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273 instead of 273.15 is a rounded approximation, not the exact Celsius-to-kelvin conversion. The precise conversion requires adding 273.15."},{"text":"Subtract 273.15","correct":false,"explanation":"Subtracting 273.15 reverses the conversion and would move from kelvin to Celsius. Celsius temperature must have 273.15 added to give kelvin."},{"text":"Add 274.15","correct":false,"explanation":"The offset between the Celsius and kelvin scales is 273.15, so adding 274.15 introduces an extra kelvin and gives an incorrect absolute temperature."}]},{"id":"0f5aab07-fe4f-40c3-8d68-8edc4dce423e","cardType":"cloze","points":null,"front":"The number of molecules $N$ in a sample is calculated by $N = \\text{[...]} where $n$ is the amount in moles.","back":"The number of molecules $N$ in a sample is calculated by $N = \\boxed{nN_A} where $n$ is the amount in moles.","mcqOptions":[{"text":"$$nN_A$","correct":true,"explanation":"Each mole contains $N_A$ molecules, so multiplying the amount of substance $n$ by Avogadro’s constant gives the total number of molecules: $N=nN_A$."},{"text":"$$nR$","correct":false,"explanation":"The gas constant $R$ is not the number of molecules per mole. The required conversion from moles to molecules uses Avogadro’s constant, giving $N=nN_A$."},{"text":"$$rac{N_A}{n}$","correct":false,"explanation":"Dividing by Avogadro’s constant reverses the required conversion and produces a quantity far smaller than the number of molecules. The correct relationship is $N=nN_A$."},{"text":"$$rac{n}{N_A}$","correct":false,"explanation":"Dividing the amount by Avogadro’s constant has the wrong relationship and units for counting molecules. Multiplication is required: $N=nN_A$."}]},{"id":"4c5dc07f-9113-4e36-8351-2684febcbbf7","cardType":"cloze","points":null,"front":"The number of molecules $N$ in a sample is calculated by $N = nN_A where $n$ is the \\text{[...]}.","back":"The number of molecules $N$ in a sample is calculated by $N = nN_A where $n$ is the \\boxed{amount in moles}.","mcqOptions":[{"text":"amount in moles","correct":true,"explanation":"The symbol $n$ denotes the amount of substance, measured in moles. Multiplying it by Avogadro’s constant gives the number of molecules."},{"text":"number of molecules","correct":false,"explanation":"The number of molecules is represented by $N$, not $n$. The symbol $n$ is the amount in moles, which is then multiplied by $N_A$."},{"text":"mass in kilograms","correct":false,"explanation":"Mass in kilograms is a different physical quantity and is not what $n$ represents in this equation. Here, $n$ is the amount in moles."},{"text":"volume in cubic metres","correct":false,"explanation":"Volume in cubic metres describes the space occupied by a sample, not its amount of substance. In $N=nN_A$, $n$ means amount in moles."}]},{"id":"faddd1e9-3c77-40d9-b39b-116d7dbcdcb5","cardType":"basic","points":1,"front":"How is the molar mass ($M$) in $\\mathrm{kg\\,mol^{-1}}$ calculated from the relative molecular mass ($M_r$)?","back":"$$$M = \\frac{M_r}{1000}$$","mcqOptions":[{"text":"$$$M = \\frac{M_r}{1000}$$","correct":true,"explanation":"Relative molecular mass is compared with one-twelfth of a carbon-12 atom’s mass and has no units. Dividing by 1000 gives the molar mass in kilograms per mole."},{"text":"$$$M = 1000M_r$$","correct":false,"explanation":"Multiplying by 1000 converts a numerical value in kilograms per mole into grams per mole, rather than converting relative molecular mass to kilograms per mole. The required operation is division by 1000."},{"text":"$$$M = \\frac{M_r}{100}$$","correct":false,"explanation":"Dividing by 100 gives the wrong unit conversion factor. Since $1\\ \\mathrm{kg}=1000\\ \\mathrm{g}$, the numerical molar mass in kilograms per mole is $M_r/1000$."},{"text":"$$$M = \\frac{M_r}{10000}$$","correct":false,"explanation":"Dividing by 10 000 makes the molar mass ten times too small compared with the required grams-to-kilograms conversion. The correct divisor is 1000."}]},{"id":"30bc30f0-2049-46f9-9c11-bfb4f5e1b539","cardType":"cloze","points":null,"front":"The Boltzmann constant $k$ is defined by the ratio $\\text{[...]}$.","back":"The Boltzmann constant $k$ is defined by the ratio $\\boxed{\\frac{R}{N_A}}$.","mcqOptions":[{"text":"$$\\frac{R}{N_A}$","correct":true,"explanation":"The molar gas constant applies per mole, whereas the Boltzmann constant applies per particle. Dividing $R$ by Avogadro’s constant converts between these scales: $k=R/N_A$."},{"text":"$$\\frac{N_A}{R}$","correct":false,"explanation":"This is the reciprocal relationship and would have incompatible physical meaning for the Boltzmann constant. The correct ratio is the molar gas constant divided by Avogadro’s constant."},{"text":"$$\\frac{R}{1000}$","correct":false,"explanation":"The factor 1000 is used in some mass-unit conversions, not in the relationship between the gas constant and Avogadro’s constant. The definition is $k=R/N_A$."},{"text":"$$RN_A$","correct":false,"explanation":"Multiplying the constants does not convert a per-mole quantity into a per-particle quantity. Division by $N_A$ is required, so $k=R/N_A$."}]},{"id":"c4b70165-cd67-4da3-8d58-a937b25a828f","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":"At low pressure, molecules are far apart so intermolecular forces are negligible, and at high temperature their kinetic energy makes those forces less significant. These conditions favour ideal behaviour."},{"text":"

