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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/chemistry/edexcel/gas-volumes-and-ideal-gas/flashcards","label":"Gas volumes and ideal gas","subjectCode":"a_level_chemistry","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"3966403a-fa8c-4c48-aa92-3314fa17c5c1","title":"Gas volumes and ideal gas Flashcards","topicId":"a12a0b33-0819-5d4b-bf4b-207bc3633346","topicSlug":"gas-volumes-and-ideal-gas","cards":[{"id":"2e1c6699-8152-4bbb-993e-3e1f688bf446","cardType":"cloze","points":null,"front":"The equation for the amount of gas in moles at room temperature and pressure (RTP) is $n = \\frac{V}{\\text{[...]}}$ when $V$ is in $\\text{dm}^3$.","back":"The equation for the amount of gas in moles at room temperature and pressure (RTP) is $n = \\frac{V}{\\boxed{24.0}}$ when $V$ is in $\\text{dm}^3$.","mcqOptions":[{"text":"$$24.0$","correct":true,"explanation":"At RTP, one mole of gas occupies $24.0\\text{ dm}^3$, so dividing the volume in $\\text{dm}^3$ by $24.0$ gives the amount in moles."},{"text":"$$22.4$","correct":false,"explanation":"The value $22.4$ is the approximate molar gas volume at standard temperature and pressure, not RTP. At RTP, divide the volume by $24.0$."},{"text":"$$24000$","correct":false,"explanation":"The value $24000$ applies when the gas volume is measured in $\\text{cm}^3$. For a volume in $\\text{dm}^3$, the required denominator is $24.0$."},{"text":"$$0.0240$","correct":false,"explanation":"The value $0.0240$ is the molar gas volume expressed in $\\text{m}^3$. Since the volume here is in $\\text{dm}^3$, the denominator must be $24.0$."}]},{"id":"e7d4042a-e66f-4e91-8984-5c7f17d843ff","cardType":"cloze","points":null,"front":"The equation for the amount of gas in moles at room temperature and pressure (RTP) is $n = \\frac{V}{24.0}$ when $V$ is in $\\text{[...]}$.","back":"The equation for the amount of gas in moles at room temperature and pressure (RTP) is $n = \\frac{V}{24.0}$ when $V$ is in $\\boxed{\\text{dm}^3}$.","mcqOptions":[{"text":"$$\\text{dm}^3$","correct":true,"explanation":"At RTP, the molar gas volume is $24.0\\text{ dm}^3\\text{ mol}^{-1}$, so $V$ must be measured in $\\text{dm}^3$ when using this form of the equation."},{"text":"$$\\text{cm}^3$","correct":false,"explanation":"Using $\\text{cm}^3$ with a denominator of $24.0$ creates a factor-of-1000 error. Volumes in $\\text{cm}^3$ require division by $24000$, while this equation requires $\\text{dm}^3$."},{"text":"$$\\text{m}^3$","correct":false,"explanation":"Using $\\text{m}^3$ with $24.0$ confuses cubic metres with cubic decimetres. For $\\text{m}^3$, the denominator is $0.0240$; here $V$ must be in $\\text{dm}^3$."},{"text":"$$\\text{mm}^3$","correct":false,"explanation":"Cubic millimetres are far smaller than cubic decimetres, so using them with $24.0$ causes a major conversion error. This equation requires $V$ in $\\text{dm}^3$."}]},{"id":"ce004695-7d4f-4feb-9e23-10dad3f22993","cardType":"basic","points":1,"front":"What volume does 1 mole of any gas occupy at room temperature and pressure (RTP)?","back":"24.0 dm³ (or 24000 cm³).","mcqOptions":[{"text":"24.0 dm³ (or 24000 cm³).","correct":true,"explanation":"At room temperature and pressure, the accepted molar gas volume is $24.0\\text{ dm}^3\\text{ mol}^{-1}$, equivalent to $24000\\text{ cm}^3\\text{ mol}^{-1}$."},{"text":"22.4 dm³ (or 22400 cm³).","correct":false,"explanation":"The value $22.4\\text{ dm}^3$ is associated with standard temperature and pressure rather than RTP. At RTP, one mole occupies $24.0\\text{ dm}^3$, or $24000\\text{ cm}^3$."},{"text":"240 dm³ (or 240000 cm³).","correct":false,"explanation":"This is ten times too large, usually from a misplaced decimal point. One mole at RTP occupies $24.0\\text{ dm}^3$, which is $24000\\text{ cm}^3$."},{"text":"2.40 dm³ (or 2400 cm³).","correct":false,"explanation":"This is ten times too small, usually from an incorrect decimal conversion. One mole at RTP occupies $24.0\\text{ dm}^3$, equivalent to $24000\\text{ cm}^3$."