Unit 3: Amount of Substance - Gases

Exploring the relationships between volume, pressure, temperature, and moles in gaseous systems.

3.14 Avogadro's Law

Avogadro's Law establishes a fundamental relationship between the volume of a gas and the amount of gas present. It states that at a constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas. The reason for this is that in a gas, the individual particles are very far apart, so their individual sizes are negligible compared to the total volume they occupy. Consequently, it doesn't matter if the gas is hydrogen (H₂) or uranium hexafluoride (UF₆); one mole of any gas will occupy the same volume under the same conditions.

Key Formula: $$ \frac{V_1}{n_1} = \frac{V_2}{n_2} $$ Where $V_1$ and $n_1$ are the initial volume and moles, and $V_2$ and $n_2$ are the final volume and moles. This shows that the ratio of volume to moles is constant as long as temperature and pressure do not change.
Solved Examples:
  1. If 0.5 moles of a gas occupy 12 dm³, what volume will 1.5 moles occupy under the same conditions?
    Solution: The number of moles is tripled (1.5 mol / 0.5 mol = 3). Therefore, the volume will also triple: $12 \, \text{dm}^3 \times 3 = 36 \, \text{dm}^3$.
  2. A balloon contains 44 g of CO₂ (1 mole) and has a volume of 24 dm³. If 22 g of CO₂ leak out, what is the new volume?
    Solution: 22 g of CO₂ is 0.5 moles. The total moles are now 0.5 mol. Since the moles halved, the volume also halves: $24 \, \text{dm}^3 / 2 = 12 \, \text{dm}^3$.
  3. If 3 moles of nitrogen gas are in a 60 L container, how many moles are in a 40 L container under identical conditions?
    Solution: Using the formula $\frac{V_1}{n_1} = \frac{V_2}{n_2}$: $\frac{60 \, \text{L}}{3 \, \text{mol}} = \frac{40 \, \text{L}}{n_2}$.
    $n_2 = \frac{40 \, \text{L} \times 3 \, \text{mol}}{60 \, \text{L}} = 2 \, \text{mol}$.
  4. A sample of 0.2 moles of helium gas has a volume of 5 dm³. What is the volume of 0.2 moles of argon gas at the same temperature and pressure?
    Solution: According to Avogadro's Law, equal moles of any gas at the same temperature and pressure occupy the same volume. The volume will still be 5 dm³.
  5. A chemical reaction produces 0.05 moles of H₂ gas. If 1 mole of gas under these conditions occupies 24 dm³, what volume of H₂ is produced?
    Solution: Volume = $0.05 \, \text{mol} \times 24 \, \text{dm}^3/\text{mol} = 1.2 \, \text{dm}^3$.
  6. A flexible container holds 5.0 dm³ of gas with 'n' moles. If 0.25 moles of gas are added, the volume becomes 6.0 dm³. Find the initial number of moles, n.
    Solution: $\frac{5.0}{n} = \frac{6.0}{n + 0.25}$. Cross-multiply: $5.0(n + 0.25) = 6.0n$.
    $5.0n + 1.25 = 6.0n \implies 1.25 = 1.0n$. So, $n = 1.25$ moles.
  7. Which sample contains more molecules: 10 L of O₂ or 12 L of CH₄ at the same temperature and pressure?
    Solution: Since volume is proportional to moles (and therefore molecules), the sample with the larger volume contains more molecules. 12 L of CH₄ contains more molecules.
  8. A reaction vessel can hold a maximum of 50 dm³. If it already contains 2 moles of argon, can you add 0.5 moles of neon without it bursting, assuming 1 mole occupies 22 dm³?
    Solution: Total moles would be $2 + 0.5 = 2.5$ moles. Total volume required = $2.5 \, \text{mol} \times 22 \, \text{dm}^3/\text{mol} = 55 \, \text{dm}^3$. This exceeds the 50 dm³ capacity, so it would burst.
  9. If a gas sample at a certain volume is heated, causing its volume to double, what must have happened to the number of moles if pressure is constant and some gas was added?
    Solution: If the volume doubled solely due to adding more gas (at constant T & P), the number of moles must have also doubled.
  10. A container with a movable piston has 1 mole of gas at 10 dm³. If the piston is moved to 25 dm³, how many moles of gas were added?
    Solution: $\frac{10 \, \text{dm}^3}{1 \, \text{mol}} = \frac{25 \, \text{dm}^3}{n_2}$. $n_2 = 2.5$ moles. Moles added = $2.5 - 1.0 = 1.5$ moles.

3.15 The Ideal Gas Equation

The Ideal Gas Equation is a comprehensive formula that merges the simple gas laws (Boyle's Law, Charles's Law, and Avogadro's Law) into a single, powerful relationship that describes the state of an ideal gas. It is essential for situations where multiple variables (pressure, volume, and temperature) are changing or when you need to find one of these properties from the others. Mastering the units is critical for its correct application.