High pressure
High temperature

","correct":false,"explanation":"High pressure brings molecules closer together, making their finite volume and intermolecular forces more important. Ideal gases are favoured by low pressure and high temperature."},{"text":"

Low pressure
Low temperature

","correct":false,"explanation":"Low temperature reduces molecular kinetic energy, so attractive intermolecular forces have a greater effect. Real gases most closely approach ideal behaviour at low pressure and high temperature."},{"text":"

High pressure
Low temperature

","correct":false,"explanation":"High pressure and low temperature both increase the importance of molecular size and intermolecular attractions. The ideal-gas approximation is best at low pressure and high temperature."}]},{"id":"0def9542-a0e0-4ca2-a288-927ddbf9aea2","cardType":"basic","points":1,"front":"What is the gradient of a graph of pressure ($p$) against the reciprocal of volume ($\\frac{1}{V}$) for $n$ moles of an ideal gas?","back":"$$nRT$","mcqOptions":[{"text":"$$nRT$","correct":true,"explanation":"From $pV=nRT$, rearranging gives $p=nRT(1/V)$. Therefore, a graph of $p$ against $1/V$ has gradient $nRT$."},{"text":"$$\\frac{RT}{n}$","correct":false,"explanation":"$$RT/n$ reverses the dependence on the amount of gas. Rearranging $pV=nRT$ shows that the coefficient of $1/V$ is $nRT$."},{"text":"$$\\frac{nR}{T}$","correct":false,"explanation":"Although $R$ and $T$ appear in the equation, dividing by temperature is not justified. Writing $p=nRT(1/V)$ gives the gradient $nRT$."},{"text":"$$\\frac{nT}{R}$","correct":false,"explanation":"Dividing by the gas constant does not follow from the ideal gas equation. The coefficient multiplying $1/V$ after rearrangement is the product $nRT$."}]},{"id":"30611bfd-c110-4a72-9aab-208c5dc0b069","cardType":"basic","points":2,"front":"Why might the ideal gas equation $pV = \\text{constant}$ fail if a gas is compressed very quickly?","back":"The process is not isothermal (temperature rises during rapid compression).","mcqOptions":[{"text":"The process is not isothermal (temperature rises during rapid compression).","correct":true,"explanation":"Rapid compression leaves little time for thermal energy to transfer to the surroundings, so the temperature rises. The process is therefore not isothermal, and $pV$ does not remain constant."},{"text":"The process is isothermal (temperature remains constant during rapid compression).","correct":false,"explanation":"Rapid compression does not remain isothermal because the gas temperature rises when work is done on it faster than heat can escape. Thus the ideal relation with constant $pV$ fails."},{"text":"The process is not isothermal (temperature falls during rapid compression).","correct":false,"explanation":"Compression rapidly transfers work to the gas, increasing its internal energy and temperature rather than lowering it. The process is not isothermal because temperature rises."},{"text":"The process is not isothermal (pressure falls during rapid compression).","correct":false,"explanation":"The pressure generally increases during compression, but that is not the reason the constant-$pV$ relation fails. The key issue is that rapid compression makes the temperature rise."}]},{"id":"1047616e-d128-4778-b4d7-9c717cd79d53","cardType":"basic","points":1,"front":"What is the kinetic theory assumption regarding the volume of gas molecules?","back":"Molecular volume is negligible compared to the volume of the container.","mcqOptions":[{"text":"Molecular volume is negligible compared to the volume of the container.","correct":true,"explanation":"Treating molecules as point particles allows their total volume to be ignored relative to the container, so the gas volume is effectively the container’s volume."