}]},{"id":"3d8d4e8a-06d8-406d-8382-48d54e6cad7b","cardType":"basic","points":2,"front":"State Avogadro’s law regarding gas volumes.","back":"At the same temperature and pressure, equal volumes of gases contain equal numbers of particles.","mcqOptions":[{"text":"At the same temperature and pressure, equal volumes of gases contain equal numbers of particles.","correct":true,"explanation":"Avogadro’s law states that equal gas volumes contain equal numbers of particles when temperature and pressure are unchanged, regardless of the gases’ chemical identities."},{"text":"At the same temperature and pressure, equal volumes of gases contain equal masses of particles.","correct":false,"explanation":"Equal volumes contain equal numbers of particles, not necessarily equal masses, because different gases have different molar masses. Temperature and pressure must be the same."},{"text":"At the same temperature and pressure, equal volumes of gases contain different numbers of particles.","correct":false,"explanation":"This reverses Avogadro’s law: equal volumes at the same temperature and pressure contain equal, not different, numbers of particles."},{"text":"At the same temperature and pressure, equal masses of gases contain equal numbers of particles.","correct":false,"explanation":"Equal masses of different gases generally represent different amounts in moles because their molar masses differ. Avogadro’s law instead relates equal volumes to equal particle numbers."}]},{"id":"cd6eb7fd-db7d-4441-bccc-69242778897d","cardType":"basic","points":1,"front":"When can you use gas volumes directly as mole ratios in a chemical equation?","back":"When all gases are at the same temperature and pressure.","mcqOptions":[{"text":"When all gases are at the same temperature and pressure.","correct":true,"explanation":"At a common temperature and pressure, gas volume is directly proportional to amount in moles. The balanced equation’s mole ratios can therefore be used as volume ratios."},{"text":"When all gases are at the same temperature but different pressures.","correct":false,"explanation":"Equal temperature alone is insufficient because changing pressure changes a gas’s volume. Direct volume ratios require every gas to be at the same temperature and pressure."},{"text":"When all gases are at the same pressure but different temperatures.","correct":false,"explanation":"Equal pressure alone is insufficient because changing temperature changes gas volume. Direct volume ratios require every gas to be at the same temperature and pressure."},{"text":"When all gases are at any temperature and pressure.","correct":false,"explanation":"Gas volumes depend on both temperature and pressure, so they cannot always represent mole ratios directly. All gases must be measured at the same temperature and pressure."}]},{"id":"b768f8cb-1c7a-4795-9129-ebfd5c548076","cardType":"basic","points":1,"front":"Equation: $N_2(g) + 3H_2(g) \\rightarrow 2NH_3(g)$.
Calculate the volume of $NH_3$ produced from $30\\text{ cm}^3$ of $N_2$ (excess $H_2$).","back":"60 cm³ (ratio 1:2).","mcqOptions":[{"text":"60 cm³ (ratio 1:2).","correct":true,"explanation":"The equation shows that one volume of $N_2$ forms two volumes of $NH_3$ under identical conditions. Therefore, $30\\text{ cm}^3$ produces $60\\text{ cm}^3$."},{"text":"90 cm³ (ratio 1:3).","correct":false,"explanation":"The coefficient 3 belongs to $H_2$, so multiplying the nitrogen volume by three uses the wrong ratio. The $N_2:NH_3$ ratio is 1:2, giving $60\\text{ cm}^3$."},{"text":"30 cm³ (ratio 1:1).","correct":false,"explanation":"A 1:1 ratio ignores the coefficient 2 before $NH_3$. One volume of nitrogen produces two volumes of ammonia, so the correct volume is $60\\text{ cm}^3$."},{"text":"15 cm³ (ratio 2:1).","correct":false,"explanation":"This reverses the $N_2:NH_3$ ratio and halves the nitrogen volume. The equation gives a 1:2 ratio, so $30\\text{ cm}^3$ produces $60\\text{ cm}^3$."