Higher Tier, Chemistry-only: gas volumes at room temperature and pressure
| English | Français |
|---|---|
| room temperature and pressure | room temperature and pressure |
| molar gas volume/ˈməʊlə ɡæs ˈvɒljuːm/ | molar gas volume |
What would explain this observation?
- Higher Tier: One mole of hydrogen and one mole of carbon dioxide occupy the same gas volume at the same temperature and pressure. Their different masses do not change that equal-amount volume rule.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- At the specification’s room temperature and pressure 室温和大气压, 20 °C and 1 atmosphere, one mole of gas occupies 24 dm³. Calculate V=24n in dm³, or n=V/24. For a gas mass, first calculate n=m/M, then volume. Conversely, a gas volume gives amount and then mass using its molar mass. This fixed molar-volume value is tied to the stated conditions.
- molar gas volume 气体摩尔体积: The volume occupied by one mole of gas at specified conditions; room temperature and pressure: Here the specified reference conditions of 20 °C and 1 atmosphere.
When do balanced coefficients give gas-volume ratios?
Equal gas amounts occupy equal volumes at identical conditions, so balanced coefficients give gaseous volume ratios. N₂(g) + 3H₂(g) → 2NH₃(g) gives ratio 1:3:2 for gas volumes measured at the same temperature and pressure. This ratio does not apply to a liquid or solid volume. A predicted product volume assumes sufficient reactants and the stated complete-conversion model; real equilibrium can reduce conversion.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- Equal gas amounts occupy equal volumes at identical conditions, so balanced coefficients give gaseous volume ratios. N₂(g) + 3H₂(g) → 2NH₃(g) gives ratio 1:3:2 for gas volumes measured at the same temperature and pressure. This ratio does not apply to a liquid or solid volume. A predicted product volume assumes sufficient reactants and the stated complete-conversion model; real equilibrium can reduce conversion.
- Convert cm³ to dm³ consistently and label every volume’s conditions. Do not use 22.4 dm³ per mol, a value associated with different reference conditions, in this RTP task. Use teacher-provided gas data or approved low-risk gas collection under supervision; do not propose making ammonia or hydrogen in an improvised apparatus. Water-vapour and collection losses can affect real measurements.
Which two habits make the investigation or model in this case more defensible?
Convert cm³ to dm³ consistently and label every volume’s conditions. Do not use 22.4 dm³ per mol, a value associated with different reference conditions, in this RTP task. Use teacher-provided gas data or approved low-risk gas collection under supervision; do not propose making ammonia or hydrogen in an improvised apparatus. Water-vapour and collection losses can affect real measurements.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: 4.4 g CO₂ with molar mass 44 $\dfrac{\text{g}}{\text{mol}}$ gives 0.100 mol, hence 2.40 dm³ at RTP. In the ammonia equation, 6.0 dm³ H₂ can theoretically react with 2.0 dm³ N₂ to form 4.0 dm³ NH₃ at identical conditions. There are four reactant gas volumes for two product volumes; total gas volume need not be conserved even though atoms and mass are.
At RTP, find the volume of 8.8 g CO₂ using molar mass 44 $\dfrac{\text{g}}{\text{mol}}$ and 24 dm³ per mol. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
At RTP, find the volume of 8.8 g CO₂ using molar mass 44 g per mol and 24 dm³ per mol.
The result is 4.8 dm³. Known: 4.4 g CO₂ with molar mass 44 g per mol gives 0.100 mol, hence 2.40 dm³ at RTP. In the ammonia equation, 6.0 dm³ H₂ can theoretically react with 2.0 dm³ N₂ to form 4.0 dm³ NH₃ at identical conditions. There are four reactant gas volumes for two product volumes; total gas volume need not be conserved even though atoms and mass are.
Check the conclusion and its limits
- The number 24 is not a universal gas volume at every temperature and pressure. Coefficients concern gas volume only under matched conditions. Equal gas volumes can have different masses, and a reduction in total gas volume does not mean atoms disappeared.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Total gas volume is always conserved in a balanced chemical reaction. This claim is false: The number 24 is not a universal gas volume at every temperature and pressure. Coefficients concern gas volume only under matched conditions. Equal gas volumes can have different masses, and a reduction in total gas volume does not mean atoms disappeared.
Higher Tier, Chemistry-only: gas volumes at room temperature and pressure: Equal gas amounts occupy equal volumes at identical conditions, so balanced coefficients give gaseous volume ratios. N₂(g) + 3H₂(g) → 2NH₃(g) gives ratio 1:3:2 for gas volumes measured at the same temperature and pressure. This ratio does not apply to a liquid or solid volume. A predicted product volume assumes sufficient reactants and the stated complete-conversion model; real equilibrium can reduce conversion.
Total gas volume is always conserved in a balanced chemical reaction.
The number 24 is not a universal gas volume at every temperature and pressure. Coefficients concern gas volume only under matched conditions. Equal gas volumes can have different masses, and a reduction in total gas volume does not mean atoms disappeared.
The volume occupied by one mole of gas at specified conditions: write the technical term.
molar gas volume means The volume occupied by one mole of gas at specified conditions.