},{"text":"Molecular volume is equal to the volume of the container.","correct":false,"explanation":"Equating molecular volume with container volume treats the gas as a continuous solid. The molecular volume is negligible compared with the container’s volume."},{"text":"Molecular volume is greater than the volume of the container.","correct":false,"explanation":"A larger molecular volume would leave no space for random molecular motion. Kinetic theory assumes molecular volume is negligible compared with the container’s volume."},{"text":"Molecular volume is negligible compared to the speed of the molecules.","correct":false,"explanation":"Comparing molecular volume with molecular speed confuses a volume with a rate of motion. Molecular volume is negligible relative to the volume of the container."}]},{"id":"1aa8a540-3387-4295-9a4f-66a5d1e9e990","cardType":"basic","points":1,"front":"What is the kinetic theory assumption regarding intermolecular forces?","back":"There are no intermolecular forces except during collisions.","mcqOptions":[{"text":"There are no intermolecular forces except during collisions.","correct":true,"explanation":"Ideal-gas molecules exert no attractive or repulsive forces on one another while separated; forces act only briefly when molecules collide."},{"text":"There are attractive intermolecular forces except during collisions.","correct":false,"explanation":"Assuming continuous intermolecular attraction describes a real-gas effect rather than the ideal kinetic model. The model neglects intermolecular forces except during collisions."},{"text":"There are repulsive intermolecular forces except during collisions.","correct":false,"explanation":"Permanent repulsive forces would continually alter molecular trajectories and potential energy. Kinetic theory assumes no intermolecular forces act except during collisions."},{"text":"There are intermolecular forces only between collisions.","correct":false,"explanation":"Restricting forces to the periods between collisions reverses the assumption. Intermolecular forces are neglected between collisions and act only during collisions."}]},{"id":"ff5fda0c-5a3d-4df1-a695-3891971fb916","cardType":"basic","points":1,"front":"What is the kinetic theory assumption regarding the nature of collisions?","back":"Collisions are perfectly elastic (kinetic energy is conserved).","mcqOptions":[{"text":"Collisions are perfectly elastic (kinetic energy is conserved).","correct":true,"explanation":"In a perfectly elastic collision, total kinetic energy and momentum are conserved, so collisions do not produce a net loss of molecular kinetic energy."},{"text":"Collisions are perfectly inelastic (kinetic energy is not conserved).","correct":false,"explanation":"Perfectly inelastic collisions would make molecules stick together and lose kinetic energy. Kinetic theory instead assumes perfectly elastic collisions in which kinetic energy is conserved."},{"text":"Collisions are partially elastic (kinetic energy is partly conserved).","correct":false,"explanation":"Partially elastic collisions would progressively reduce the gas molecules’ kinetic energy. The ideal model assumes collisions are perfectly elastic, conserving kinetic energy."},{"text":"Collisions are perfectly explosive (kinetic energy is increased).","correct":false,"explanation":"Explosive collisions would increase kinetic energy by releasing internal energy, which is not assumed. Molecular collisions are perfectly elastic, so kinetic energy is conserved."