}]},{"id":"d5ffef35-9da1-4b35-9d79-a0fb95dea38e","cardType":"cloze","points":null,"front":"In the ideal gas equation $pV = nRT$, the unit for pressure ($p$) must be $\\text{[...]}$ and the unit for volume ($V$) must be $\\text{m}^3$.","back":"In the ideal gas equation $pV = nRT$, the unit for pressure ($p$) must be $\\boxed{\\text{Pa}}$ and the unit for volume ($V$) must be $\\text{m}^3$.","mcqOptions":[{"text":"$$\\text{Pa}$","correct":true,"explanation":"With the standard SI gas constant $R=8.31\\text{ J mol}^{-1}\\text{ K}^{-1}$ and volume in $\\text{m}^3$, pressure must be measured in pascals."},{"text":"$$\\text{kPa}$","correct":false,"explanation":"Kilopascals are a factor of 1000 larger than pascals. With volume in $\\text{m}^3$ and the standard SI value of $R$, pressure must be converted to $\\text{Pa}$."},{"text":"$$\\text{atm}$","correct":false,"explanation":"Atmospheres can only be used with a correspondingly adjusted gas constant. With $R=8.31$ in SI units and volume in $\\text{m}^3$, pressure must be in $\\text{Pa}$."},{"text":"$$\\text{bar}$","correct":false,"explanation":"Bar is a pressure unit, but it is not the SI unit required with $R=8.31$ and volume in $\\text{m}^3$. Convert pressure to $\\text{Pa}$."}]},{"id":"3200c64b-2d4c-42cd-aae3-d06c74dc98d4","cardType":"cloze","points":null,"front":"In the ideal gas equation $pV = nRT$, the unit for pressure ($p$) must be $\\text{Pa}$ and the unit for volume ($V$) must be $\\text{[...]}$.","back":"In the ideal gas equation $pV = nRT$, the unit for pressure ($p$) must be $\\text{Pa}$ and the unit for volume ($V$) must be $\\boxed{\\text{m}^3}$.","mcqOptions":[{"text":"$$\\text{m}^3$","correct":true,"explanation":"The standard SI form uses pressure in pascals and volume in cubic metres, ensuring that $pV$ has units of joules consistent with $R=8.31\\text{ J mol}^{-1}\\text{ K}^{-1}$."},{"text":"$$\\text{dm}^3$","correct":false,"explanation":"Cubic decimetres must be divided by 1000 before substitution into the SI form of the equation. With pressure in pascals, volume must be in $\\text{m}^3$."},{"text":"$$\\text{cm}^3$","correct":false,"explanation":"Cubic centimetres are $10^6$ times smaller than cubic metres, so using them directly causes a large conversion error. With pressure in pascals, use $\\text{m}^3$."},{"text":"$$\\text{L}$","correct":false,"explanation":"Litres are equivalent to cubic decimetres and must be divided by 1000 for the standard SI equation. With pressure in pascals, volume must be in $\\text{m}^3$."}]},{"id":"2bb72749-584e-4f41-8602-517437fb1ea5","cardType":"cloze","points":null,"front":"To convert temperature from Celsius to Kelvin for gas calculations, you use the formula: $T(\\text{K}) = T(^{\\circ}\\text{C}) + \\text{[...]}$.","back":"To convert temperature from Celsius to Kelvin for gas calculations, you use the formula: $T(\\text{K}) = T(^{\\circ}\\text{C}) + \\boxed{273}$.","mcqOptions":[{"text":"273","correct":true,"explanation":"The Kelvin scale begins 273 degrees below the Celsius scale, so adding 273 converts a Celsius temperature to the absolute temperature required in gas calculations."},{"text":"0","correct":false,"explanation":"Adding zero leaves the temperature in degrees Celsius. Gas equations require absolute temperature, found by adding 273 to the Celsius value."},{"text":"100","correct":false,"explanation":"Adding 100 confuses the Celsius-Kelvin conversion with a simple base-ten adjustment. The required approximate offset between the scales is 273."},{"text":"2730","correct":false,"explanation":"This uses an offset ten times too large, probably from an incorrect place-value conversion. Celsius is converted to Kelvin by adding 273."}]},{"id":"6b442f5f-f844-4e72-9bcf-90dd51ebb5c2","cardType":"basic","points":1,"front":"What is the value and unit of the gas constant ($R$) in the ideal gas equation?","back":"8.31 J mol⁻¹ K⁻¹.","mcqOptions":[{"text":"8.31 J mol⁻¹ K⁻¹.","correct":true,"explanation":"This is the gas constant in SI units, appropriate when pressure is in pascals, volume is in cubic metres, and temperature is in kelvin."