Key Formula: $$ PV = nRT $$ Where:
  • P = Pressure of the gas, in Pascals (Pa). ($1 \, \text{kPa} = 1000 \, \text{Pa}$).
  • V = Volume of the gas, in cubic meters (m³). ($1 \, \text{m}^3 = 1000 \, \text{dm}^3$).
  • n = Amount of substance in moles (mol).
  • R = The molar gas constant (8.31 J mol⁻¹ K⁻¹). This constant links the macroscopic properties of the gas to its energy per mole per degree.
  • T = Temperature of the gas, in Kelvin (K). ($K = °C + 273$).
Solved Examples:
  1. Calculate the number of moles of a gas that occupies 250 cm³ at 25 °C and 101 kPa.
    Solution: V=250 cm³ = 0.00025 m³, T=25+273=298 K, P=101 kPa = 101000 Pa.
    $ n = \frac{PV}{RT} = \frac{101000 \times 0.00025}{8.31 \times 298} \approx 0.0102 \, \text{mol} $
  2. What volume in dm³ is occupied by 16 g of methane (CH₄, 16 g/mol) at 300 K and 150 kPa?
    Solution: n=16 g / 16 g/mol = 1 mol. P=150000 Pa.
    $ V = \frac{nRT}{P} = \frac{1 \times 8.31 \times 300}{150000} = 0.01662 \, \text{m}^3 = 16.62 \, \text{dm}^3 $
  3. Calculate the pressure (in kPa) of 0.5 moles of N₂ in a 10 dm³ cylinder at 27 °C.
    Solution: V=10 dm³ = 0.01 m³, T=27+273=300 K.
    $ P = \frac{nRT}{V} = \frac{0.5 \times 8.31 \times 300}{0.01} = 124650 \, \text{Pa} = 124.65 \, \text{kPa} $
  4. Determine the molar mass of a gas if a 0.586 g sample occupies 1.0 dm³ at 298 K and 100 kPa.
    Solution: First find moles: V=0.001 m³, P=100000 Pa.
    $n = \frac{100000 \times 0.001}{8.31 \times 298} \approx 0.0404$ mol. Molar Mass = mass/moles = $0.586 \, \text{g} / 0.0404 \, \text{mol} \approx 14.5 \, \text{g/mol}$.
  5. What temperature (in °C) is required for 2 moles of gas to exert a pressure of 200 kPa in a 25 dm³ container?
    Solution: P=200000 Pa, V=0.025 m³.
    $ T = \frac{PV}{nR} = \frac{200000 \times 0.025}{2 \times 8.31} \approx 300.8 \, \text{K} $. Temp in °C = $300.8 - 273 = 27.8 \, \text{°C}$.
  6. Calculate the density (in g/dm³) of oxygen gas (O₂, 32 g/mol) at 298 K and 101 kPa.
    Solution: Density = mass/volume. For 1 mole, mass=32 g. Find the volume of 1 mole: V=nRT/P = $(1 \times 8.31 \times 298) / 101000 \approx 0.0245 \, \text{m}^3 = 24.5 \, \text{dm}^3$.
    Density = $32 \, \text{g} / 24.5 \, \text{dm}^3 \approx 1.31 \, \text{g/dm}^3$.
  7. A 5.0 dm³ container of helium at 120 kPa is heated from 20°C to 100°C. What is the final pressure in kPa?
    Solution: Since n and V are constant, $P_1/T_1 = P_2/T_2$. T₁=293 K, T₂=373 K.
    $P_2 = P_1 \times (T_2/T_1) = 120 \, \text{kPa} \times (373/293) \approx 152.8 \, \text{kPa}$.
  8. How many atoms are in a 500 cm³ flask of neon gas at 100 kPa and 0°C?
    Solution: V=0.0005 m³, P=100000 Pa, T=273 K.
    $n = \frac{100000 \times 0.0005}{8.31 \times 273} \approx 0.022$ mol. Atoms = $0.022 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 1.33 \times 10^{22}$ atoms.
  9. A gas cylinder has a volume of 50 dm³. How many grams of N₂ (28 g/mol) does it contain if the pressure is 2000 kPa at 20°C?
    Solution: V=0.05 m³, P=2000000 Pa, T=293 K.
    $n = \frac{2000000 \times 0.05}{8.31 \times 293} \approx 41.06$ mol. Mass = $41.06 \, \text{mol} \times 28 \, \text{g/mol} \approx 1150 \, \text{g}$ or 1.15 kg.
  10. If 4.0 g of argon (Ar, 40 g/mol) is in a 2.0 dm³ container and 2.0 g of neon (Ne, 20 g/mol) is added, what is the new pressure at 298 K?
    Solution: n(Ar) = 4/40=0.1 mol. n(Ne)=2/20=0.1 mol. Total moles n=0.2 mol. V=0.002 m³.
    $P = \frac{0.2 \times 8.31 \times 298}{0.002} \approx 247758 \, \text{Pa} \approx 247.8 \, \text{kPa}$.

3.16 Molar Volume of a Gas

The Molar Volume is a convenient shortcut derived from the Ideal Gas Equation for specific, commonly used conditions. It represents the volume that one mole of any ideal gas occupies. Using molar volume simplifies calculations, as you don't need to use the full PV=nRT equation, provided the conditions match one of the standards.