}]},{"id":"4d76bcaf-9baa-402f-a521-a74b7e0cf589","cardType":"basic","points":1,"front":"What is the kinetic theory assumption regarding the duration of collisions?","back":"Collision time is negligible compared to the time between collisions.","mcqOptions":[{"text":"Collision time is negligible compared to the time between collisions.","correct":true,"explanation":"Collisions are treated as effectively instantaneous because their duration is extremely short compared with the interval of free motion between successive collisions."},{"text":"Collision time is equal compared to the time between collisions.","correct":false,"explanation":"Equal collision and free-motion times would mean molecules spend a substantial period interacting. Kinetic theory assumes collision time is negligible compared with time between collisions."},{"text":"Collision time is longer compared to the time between collisions.","correct":false,"explanation":"A collision time longer than the interval between collisions would invalidate the model of brief, separate interactions. Collision time is negligible compared with time between collisions."},{"text":"Collision time is negligible compared to the volume of the container.","correct":false,"explanation":"Comparing collision time with container volume mixes quantities with different dimensions. Collision time is negligible relative to the time between successive collisions."}]},{"id":"1214c401-bbf2-44cd-95b9-8413082ac07b","cardType":"cloze","points":null,"front":"The average used in kinetic theory is the [ ], denoted by the symbol $\\overline{c^2}$.","back":"The average used in kinetic theory is the mean square speed, denoted by the symbol $\\overline{c^2}$.","mcqOptions":[{"text":"mean square speed","correct":true,"explanation":"Mean square speed is found by squaring each molecule’s speed and then averaging, giving the quantity represented by $\\overline{c^2}$."},{"text":"mean speed","correct":false,"explanation":"Mean speed averages the speeds before any squaring and is represented by $\\overline{c}$. The quantity $\\overline{c^2}$ denotes mean square speed."},{"text":"root mean square speed","correct":false,"explanation":"Root mean square speed is $\\sqrt{\\overline{c^2}}$, obtained after taking the square root. The unsquared quantity $\\overline{c^2}$ is mean square speed."},{"text":"most probable speed","correct":false,"explanation":"Most probable speed is the peak of the molecular speed distribution, not the average of squared speeds. The symbol $\\overline{c^2}$ denotes mean square speed."}]},{"id":"a544050f-2919-42fd-b1c1-6f7892d1f1f9","cardType":"cloze","points":null,"front":"The average used in kinetic theory is the mean square speed, denoted by the symbol [ ].","back":"The average used in kinetic theory is the mean square speed, denoted by the symbol $\\overline{c^2}$.","mcqOptions":[{"text":"$$\\overline{c^2}$","correct":true,"explanation":"The bar indicates an average and encloses $c^2$, showing that each molecular speed is squared before the values are averaged."},{"text":"$$\\overline{c}$","correct":false,"explanation":"The symbol $\\overline{c}$ denotes mean speed, without squaring individual speeds. Mean square speed is written $\\overline{c^2}$."},{"text":"$$\\left(\\overline{c}\\right)^2$","correct":false,"explanation":"This squares the mean speed rather than averaging the squares; these differ because molecular speeds vary. Mean square speed is denoted by $\\overline{c^2}$."},{"text":"$$\\sqrt{\\overline{c^2}}$","correct":false,"explanation":"This is root mean square speed, whose units are speed rather than speed squared. Mean square speed itself is denoted by $\\overline{c^2}$."}]},{"id":"ca0f57b0-6442-471e-866a-affae664e07f","cardType":"basic","points":2,"front":"How do gas molecules exert pressure on the walls of a container?","back":"