},{"text":"8.31 kJ mol⁻¹ K⁻¹.","correct":false,"explanation":"Using kilojoules introduces a factor-of-1000 unit error while retaining the numerical value 8.31. The SI gas constant is 8.31 J mol⁻¹ K⁻¹."},{"text":"0.00831 J mol⁻¹ K⁻¹.","correct":false,"explanation":"This value is 1000 times too small, arising from an incorrect conversion between joules and kilojoules. The required value is 8.31 J mol⁻¹ K⁻¹."},{"text":"8.31 J mol⁻¹ °C⁻¹.","correct":false,"explanation":"Ideal gas calculations use absolute temperature, so the temperature unit in the gas constant must be kelvin rather than degrees Celsius. The correct unit is J mol⁻¹ K⁻¹."}]},{"id":"e83a6455-812e-46d2-913f-a2f5c89f8d32","cardType":"basic","points":1,"front":"Convert $150\\text{ kPa}$ into the standard unit for the ideal gas equation.","back":"150,000 Pa.","mcqOptions":[{"text":"150,000 Pa.","correct":true,"explanation":"One kilopascal equals 1000 pascals, so multiplying 150 by 1000 gives 150,000 Pa, the SI pressure needed with the standard gas constant."},{"text":"15,000 Pa.","correct":false,"explanation":"This multiplies by only 100, underestimating the pressure by a factor of ten. Since 1 kPa equals 1000 Pa, the answer is 150,000 Pa."},{"text":"1,500,000 Pa.","correct":false,"explanation":"This multiplies by 10,000 rather than 1000, making the pressure ten times too large. Converting 150 kPa correctly gives 150,000 Pa."},{"text":"150 Pa.","correct":false,"explanation":"Keeping the numerical value unchanged overlooks the kilo- prefix. Each kilopascal contains 1000 pascals, so 150 kPa equals 150,000 Pa."}]},{"id":"0676f008-52db-4030-b674-4aab02bbf63d","cardType":"basic","points":1,"front":"What is the conversion factor to turn $\\text{dm}^3$ into $\\text{m}^3$?","back":"Divide by 1000 (or $\\times 10^{-3}$).","mcqOptions":[{"text":"Divide by 1000 (or $\\times 10^{-3}$).","correct":true,"explanation":"Because 1 dm equals 0.1 m, cubing the conversion gives 1 dm³ = 0.001 m³. Therefore, divide the volume by 1000."},{"text":"Divide by 10 (or $\\times 10^{-1}$).","correct":false,"explanation":"Dividing by ten applies the linear decimetre-to-metre conversion without cubing it. Volume requires cubing 0.1, so divide by 1000 instead."},{"text":"Divide by 100 (or $\\times 10^{-2}$).","correct":false,"explanation":"Dividing by 100 incorrectly squares the linear conversion factor rather than cubing it. Cubic units require division by 1000."},{"text":"Divide by 1,000,000 (or $\\times 10^{-6}$).","correct":false,"explanation":"Division by one million is the conversion from cm³ to m³, not from dm³. A dm³ volume should be divided by 1000."}]},{"id":"59e9afb5-6279-42d3-9489-854f6fdff3ba","cardType":"basic","points":1,"front":"What is the conversion factor to turn $\\text{cm}^3$ into $\\text{m}^3$?","back":"Divide by 1,000,000 (or $\\times 10^{-6}$).","mcqOptions":[{"text":"Divide by 1,000,000 (or $\\times 10^{-6}$).","correct":true,"explanation":"One centimetre is $10^{-2}$ metres, and cubing this gives $10^{-6}$. Therefore, converting cm³ to m³ requires division by 1,000,000."},{"text":"Divide by 100 (or $\\times 10^{-2}$).","correct":false,"explanation":"Dividing by 100 uses the linear centimetre-to-metre conversion without cubing it. For volume, the factor is $100^3$, or 1,000,000."},{"text":"Divide by 1000 (or $\\times 10^{-3}$).","correct":false,"explanation":"Division by 1000 is appropriate for converting dm³ to m³. Since centimetres are smaller, cm³ must be divided by 1,000,000."},{"text":"Divide by 10,000 (or $\\times 10^{-4}$).","correct":false,"explanation":"This squares the linear conversion factor instead of cubing it, leaving the result 100 times too large. The correct divisor is 1,000,000."}]},{"id":"48ed5c3c-9745-419f-971c-74259697a946","cardType":"basic","points":2,"front":"State two assumptions made about the particles in an ideal gas.","back":"

