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IB DP Chemistry – S1.5.4 The ideal gas equation – Study Notes

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

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