IB DP Chemistry S1.5.4 The ideal gas equation - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S1.5.4 The ideal gas equation – Study Notes – New Syllabus
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Structure 1.5.4 — Ideal Gas Equation and Combined Gas Law
Structure 1.5.4 — Ideal Gas Equation and Combined Gas Law
Ideal Gas Equation
The behavior of an ideal gas can be described using the ideal gas law:
\( PV = nRT \)
Where:
- \( P \) = pressure in pascals (Pa)
- \( V \) = volume in cubic meters (m³)
- \( n \) = amount of substance in moles (mol)
- \( R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1} \) (universal gas constant)
- \( T \) = temperature in kelvin (K)
Unit Conversions:
- Volume: cm³ → m³: divide by \( 10^6 \)
- Volume: dm³ → m³: divide by \( 1000 \)
- Temperature: °C → K: add 273.15
- Pressure: kPa → Pa: multiply by 1000
Rearranged Forms of the Ideal Gas Equation:
- \( n = \frac{PV}{RT} \)
- \( V = \frac{nRT}{P} \)
- \( P = \frac{nRT}{V} \)
- \( T = \frac{PV}{nR} \)
Combined Gas Law
The combined gas law relates pressure, volume, and temperature before and after a change:
\( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)
- Use when the number of moles remains constant.
- All temperatures must be converted to kelvin (K).
- All pressures must be in pascals (Pa) and volumes in m³.
Molar Volume at STP
- Standard Temperature and Pressure (STP):
- Temperature = 273 K (0°C)
- Pressure = \( 1.00 \times 10^5 \, \text{Pa} \)
- Molar volume of an ideal gas = 22.7 dm³ mol⁻¹
So at STP:
\( V = n \times 22.7 \quad \text{(in dm}^3\text{)} \quad \text{or} \quad n = \frac{V}{22.7} \)
Ensure volume is in dm³ when using this shortcut equation.
Example
What is the volume, in dm³, occupied by 0.30 mol of an ideal gas at STP?
▶️Answer
At STP, molar volume = 22.7 dm³ mol⁻¹
\( V = n \times 22.7 = 0.30 \times 22.7 = 6.81 \, \text{dm}^3 \)
Correct answer: 6.81 dm³
Example
A 0.050 mol sample of nitrogen gas is placed in a 0.0025 m³ container at a temperature of 298 K. Calculate the pressure of the gas in pascals.
▶️Answer/Working
Use ideal gas law: \( PV = nRT \)
\( P = \frac{nRT}{V} = \frac{0.050 \times 8.31 \times 298}{0.0025} \)
\( P = \frac{123.768}{0.0025} = 49,507.2 \, \text{Pa} \)
Final answer: \( 4.95 \times 10^4 \, \text{Pa} \)
Example
Calculate the volume, in dm³, occupied by 0.800 mol of nitrogen gas at 298 K and 100 kPa. Assume ideal gas behavior.
▶️ Answer/Explanation
Step 1: Use the ideal gas equation:
\( PV = nRT \)
Convert pressure into Pascals:
\( 100 \, \text{kPa} = 100 \times 10^3 \, \text{Pa} \)
Now solve for volume:
\( V = \frac{nRT}{P} = \frac{(0.800)(8.31)(298)}{100000} = 0.0198 \, \text{m}^3 \)
Convert to dm³:
\( 0.0198 \, \text{m}^3 = 19.8 \, \text{dm}^3 \)
Final Answer: 19.8 dm³
Example
A fixed amount of gas occupies 500 cm³ at 300 K and 101 kPa. What will be its volume at 400 K and 101 kPa?
A. 375 cm³
B. 600 cm³
C. 667 cm³
D. 750 cm³
▶️ Answer/Explanation
Use the combined gas law since pressure is constant:
\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow V_2 = V_1 \times \frac{T_2}{T_1} \)
\( V_2 = 500 \times \frac{400}{300} = 667 \, \text{cm}^3 \)
Correct Answer: C. 667 cm³
Example
A gas occupies 1.5 dm³ at 298 K and 101325 Pa. How many moles of gas are present?
▶️ Answer/Explanation
Convert volume to m³:
\( V = \frac{1.5}{1000} = 0.0015 \, \text{m}^3 \)
Use \( n = \frac{PV}{RT} \)
\( n = \frac{(101325)(0.0015)}{(8.31)(298)} \approx 0.061 \, \text{mol} \)
Answer: 0.061 mol
Example
A student plots a graph of pressure vs. 1/volume for a fixed mass of gas at constant temperature. The result is a straight line. Which gas law does this verify?
A. Charles’ Law
B. Boyle’s Law
C. Avogadro’s Law
D. Gay-Lussac’s Law
▶️ Answer/Explanation
\( P \propto \frac{1}{V} \) at constant temperature is Boyle’s Law.
Correct Answer: B. Boyle’s Law