Particles have negligible volume
No intermolecular forces between particles

","mcqOptions":[{"text":"

Particles have negligible volume
No intermolecular forces between particles

","correct":true,"explanation":"The ideal gas model treats particles as point-like, so their own volume is insignificant, and assumes that they exert no attractive or repulsive forces on one another."},{"text":"

Particles have significant volume
No intermolecular forces between particles

","correct":false,"explanation":"Assigning significant volume to particles describes a source of real-gas deviation. Ideal particles have negligible volume and experience no intermolecular forces."},{"text":"

Particles have negligible volume
Strong intermolecular forces between particles

","correct":false,"explanation":"Strong intermolecular forces contradict the ideal gas model and become important in real gases. Ideal particles have negligible volume and no intermolecular forces."},{"text":"

Particles have negligible mass
Balanced intermolecular forces between particles

","correct":false,"explanation":"Ideal particles must possess mass because their motion produces pressure; it is their volume that is treated as negligible. Intermolecular forces are also assumed absent."}]},{"id":"37504276-05b0-470e-b9b4-e8f2e40304f5","cardType":"basic","points":2,"front":"Under what two physical conditions do real gases behave most like an ideal gas?","back":"

Low pressure
High temperature

","mcqOptions":[{"text":"

Low pressure
High temperature

","correct":true,"explanation":"Low pressure keeps particles far apart, reducing the importance of their volume, while high temperature makes intermolecular attractions less significant relative to kinetic energy."},{"text":"

Low pressure
Low temperature

","correct":false,"explanation":"Low temperature makes attractive intermolecular forces more influential and can lead towards condensation. Real gases are most ideal at low pressure and high temperature."},{"text":"

