IB DP Chemistry -S1.5.3 Molar volume of an ideal gas - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S1.5.3 Molar volume of an ideal gas – Study Notes – New Syllabus
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Structure 1.5.3 — Molar Volume of an Ideal Gas & Gas Laws
Structure 1.5.3 — Molar Volume of an Ideal Gas & Gas Laws
The molar volume of an ideal gas is the volume occupied by 1 mole of gas at a specific temperature and pressure.
At standard temperature and pressure (STP: 0°C and 1 atm), 1 mole of any ideal gas occupies 22.7 dm³.
Formula:
\( V = n \times 22.7 \)
or rearranged,
\( n = \frac{V}{22.7} \)
- \( V \) = volume of gas (in dm³)
- \( n \) = number of moles of gas
- 22.7 dm³ = molar volume of gas at STP (from Data Booklet)
Important: Volume must be in dm³ for this formula to work.
According to the IBDP Chemistry Data Booklet, at standard temperature and pressure (STP):
- Temperature = 273 K
- Pressure = 100 kPa
- Molar Volume = \( 22.7 \, \text{dm}^3\text{/mol} \)
At room temperature and pressure (RTP), a common approximation is:
- Temperature = 298 K
- Pressure = 100 kPa
- Molar Volume = \( 24.0 \, \text{dm}^3\text{/mol} \)
This value is useful when performing calculations involving gases without using the full ideal gas equation.
Ideal Gas Equation:
\( PV = nRT \)
- \( P \) = pressure (Pa)
- \( V \) = volume (m3)
- \( n \) = number of moles (mol)
- \( R \) = ideal gas constant = \( 8.31 \, \text{J mol}^{-1}\text{K}^{-1} \)
- \( T \) = temperature (K)
Relationship Between Variables (Gas Laws)
This shows how ideal gases behave under different conditions of pressure, temperature, volume, and amount (mol). Each law isolates two changing variables while keeping the others constant.
1. Boyle’s Law (pressure ↔ volume, constant T and n):
\( P_1V_1 = P_2V_2 \)
- When volume decreases, pressure increases — and vice versa.
- In the diagram: the piston is pushed down (volume ↓), particles collide more with walls (pressure ↑).
- Graph: Pressure vs. Volume is a hyperbola; Pressure vs. 1/Volume is a straight line.
2. Charles’ Law (volume ↔ temperature, constant P and n):
\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
- As temperature (K) increases, gas particles move faster and spread out → volume increases.
- In the diagram: more heat causes piston to rise as volume expands.
- Graph: Volume vs. Temperature (K) is a straight line.
3. Gay-Lussac’s Law (pressure ↔ temperature, constant V and n):
\( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)
- As temperature increases, particles move faster → more collisions → pressure increases.
- In the diagram: heating sealed container (fixed volume) increases pressure.
- Graph: Pressure vs. Temperature (K) is a straight line.
4. Avogadro’s Law (volume ↔ moles, constant P and T):
\( \frac{V_1}{n_1} = \frac{V_2}{n_2} \)
- More moles of gas take up more space at the same temperature and pressure.
- In the diagram: increasing number of moles causes the piston to move upward → volume increases.
- Graph: Volume vs. Moles is a straight line.
Key Points:
- Temperature must always be in Kelvin when using gas equations.
- Convert cm3 to m3 by dividing by \( 10^6 \); and dm3 to m3 by dividing by \( 10^3 \).
- Standard units: \( P \) in Pa, \( V \) in m3, \( T \) in K.
Example
Calculate the volume occupied by 2.50 mol of an ideal gas at STP.
▶️Answer/Explanation
At STP (273 K, 100 kPa), molar volume = \( 22.7 \, \text{dm}^3/\text{mol} \)
Using:
\( \text{Volume} = \text{moles} \times \text{molar volume} \)
\( \text{Volume} = 2.50 \times 22.7 = 56.75 \, \text{dm}^3 \)
Final Answer: \( \boxed{56.75 \, \text{dm}^3} \)
Example
A gas has a volume of 600 cm³ at a pressure of 100 kPa. What is its volume when the pressure is increased to 150 kPa (at constant temperature)?
▶️Answer/Explanation
Using Boyle’s Law: \( P_1V_1 = P_2V_2 \)
\( 100 \times 600 = 150 \times V_2 \)
\( V_2 = \frac{100 \times 600}{150} = 400 \, \text{cm}^3 \)
Final Answer: \( \boxed{400 \, \text{cm}^3} \)
Example
A gas occupies 250 cm³ at 273 K. What is the volume at 373 K (at constant pressure)?
▶️Answer/Explanation
Using Charles’ Law: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
\( \frac{250}{273} = \frac{V_2}{373} \)
\( V_2 = \frac{250 \times 373}{273} \approx 341.2 \, \text{cm}^3 \)
Final Answer: \( \boxed{341.2 \, \text{cm}^3} \)
Example
A gas is kept in a sealed container at 100 kPa and 300 K. What will the pressure be if the temperature is increased to 360 K?
▶️Answer/Explanation
Using Gay-Lussac’s Law: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)
\( \frac{100}{300} = \frac{P_2}{360} \)
\( P_2 = \frac{100 \times 360}{300} = 120 \, \text{kPa} \)
Final Answer: \( \boxed{120 \, \text{kPa}} \)
Example
Calculate the volume occupied by 0.500 mol of gas at 298 K and 101325 Pa.
▶️Answer/Explanation
Given: \( n = 0.500 \, \text{mol}, T = 298 \, \text{K}, P = 101325 \, \text{Pa}, R = 8.31 \, \text{J mol}^{-1}\text{K}^{-1} \)
Using: \( PV = nRT \Rightarrow V = \frac{nRT}{P} \)
\( V = \frac{0.500 \times 8.31 \times 298}{101325} \approx 0.0122 \, \text{m}^3 = 12.2 \, \text{dm}^3 \)
Final Answer: \( \boxed{12.2 \, \text{dm}^3} \)