Molecules collide with walls and undergo a change in momentum.
The wall exerts a force on the molecule; by Newton's third law, the molecule exerts an equal force on the wall.

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

Molecules collide with walls and undergo a change in momentum.
The wall exerts a force on the molecule; by Newton's third law, the molecule exerts an equal force on the wall.

","correct":true,"explanation":"A molecule’s momentum changes when it rebounds, so the wall exerts a force on it. Newton's third law gives an equal and opposite force on the wall."},{"text":"

Molecules collide with walls and undergo no change in momentum.
The wall exerts a force on the molecule; by Newton's third law, the molecule exerts an equal force on the wall.

","correct":false,"explanation":"Elastic wall collisions conserve kinetic energy but reverse the perpendicular momentum component, so momentum changes. Newton's third law then gives the force exerted on the wall."},{"text":"

Molecules collide with walls and undergo a change in momentum.
The wall exerts a force on the molecule; by Newton's second law, the molecule exerts an equal force on the wall.

","correct":false,"explanation":"Newton's second law links force to rate of change of momentum, but the equal force on the wall follows specifically from Newton's third law."},{"text":"

Molecules collide with walls and undergo a change in mass.
The wall exerts a force on the molecule; by Newton's first law, the molecule exerts an unequal force on the wall.

","correct":false,"explanation":"A rebound changes momentum rather than molecular mass, and action-reaction forces are equal, not unequal. Newton's third law therefore explains the force on the wall."}]},{"id":"8a375e7a-2456-4147-9539-c965e63e1830","cardType":"cloze","points":null,"front":"The pressure of a gas with density $\\rho$ is given by $p = \\text{[...]}$.","back":"The pressure of a gas with density $\\rho$ is given by $p = \\boxed{\\frac{1}{3}\\rho \\overline{c^2}}$.","mcqOptions":[{"text":"$$\\frac{1}{3}\\rho \\overline{c^2}$","correct":true,"explanation":"Kinetic theory relates pressure to density and mean square speed by $p=\\frac{1}{3}\\rho\\overline{c^2}$, with the factor $\\frac{1}{3}$ arising from three-dimensional random motion."},{"text":"$$\\frac{1}{2}\\rho \\overline{c^2}$","correct":false,"explanation":"Using $\\frac{1}{2}$ confuses the pressure equation with kinetic energy per unit volume. The correct kinetic-theory relation is $p=\\frac{1}{3}\\rho\\overline{c^2}$."},{"text":"$$\\rho \\overline{c^2}$","correct":false,"explanation":"Omitting the factor $\\frac{1}{3}$ ignores that only one of three perpendicular velocity components contributes to pressure on a particular wall. The correct expression includes $\\frac{1}{3}$."},{"text":"$$3\\rho \\overline{c^2}$","correct":false,"explanation":"Multiplying by $3$ reverses the three-dimensional averaging factor and makes the pressure nine times too large. The required coefficient is $\\frac{1}{3}$."}]},{"id":"2a90b0a0-06c5-42d6-a3e9-d5d6a610937f","cardType":"basic","points":2,"front":"In the equation $p = \\frac{1}{3}\\rho \\overline{c^2}$, why is the factor of $\\frac{1}{3}$ necessary?","back":"Motion is random in three dimensions; on average, only 1/3 of the mean square speed acts in any one direction.","mcqOptions":[{"text":"Motion is random in three dimensions; on average, only 1/3 of the mean square speed acts in any one direction.","correct":true,"explanation":"Random molecular motion has equal mean square velocity components in three perpendicular directions, so the component normal to any particular wall contributes one-third of the total mean square speed."},{"text":"Collisions are perfectly elastic at the walls; on average, molecules retain only 1/3 of their mean square speed after each collision.","correct":false,"explanation":"The factor does not arise because collisions reduce molecular speed. Elastic collisions conserve kinetic energy, while the correct factor comes from sharing mean square speed equally among three dimensions."