Notes on Gases: Enhanced Greenhouse Effect, SLC, and the Ideal Gas Equation
15.2 GASES AND THE ENHANCED GREENHOUSE EFFECT
- Greenhouse gases absorb infrared (IR) radiation and contribute to the greenhouse effect. Major gases include CO₂, CH₄ and H₂O.
- Natural greenhouse effect vs. enhanced greenhouse effect:
- Natural greenhouse effect keeps Earth warmer than it would be otherwise, enabling life-supporting temperatures.
- Enhanced greenhouse effect results from increased concentrations of greenhouse gases due to human activities, leading to more heat being trapped and global warming/climate change.
- Global warming vs. climate change:
- Global warming: Earth’s rising average surface temperature, driven largely by higher greenhouse gas concentrations.
- Climate change: Broad, long-term changes in climate measures (temperature, precipitation, extreme weather, etc.) resulting from global warming and other factors.
- Greenhouse gases of concern and their role:
- CO₂, CH₄, H₂O absorb infrared radiation and contribute to heat retention in the atmosphere.
- Water vapour is the most abundant greenhouse gas but its atmospheric concentration is driven largely by natural processes and feedbacks, not directly by human emissions.
- Global Warming Potential (GWP): a way to compare the energy absorption and atmospheric lifetime of a gas relative to CO₂.
- 1 tonne of CO₂ has a GWP of 1 over a 100-year period.
- CH₄ has a GWP > 20 (more potent per tonne than CO₂).
- N₂O has a GWP > 200.
- Despite high GWP values for some gases, CO₂ remains the most significant greenhouse gas overall because of large global emissions.
- CO₂ sources and role in life:
- Major human sources: transportation, industrial processes, land-use change, energy production.
- CO₂ is essential for photosynthesis in plants.
- Historically, Earth’s atmospheric CO₂ levels remained relatively constant for thousands of years, but human activities are increasing CO₂ faster than it can be absorbed by natural sinks.
- Figures (conceptual):
- Figure 15.5: Greenhouse effect allows some heat to be trapped, maintaining a relatively constant temperature.
- Figure 15.6: Excess greenhouse gases lead to more heat retention and higher average Earth temperatures.
- Figure 15.7: Global temperature anomaly vs. atmospheric CO₂ concentration over time demonstrates a relationship between CO₂ levels and temperature.
- Practice/exercise themes (linked to Page 9):
- Describe natural vs. enhanced greenhouse effects.
- Discuss negative effects of agriculture on greenhouse gas emissions and mitigation approaches.
- Compare electric vs. gas appliances in terms of greenhouse emissions.
- Distinguish between global warming and climate change.
- Compare land-use implications of growing crops vs. rearing cattle for greenhouse gas emissions.
15.3 GASES AT STANDARD LABORATORY CONDITIONS (SLC)
- Kinetic Molecular Theory (KMT) – core ideas:
- 1) Gases are composed of particles moving constantly and randomly.
- 2) Gas particles are far apart; particle size is negligible relative to volume.
- 3) No attractive/repulsive forces between gas particles (idealisation).
- 4) Gas particles collide with container walls; collisions are elastic and do not lose kinetic energy.
- 5) Higher temperature => faster particle motion due to higher kinetic energy.
- Pressure: defined as force per unit area. SI unit is the pascal (Pa).
- Common units of gas pressure (with their conversions):
- 1\ \text{Pa} = 1\ \text{N m}^{-2}
- 1\ \text{kPa} = 1\times 10^{3}\ \text{Pa}
- 1\ \text{atm} = 101.3\ \text{kPa} = 1.013\times 10^{5}\ \text{N m}^{-2}
- 1\ \text{bar} = 100\ \text{kPa} = 1.00\times 10^{5}\ \text{N m}^{-2}
- 760\ \text{mmHg} = 1\ \text{atm}
- Additional conversion relationships:
- 1000\ \text{Pa} = 1\ \text{kPa}
- 100\ \text{kPa} = 0.987\ \text{atm}
- Standard Laboratory Conditions (SLC): a reference state for gas comparisons
- Temperature: T = 25\,^{\circ}\text{C} = 298\ \text{K}
- Pressure: P = 100\ \text{kPa}
- Molar volume and molar amount at SLC:
- Molar volume: the volume occupied by one mole of an ideal gas at a given T and P.
- At SLC: V_m = 24.8\ \text{L mol}^{-1}
- Relationship: n = \frac{V}{V_m} \quad (V\text{ in L},\ n\text{ in mol})
- Worked example (Volume from amount at SLC):
- If n = 0.24\ \text{mol} of gas at SLC, then
- V = n\,V_m = 0.24 \times 24.8 = 5.952\ \text{L} \approx 6.0\ \text{L}
- Worked example (Mass from volume at SLC):
- Given V = 1.5565\ \text{L} of H₂ at SLC, find mass.