High pressure
High temperature

","correct":false,"explanation":"High pressure forces particles close together, making particle volume and intermolecular interactions significant. Ideal behaviour is favoured by low pressure and high temperature."},{"text":"

High pressure
Low temperature

","correct":false,"explanation":"High pressure and low temperature both strengthen deviations from ideal behaviour by crowding particles and increasing the relative effect of attractions. The ideal conditions are the reverse."}]},{"id":"0e43a8d3-38f5-408c-92f7-78b0b564dee3","cardType":"basic","points":2,"front":"Outline the two-step process to find molar mass ($M$) from gas volume, mass, pressure, and temperature.","back":"

Find moles ($n$) using $n = \\frac{pV}{RT}$
Calculate $M$ using $M = \\frac{m}{n}$

","mcqOptions":[{"text":"

Find moles ($n$) using $n = \\frac{pV}{RT}$
Calculate $M$ using $M = \\frac{m}{n}$

","correct":true,"explanation":"Rearranging $pV=nRT$ gives the amount in moles. Molar mass is then obtained by dividing the measured mass by that amount."},{"text":"

Find moles ($n$) using $n = \\frac{RT}{pV}$
Calculate $M$ using $M = \\frac{m}{n}$

","correct":false,"explanation":"The first expression is the reciprocal of the correct rearrangement and therefore does not calculate moles. Use $n=pV/RT$, then calculate $M=m/n$."},{"text":"

Find moles ($n$) using $n = \\frac{pV}{RT}$
Calculate $M$ using $M = \\frac{n}{m}$

","correct":false,"explanation":"The mole calculation is correct, but dividing moles by mass gives the reciprocal of molar mass. Molar mass must be calculated using $M=m/n$."},{"text":"

Find moles ($n$) using $n = \\frac{pV}{RT}$
Calculate $M$ using $M = m \\times n$