},{"text":"Pressure acts over a two-dimensional surface; on average, only 1/3 of the container area experiences molecular collisions.","correct":false,"explanation":"Although pressure is measured over a surface, dividing by three is not an area effect. It arises because random motion has three equal velocity components, only one of which is normal to a wall."},{"text":"Momentum is reversed during each collision; on average, only 1/3 of a molecule’s momentum is transferred to the wall.","correct":false,"explanation":"A molecule’s momentum change at a wall involves a factor of two, not one-third. The one-third factor instead comes from resolving random molecular motion into three equal spatial components."}]},{"id":"5351f743-b34d-411e-a7dc-69daf601618f","cardType":"basic","points":1,"front":"Define root mean square speed ($c_{\\mathrm{rms}}$) in terms of mean square speed ($\\overline{c^2}$).","back":"$$$c_{\\mathrm{rms}} = \\sqrt{\\overline{c^2}}$$","mcqOptions":[{"text":"$$$c_{\\mathrm{rms}} = \\sqrt{\\overline{c^2}}$$","correct":true,"explanation":"Root mean square speed is obtained by taking the square root of the mean of the squared molecular speeds, producing a quantity with the dimensions of speed."},{"text":"$$$c_{\\mathrm{rms}} = \\left(\\overline{c}\\right)^2$$","correct":false,"explanation":"This squares the mean speed rather than taking the root of the mean square. The correct definition is $c_{\\mathrm{rms}}=\\sqrt{\\overline{c^2}}$."},{"text":"$$$c_{\\mathrm{rms}} = \\sqrt{\\left(\\overline{c}\\right)^2}$$","correct":false,"explanation":"The mean speed is generally not equal to the root mean square speed because averaging and squaring do not commute. The correct expression is $c_{\\mathrm{rms}}=\\sqrt{\\overline{c^2}}$."},{"text":"$$$c_{\\mathrm{rms}} = \\overline{c^2}$$","correct":false,"explanation":"The mean square speed has units of speed squared, so it cannot itself equal a speed. Taking its square root gives $c_{\\mathrm{rms}}=\\sqrt{\\overline{c^2}}$."}]},{"id":"9548120e-511a-4a1d-9c1c-e1492874272a","cardType":"cloze","points":null,"front":"Absolute temperature $T$ is directly proportional to the [ ] of the gas molecules.","back":"Absolute temperature $T$ is directly proportional to the mean translational kinetic energy of the gas molecules.","mcqOptions":[{"text":"mean translational kinetic energy","correct":true,"explanation":"For an ideal gas, the average translational kinetic energy of one molecule is $\\frac{3}{2}kT$, so it is directly proportional to absolute temperature."},{"text":"mean translational momentum","correct":false,"explanation":"Mean molecular momentum is zero in a stationary gas because molecular directions are random. Absolute temperature is instead proportional to the mean translational kinetic energy."},{"text":"mean translational speed","correct":false,"explanation":"Mean speed increases in proportion to $\\sqrt{T}$, not directly with $T$. Absolute temperature is directly proportional to mean translational kinetic energy, which depends on mean square speed."},{"text":"mean intermolecular potential energy","correct":false,"explanation":"Individual molecular potential energy is not what defines ideal-gas temperature, since intermolecular forces are neglected. Temperature is directly proportional to mean translational kinetic energy."}]},{"id":"c1c77f2c-a6ed-4720-a20c-d52f812e09ef","cardType":"basic","points":1,"front":"State the formula for the mean translational kinetic energy ($E_{\\mathrm{k,mean}}$) of a single gas molecule.","back":"$$$\\frac{3}{2}kT$$","mcqOptions":[{"text":"$$$\\frac{3}{2}kT$$","correct":true,"explanation":"A molecule has three translational degrees of freedom, each contributing $\\frac{1}{2}kT$ by equipartition, giving a mean translational kinetic energy of $\\frac{3}{2}kT$."},{"text":"$$$\\frac{1}{2}kT$$","correct":false,"explanation":"This includes only one translational degree of freedom. A molecule moves independently in three dimensions, so equipartition gives $3\\times\\frac{1}{2}kT=\\frac{3}{2}kT$."},{"text":"$$$\\frac{3}{2}RT$$","correct":false,"explanation":"This uses the molar gas constant $R$, so it describes energy on a per-mole basis. For a single molecule, the required constant is $k$, giving $\\frac{3}{2}kT$."