- First, n = \frac{V}{V_m} = \frac{1.5565}{24.8} \approx 0.0628\ \text{mol}
- M(H₂) = 2.0\ \text{g mol}^{-1}\n - Mass: m = n\,M = 0.0628\times 2.0 \approx 0.126\ \text{g}
- Report to 2 s.f. ⇒ m \approx 0.13\ \text{g}
- General notes on conversions and common pitfalls:
- When using the ideal gas equation at non-standard conditions, convert all quantities to the correct SI units: P\to\text{Pa} (\text{or kPa}),\ V\to\text{L},\ T\to\text{K},\ n\to\text{mol}
- Use the correct value of R depending on the units chosen:
- R = 8.31\ \text{J K}^{-1}\text{mol}^{-1} (SI units: Pa, m³, etc.)
- Equivalently, with P in kPa and V in L: R \approx 8.31\ \text{kPa L mol}^{-1}\text{K}^{-1}
- Sample/conceptual conversions (quick review):
- Pressure unit conversions and common pitfalls when using PV = nRT, especially ensuring T is in K and V in L.
15.4 CALCULATIONS INVOLVING THE IDEAL GAS EQUATION AND STOICHIOMETRY
- The universal gas equation, combining Boyle’s law, Charles’ law and Avogadro’s hypothesis:
- PV = nRT
- In data book notation, with the standard SI-consistent choices: P in kPa, V in L, T in K, n in mol, and R ≈ 8.31\,\text{L kPa mol}^{-1}\text{K}^{-1}
- Alternatively, using SI base units: PV = nRT with R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} (and 1\text{ J} = 1\text{ Pa m}^{3}).
- At SLC, the molar volume is Vm = 24.8\ \text{L mol}^{-1} and the equation becomes particularly convenient: V = n\,Vm\quad \text{(at SLC)}
- Examples and worked problems (highlights):
- Worked example 3.2.2: Volume from amount at SLC
- Given: n = 0.24\ \text{mol} and at SLC, V_m = 24.8\ \text{L mol}^{-1}
- Calculation: V = n V_m = 0.24\times 24.8 = 5.952\ \text{L} \Rightarrow 6.0\ \text{L}
- Worked example 3.2.3 (universal gas equation): Volume from moles with non-SLC conditions
- Example: Solve for V when n = 2.24\ \text{mol}, P = 200\ \text{kPa}, T = 50^{\circ}\text{C}
- Convert temperature: T = 50 + 273 = 323\ \text{K}
- Use: V = \dfrac{n R T}{P} with R = 8.31\ \text{L kPa mol}^{-1}\text{K}^{-1}
- Compute: V = \dfrac{2.24 \times 8.31 \times 323}{200} \approx 30.1\ \text{L}
- Worked example: 13.0 mol CO₂ at 250 kPa and 75.0°C
- Temperature: T = 75.0^{\circ}\text{C} + 273 = 348\ \text{K}
- V = \dfrac{nRT}{P} = \dfrac{13.0 \times 8.31 \times 348}{250} \approx 150\ \text{L} (rounding to appropriate sig figs)
- Worked example: 6.30 mol CO₂ at 23.0°C and 550 kPa
- Temperature: T = 23.0 + 273 = 296\ \text{K}
- V = \dfrac{nRT}{P} = \dfrac{6.30 \times 8.31 \times 296}{550} \approx 28.2\ \text{L}
- Worked example: 12.8 mol CH₄ at 9.87 atm and 60°C
- Convert pressure to kPa: P = 9.87\ \text{atm} \approx 999\ \text{kPa}
- Temperature: T = 60 + 273 = 333\ \text{K}
- Use V = \dfrac{nRT}{P} with R = 8.31\ \text{L kPa mol}^{-1}\text{K}^{-1}
- Compute: V \approx \frac{12.8 \times 8.31 \times 333}{999} \approx 35.4\ \text{L}
- Worked example: 6.0 L container with 13.2 mol CO₂ at 27°C
- Temperature: T = 27^{\circ}\text{C} + 273 = 300\ \text{K}
- P = \frac{nRT}{V} = \frac{13.2 \times 8.31 \times 300}{6.0} \approx 5.48\times 10^{3}\ \text{kPa} (≈ 54 atm)
- Worked example: 1.60 mol CO₂ in 5300 mL at 293 K
- Volume: V = 1.556\ \text{L}
- P = \dfrac{nRT}{V} = \dfrac{1.60 \times 8.31 \times 293}{1.556} \approx 7.35\times 10^{5}\ \text{Pa} \approx 735\ \text{kPa}
- Practice Problem 4 (pressure in Pa): 1.60 mol CO₂ in 5.3 L at 293 K
- P = \dfrac{nRT}{V} = \dfrac{1.60 \times 8.31 \times 293}{5.3} \approx 7.35\times 10^{2}\ \text{kPa} = 7.35\times 10^{5}\ \text{Pa}
- Stoichiometry with gases at SLC and beyond:
- Mass–volume calculations for gas reactions:
- Step 1: Write/identify a balanced equation; determine known and unknown quantities in consistent units.