  • Standard Temperature and Pressure (STP):
    • Conditions: A temperature of 273 K (0°C) and a pressure of 101 kPa (or 1 atm).
    • Molar Volume: At STP, one mole of any gas occupies 22.4 dm³.
  • Room Temperature and Pressure (RTP):
    • Conditions: A temperature of 298 K (25°C) and a pressure of 101 kPa (or 1 atm).
    • Molar Volume: At RTP, one mole of any gas occupies 24.4 dm³ (often approximated as 24 dm³ for simpler calculations).
Key Formula: $$ \text{Number of moles (n)} = \frac{\text{Volume of gas (dm³)}}{\text{Molar volume (dm³/mol)}} $$
Solved Examples:
  1. What is the volume of 4.4 g of CO₂ (44 g/mol) at STP?
    Solution: Moles = 4.4 g / 44 g/mol = 0.1 mol. Volume = $0.1 \, \text{mol} \times 22.4 \, \text{dm}^3/\text{mol} = 2.24 \, \text{dm}^3$.
  2. How many moles of gas are in a 61 dm³ container at RTP?
    Solution: Moles = $\frac{61 \, \text{dm}^3}{24.4 \, \text{dm}^3/\text{mol}} = 2.5 \, \text{mol}$.
  3. Calculate the mass of 10 dm³ of NH₃ gas (17 g/mol) at RTP.
    Solution: Moles = $\frac{10 \, \text{dm}^3}{24.4 \, \text{dm}^3/\text{mol}} \approx 0.41$ mol. Mass = $0.41 \, \text{mol} \times 17 \, \text{g/mol} \approx 6.97 \, \text{g}$.
  4. A reaction produces 560 cm³ of O₂ gas at STP. How many moles were produced?
    Solution: Volume = 560 cm³ = 0.56 dm³. Moles = $\frac{0.56 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} = 0.025 \, \text{mol}$.
  5. Which has a greater mass: 5 dm³ of SO₂ (64.1 g/mol) or 5 dm³ of CO₂ (44 g/mol) at RTP?
    Solution: Since both have the same volume at the same conditions, they have the same number of moles. Therefore, the gas with the higher molar mass (SO₂) will have the greater mass.
  6. What is the density of neon gas (Ne, 20.2 g/mol) in g/dm³ at STP?
    Solution: At STP, 1 mole has a mass of 20.2 g and occupies 22.4 dm³. Density = Mass/Volume = $20.2 \, \text{g} / 22.4 \, \text{dm}^3 \approx 0.902 \, \text{g/dm}^3$.
  7. A container holds 3 moles of gas at STP. What would its volume be at RTP?
    Solution: The number of moles remains 3 mol. Volume at RTP = $3 \, \text{mol} \times 24.4 \, \text{dm}^3/\text{mol} = 73.2 \, \text{dm}^3$.
  8. How many molecules are present in 112 cm³ of H₂ gas at STP?
    Solution: Volume = 0.112 dm³. Moles = $0.112 / 22.4 = 0.005$ mol. Molecules = $0.005 \, \text{mol} \times 6.022 \times 10^{23} \approx 3.01 \times 10^{21}$ molecules.
  9. If 2.0 g of an unknown gas occupies 1.22 dm³ at RTP, what is its molar mass?
    Solution: Moles = $1.22 \, \text{dm}^3 / 24.4 \, \text{dm}^3/\text{mol} = 0.05$ mol. Molar Mass = mass/moles = $2.0 \, \text{g} / 0.05 \, \text{mol} = 40 \, \text{g/mol}$.
  10. The decomposition of H₂O₂ produces 4.88 dm³ of O₂ gas at RTP. How many moles of O₂ were formed?
    Solution: Moles = $\frac{4.88 \, \text{dm}^3}{24.4 \, \text{dm}^3/\text{mol}} = 0.2 \, \text{mol}$.

End of Topic Questions

Answer: The volume of a gas is directly proportional to the number of moles at constant temperature and pressure.

Answer: PV = nRT.

Answer: 22.4 dm³/mol.

Answer: 6.0 L.

Answer: Kelvin (K).

Answer: Room Temperature and Pressure; 298 K (25°C) and 101 kPa.

Answer: 6 dm³ / 24.4 dm³/mol ≈ 0.246 mol.

Answer: Avogadro's Law states equal moles of any gas at the same temperature and pressure occupy the same volume.

Answer: 8.31 J mol⁻¹ K⁻¹.

Answer: 3 mol × 22.4 dm³/mol = 67.2 dm³.

Answer: The volume is halved.

Answer: 50,000 Pa.

Answer: 1 mole.

Answer: Cubic meters (m³).

Answer: 5 L.

Answer: 0.5 moles.

Answer: 0.25 mol × 24.4 dm³/mol = 6.1 dm³.

Answer: 0.5 dm³.

Answer: n = (200000 Pa × 0.002 m³) / (8.31 J mol⁻¹ K⁻¹ × 300 K) ≈ 0.16 mol.

Answer: Moles = 44.8/22.4 = 2 mol. Mass = 2 mol × 32.0 g/mol = 64.0 g.
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