","correct":false,"explanation":"Multiplying mass by moles does not give mass per mole and produces unsuitable units. After finding $n=pV/RT$, divide mass by moles using $M=m/n$."}]},{"id":"9585f0c6-b8a6-4fcd-ac30-1dff6d600875","cardType":"basic","points":1,"front":"Why is a gas syringe often preferred over collecting gas over water?","back":"It prevents gas loss from dissolving in the water.","mcqOptions":[{"text":"It prevents gas loss from dissolving in the water.","correct":true,"explanation":"A gas syringe collects the gas directly, so gases that are soluble in water are not lost by dissolving. This gives a more accurate measured gas volume."},{"text":"It prevents gas loss from reacting with the water.","correct":false,"explanation":"The main concern is dissolution rather than chemical reaction with the water. A gas syringe avoids gas dissolving, so the correct explanation is that it prevents gas loss from dissolving in the water."},{"text":"It prevents gas gain from water evaporating into the gas.","correct":false,"explanation":"Water vapour can affect the measured pressure, but this is not the main reason for preferring a gas syringe. The correct reason is preventing gas loss through dissolution in water."},{"text":"It prevents gas loss from escaping through the bung.","correct":false,"explanation":"A faulty bung can cause leakage with any collection method, but it is not the specific advantage of a gas syringe. Its key benefit is preventing gas from dissolving in water."}]},{"id":"502c0fe5-9ea1-47c3-b818-23a9a5090bdf","cardType":"basic","points":1,"front":"What calculation correction is needed if a gas is collected over water and vapour pressure is known?","back":"Subtract the water vapour pressure from the total pressure ($p_{\\text{gas}} = p_{\\text{total}} - p_{\\text{water}}$).","mcqOptions":[{"text":"Subtract the water vapour pressure from the total pressure ($p_{\\text{gas}} = p_{\\text{total}} - p_{\\text{water}}$).","correct":true,"explanation":"The total pressure is the sum of the gas pressure and water vapour pressure. Therefore, subtracting the water vapour pressure isolates the pressure of the gas being measured."},{"text":"Add the water vapour pressure to the total pressure ($p_{\\text{gas}} = p_{\\text{total}} + p_{\\text{water}}$).","correct":false,"explanation":"Adding the water vapour pressure counts its contribution twice, making the gas pressure too high. The correct correction subtracts water vapour pressure from the total pressure."},{"text":"Subtract the total pressure from the water vapour pressure ($p_{\\text{gas}} = p_{\\text{water}} - p_{\\text{total}}$).","correct":false,"explanation":"Reversing the subtraction gives a negative value because the total pressure is greater than the water vapour pressure. The correct relationship is total pressure minus water vapour pressure."},{"text":"Divide the total pressure by the water vapour pressure ($p_{\\text{gas}} = p_{\\text{total}} / p_{\\text{water}}$).","correct":false,"explanation":"Pressure contributions are combined by subtraction here, not by taking a ratio. The gas pressure is found by removing the water vapour contribution from the total pressure."}]},{"id":"587917f2-58ca-4e72-896b-7a8b07c8148d","cardType":"cloze","points":null,"front":"Use 24.0 dm³ mol⁻¹ for calculations only when the gas is at $\\text{[...]}$; otherwise, use the $\\text{ideal gas equation} ($pV=nRT$).","back":"Use 24.0 dm³ mol⁻¹ for calculations only when the gas is at $\\boxed{\\text{RTP}}$; otherwise, use the $\\text{ideal gas equation} ($pV=nRT$).","mcqOptions":[{"text":"$$\\text{RTP}$","correct":true,"explanation":"The molar gas volume is approximately 24.0 dm³ mol⁻¹ at room temperature and pressure. At other conditions, pressure and temperature must be included using the ideal gas equation."},{"text":"$$\\text{STP}$","correct":false,"explanation":"STP is commonly associated with a molar gas volume of about 22.4 dm³ mol⁻¹, not 24.0 dm³ mol⁻¹. The 24.0 value applies at room temperature and pressure."},{"text":"$$\\text{standard pressure only}$","correct":false,"explanation":"Molar gas volume depends on both temperature and pressure, so specifying pressure alone is insufficient. The value 24.0 dm³ mol⁻¹ is used at room temperature and pressure."},{"text":"$$\\text{boiling point}$","correct":false,"explanation":"At a substance’s boiling point, the temperature is not generally room temperature, so the molar volume will not normally be 24.0 dm³ mol⁻¹. RTP is the required condition."}]},{"id":"1779747c-1fd4-40ab-923d-540f3e538430","cardType":"cloze","points":null,"front":"Use 24.0 dm³ mol⁻¹ for calculations only when the gas is at $\\text{RTP}$; otherwise, use the $\\text{[...]} ($pV=nRT$).","back":"Use 24.0 dm³ mol⁻¹ for calculations only when the gas is at $\\text{RTP}$; otherwise, use the $\\boxed{\\text{ideal gas equation}} ($pV=nRT$).","mcqOptions":[{"text":"$$\\text{ideal gas equation}$","correct":true,"explanation":"When the gas is not at room temperature and pressure, its volume must be calculated using pressure, volume, amount and temperature in $pV=nRT$, rather than a fixed molar volume."},{"text":"$$\\text{molar mass equation}$","correct":false,"explanation":"Molar mass calculations relate mass to amount of substance and do not account for gas pressure and temperature. The required relationship here is the ideal gas equation, $pV=nRT$."},{"text":"$$\\text{concentration equation}$","correct":false,"explanation":"Concentration equations describe amount per volume, usually for solutions, but do not correct gas volume for pressure and temperature. The correct method is the ideal gas equation."},{"text":"$$\\text{thermal decomposition equation}$","correct":false,"explanation":"A thermal decomposition equation describes a reaction caused by heating and cannot calculate gas volume under changing conditions. The appropriate calculation uses the ideal gas equation, $pV=nRT$."}]},{"id":"01dbf652-1675-4345-8520-276716ca0967","cardType":"basic","points":2,"front":"List two practical steps to improve the reliability of gas volume measurements.","back":"