},{"text":"$$$3kT$$","correct":false,"explanation":"This incorrectly assigns $kT$ to each translational degree of freedom. Equipartition assigns $\\frac{1}{2}kT$ to each of three degrees, giving $\\frac{3}{2}kT$."}]},{"id":"c7f1946e-80c5-48f2-abb8-167d5b8aea0b","cardType":"basic","points":1,"front":"State the formula for the total translational kinetic energy of one mole of a monatomic ideal gas.","back":"$$$\\frac{3}{2}RT$$","mcqOptions":[{"text":"$$$\\frac{3}{2}RT$$","correct":true,"explanation":"One mole contains Avogadro’s number of molecules, so multiplying the single-molecule energy $\\frac{3}{2}kT$ by $N_{\\mathrm{A}}$ and using $R=N_{\\mathrm{A}}k$ gives $\\frac{3}{2}RT$."},{"text":"$$$\\frac{3}{2}kT$$","correct":false,"explanation":"This is the mean translational kinetic energy of one molecule, not one mole. Multiplication by Avogadro’s constant replaces $k$ with $R$, giving $\\frac{3}{2}RT$."},{"text":"$$$\\frac{1}{2}RT$$","correct":false,"explanation":"This counts only one of the three translational degrees of freedom. Each contributes $\\frac{1}{2}RT$ per mole, so the total is $\\frac{3}{2}RT$."},{"text":"$$$\\frac{5}{2}RT$$","correct":false,"explanation":"The factor $\\frac{5}{2}$ includes two rotational degrees of freedom and applies to the internal energy of a diatomic ideal gas under ordinary conditions. A monatomic gas has only three translational degrees."}]},{"id":"f6b04e07-69b8-45e0-bb23-6a26b7425936","cardType":"cloze","points":null,"front":"To find the number of moles $n$ from the total mass $m_{\\text{total}}$, use the formula $n = \\text{[...]}$ where $M$ is the molar mass.","back":"To find the number of moles $n$ from the total mass $m_{\\text{total}}$, use the formula $n = \\boxed{\\frac{m_{\\text{total}}}{M}}$ where $M$ is the molar mass.","mcqOptions":[{"text":"$$\\frac{m_{\\text{total}}}{M}$","correct":true,"explanation":"Dividing the sample’s total mass by its molar mass gives the number of moles, provided both masses use consistent units."},{"text":"$$m_{\\text{total}}M$","correct":false,"explanation":"Multiplying total mass by molar mass gives incorrect dimensions and does not yield an amount in moles. The correct relationship is $n=\\frac{m_{\\text{total}}}{M}$."},{"text":"$$\\frac{M}{m_{\\text{total}}}$","correct":false,"explanation":"This reverses the required ratio, producing reciprocal moles rather than moles. Since molar mass is mass per mole, the correct expression is $\\frac{m_{\\text{total}}}{M}$."},{"text":"$$\\frac{m_{\\text{total}}}{M^2}$","correct":false,"explanation":"Squaring the molar mass makes the dimensions incorrect. Only one factor of molar mass is needed, so the number of moles is $\\frac{m_{\\text{total}}}{M}$."}]},{"id":"7dc61a38-d236-4df6-9394-f4bb18c1dee3","cardType":"cloze","points":null,"front":"To find the number of moles $n$ from the total mass $m_{\\text{total}}$, use the formula $n = \\frac{m_{\\text{total}}}{M}$ where $M$ is the [ ].","back":"To find the number of moles $n$ from the total mass $m_{\\text{total}}$, use the formula $n = \\frac{m_{\\text{total}}}{M}$ where $M$ is the molar mass.","mcqOptions":[{"text":"molar mass","correct":true,"explanation":"Molar mass is the mass per mole of a substance, so dividing total mass by $M$ gives the number of moles present."},{"text":"molecular mass","correct":false,"explanation":"Molecular mass refers to the mass of an individual molecule rather than the mass per mole used in this equation. Here, $M$ represents molar mass."},{"text":"total mass","correct":false,"explanation":"Total mass is the sample mass represented by $m_{\\text{total}}$, not by $M$. The denominator $M$ is the substance’s molar mass."},{"text":"relative atomic mass","correct":false,"explanation":"Relative atomic mass is dimensionless and applies directly to an element’s atoms, whereas this formula requires the mass per mole of the substance, called molar mass."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Component 1"},"heading":"Kinetic Theory","title":"Kinetic Theory Flashcards","topbar":{"backHref":"/a-level/physics/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"2eb88c4f-32b0-5a95-be03-526a760b6f9d","walled":"$L39"}],"$L3a","$L3b"]