- Step 2: Convert all given quantities to SI units (T in K, P in kPa, V in L).
- Step 3: Determine moles of the known quantity using stoichiometry or $n = V/V_m$ at SLC.
- Step 4: Use mole ratios from the balanced equation to find moles of the desired substance.
- Step 5: Convert moles to the desired quantity (volume at SLC via $V = nV_m$ or use $PV = nRT$ otherwise).
- Example (Ch4 to CO₂ at SLC):
- Reaction: CH₄ + 2 O₂ → CO₂ + 2 H₂O
- If 0.2 mol CH₄ is consumed, then moles CO₂ produced = 0.20 mol (1:1 ratio).
- At SLC: V{CO₂} = n{CO₂} V_m = 0.20 imes 24.8 = 4.96\ \text{L} \approx 5.0\ \text{L}
- Volume–volume calculations (Avogadro’s principle):
- If all gases are at the same T and P, molar ratios equal volume ratios.
- Example: N₂ + 3 H₂ → 2 NH₃
- If 10 mL N₂ reacts with 30 mL H₂, the resulting NH₃ volume is 20 mL (2:1 ratio in terms of moles, which matches 20 mL NH₃).
- Sample Problem 6 (combustion volumes):
- Ethene combustion: C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(g)
- If 100 L C₂H₄ reacts:
- CO₂ produced: V_{CO₂} = 2 \times 100 = 200\ \text{L}
- O₂ consumed: V_{O₂} = 3 \times 100 = 300\ \text{L}
- Practice Problem 6 (CH₄ combustion at RT&PR):
- CH₄ + 2 O₂ → CO₂ + 2 H₂O
- If 25 mL CH₄ is burned, at RT and pressure:
- CO₂ produced: 25 mL
- O₂ consumed: 50 mL
- H₂O produced: 50 mL
- 15.4.2 THE USE OF STOICHIOMETRY IN CHEMICAL REACTIONS INVOLVING GASES (summary of steps)
- Use mole ratios from the balanced equation to relate the amounts of gases involved.
- For gas reactions, volumes can be used in place of moles if gases are at the same T and P (volume–volume relationships).
- Sample Problem 7 (propane): mass of hexane required to produce a given CO₂ volume at SLC
- Reaction: C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)
- If 100 L CO₂ produced, determine the volume of C₃H₈ consumed and convert to mass.
- Steps: Use mole ratios from the balanced equation to relate CO₂ to C₃H₈, then use $V = n V_m$ and $n = m/M$ to obtain mass.
- 15.4.2 Sample Problem 5–7 (key ideas):
- 5: Ethane combustion to CO₂; determine volume of ethane given CO₂ produced.
- 6: Volume–volume with C₂H₄ and O₂ to CO₂ and H₂O.
- 7: Mass-to-volume in a hydrocarbon combustion problem; practice to compute mass of hydrocarbon from CO₂ volume.
15.5.1 Topic Summary
- Key terms and concepts to remember:
- CO₂, CH₄, H₂O as major greenhouse gases
- Global warming and climate change terminology
- Greenhouse effect vs enhanced greenhouse effect
- Kinetic Molecular Theory (five postulates)
- GWP (Global Warming Potential)
- SLC (Standard Laboratory Conditions) at T = 25^\circ\text{C} and P = 100\ \text{kPa}
- Vm (Molar Volume) at SLC: V_m = 24.8\ \text{L mol}^{-1}
- Ideal gas equation: PV = nRT
- Stoichiometry with gases: mole ratios, gas-volume relationships, and conversions
- Pressure units and conversions between Pa, kPa, atm, bar, and mmHg
- Temperature scales: Celsius, Kelvin (0 K = -273.15 °C)
- Unit consistency and conversions essential for correct gas calculations
- Quick reference values:
- R\;\approx\;8.31\ \text{J mol}^{-1}\text{K}^{-1}
- R\;\approx\;8.31\ \text{kPa L mol}^{-1}\text{K}^{-1}
- Vm\big|{SLC} = 24.8\ \text{L mol}^{-1}
- Standard conditions: T = 298\ \text{K},\ P = 100\ \text{kPa}
- Unit conversions (examples):
- 1\ \text{atm} = 101.3\ \text{kPa}
- 1\ \text{bar} = 100\ \text{kPa}
- 760\ \text{mmHg} = 1\ \text{atm}
- 0.337\ \text{bar} \rightarrow 33.7\ \text{kPa}
- 253\ \text{mmHg} \rightarrow 0.333\ \text{atm}
- 0.333\ \text{atm} \rightarrow 33.7\ \text{kPa}$$
- This set of notes covers the essential theory, key equations, worked examples, and practice problems needed to prepare for exams on gases, the greenhouse effect, SLC, and gas-phase stoichiometry.