Check apparatus is airtight
Read gas syringe at eye level (avoid parallax)

","mcqOptions":[{"text":"

Check apparatus is airtight
Read gas syringe at eye level (avoid parallax)

","correct":true,"explanation":"An airtight apparatus prevents gas escaping, while reading the syringe at eye level avoids parallax error. Together these steps improve the accuracy and reliability of the measured volume."},{"text":"

Check apparatus is open
Read gas syringe from above (increase parallax)

","correct":false,"explanation":"An open apparatus allows gas to escape, and viewing from above increases parallax error. Reliability is improved by checking the apparatus is airtight and reading at eye level."},{"text":"

Check apparatus is dry
Read gas syringe at any angle (ignore parallax)

","correct":false,"explanation":"Dry apparatus may prevent contamination but does not ensure gas is retained or eliminate reading error. The key steps are making the apparatus airtight and reading at eye level."},{"text":"

Check apparatus is pressurised
Read gas syringe at arm’s length (avoid calibration)

","correct":false,"explanation":"Pressurising the apparatus can alter the gas conditions, and distance does not remove parallax or calibration errors. Reliable measurements require an airtight setup and an eye-level reading."}]},{"id":"940896bd-569d-497a-8ed6-3014bf7a62f0","cardType":"cloze","points":null,"front":"An amount of $0.05\\text{ mol}$ of gas at RTP should have a volume of approximately $\\text{[...]} $\\text{dm}^3$.","back":"An amount of $0.05\\text{ mol}$ of gas at RTP should have a volume of approximately $\\boxed{1.2} $\\text{dm}^3$.","mcqOptions":[{"text":"$$1.2$","correct":true,"explanation":"At RTP, one mole of gas occupies approximately 24.0 dm³. Therefore, $0.05 \\times 24.0 = 1.2$ dm³."},{"text":"$$0.48$","correct":false,"explanation":"This results from multiplying by 9.6 rather than using the RTP molar volume. The correct calculation is $0.05 \\times 24.0$, giving 1.2 dm³."},{"text":"$$2.4$","correct":false,"explanation":"This value commonly comes from using 0.1 mol instead of 0.05 mol, doubling the amount of gas. Using 24.0 dm³ mol⁻¹ gives a volume of 1.2 dm³."},{"text":"$$12.0$","correct":false,"explanation":"This is ten times too large, usually from mishandling the decimal in 0.05 mol. Multiplying 0.05 by the RTP molar volume of 24.0 gives 1.2 dm³."}]},{"id":"f23487c2-2ac2-4348-a60d-a772d44c7b0c","cardType":"basic","points":1,"front":"What temperature in Kelvin corresponds to standard room temperature (25 °C)?","back":"298 K.","mcqOptions":[{"text":"298 K.","correct":true,"explanation":"Convert Celsius to Kelvin by adding 273.15: $25 + 273.15 = 298.15$ K, which rounds to 298 K at the stated precision."},{"text":"248 K.","correct":false,"explanation":"This comes from subtracting 25 from 273 rather than adding it. Kelvin temperature is found by adding 273.15 to the Celsius temperature, giving approximately 298 K."},{"text":"273 K.","correct":false,"explanation":"273 K corresponds approximately to 0 °C, not 25 °C. Adding 273.15 to 25 °C gives 298.15 K, rounded to 298 K."},{"text":"325 K.","correct":false,"explanation":"This incorrectly adds 300 instead of the Celsius-to-Kelvin conversion value of 273.15. The correct conversion is $25 + 273.15 = 298.15$ K, approximately 298 K."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"Gas volumes and ideal gas","title":"Gas volumes and ideal gas Flashcards","topbar":{"backHref":"/a-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"a12a0b33-0819-5d4b-bf4b-207bc3633346","walled":"$L39"}],"$L3a","$L3b"]