text{ dm}^3 \]
Step 1: Identify variables and convert units.
- \( m = 0.372 \text{ g} \)
- \( p = 100 \text{ kPa} = 100,000 \text{ Pa} \)
- \( V = 120 \text{ cm}^3 = 120 \times 10^{-6} \text{ m}^3 \)
- \( T = 98.0 \text{ }^\circ\text{C} = 98.0 + 273 = 371 \text{ K} \)
Step 2: Calculate the amount in moles (\( n \)) using the ideal gas equation.
\[ n = \frac{pV}{RT} = \frac{100,000 \times 120 \times 10^{-6}}{8.31 \times 371} = \frac{12.0}{3083.01} = 3.892 \times 10^{-3} \text{ mol} \]
Step 3: Calculate the molecular mass using \( M_r = \frac{m}{n} \).
\[ M_r = \frac{0.372}{3.892 \times 10^{-3}} = 95.6 \text{ g mol}^{-1} \]
Therefore, the relative molecular mass of the volatile liquid is 95.6.
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