NCERT Solutions · Class 11 Physics Kinetic Theory

Only a tiny fraction of the space a gas fills is actually taken up by matter — almost all of it is empty, which is why gases compress so easily compared with liquids or solids.The quantity wanted is a ratio: (total volume the molecules themselves would occupy, packed with no gaps) ÷ (the volume the gas actually fills).Step $\displaystyle 1$: The volume the gas actually occupies.By Avogadro's law, one mole of any gas at STP occupies the same molar volume, \[V_{\text{actual}} = 22.4\ \text{L} = 22400\ \text{cm}^3 = 2.24\times10^{-2}\ \text{m}^3 \] and contains \(\displaystyle N_A = 6.022\times10^{23}\) molecules (the Avogadro constant).Step $\displaystyle 2$: The volume the molecules themselves take up.Model each oxygen molecule as a sphere of diameter \(\displaystyle d = 3\ \text{Å} = 3\times10^{-10}\ \text{m}\), so its radius is \[r = \frac{d}{2} = 1.5\times10^{-10}\ \text{m} \] (The formula for a sphere's volume needs the radius, not the diameter — using \(\displaystyle d\) in place of \(\displaystyle r\) here overstates the volume by a factor of 8.)Volume of one molecule: \[v = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi\left(1.5\times10^{-10}\ \text{m}\right)^3 \] \[r^3 = (1.5)^3\times10^{-30}\ \text{m}^3 = 3.375\times10^{-30}\ \text{m}^3 \] \[v = 4.19\times3.375\times10^{-30}\ \text{m}^3 = 1.41\times10^{-29}\ \text{m}^3 \]One mole contains \(\displaystyle N_A\) molecules, so the volume their matter would occupy is \[V_{\text{molecular}} = N_A\,v = \left(6.022\times10^{23}\right)\left(1.41\times10^{-29}\ \text{m}^3\right) = 8.51\times10^{-6}\ \text{m}^3 \]Step $\displaystyle 3$: Take the ratio.\[\frac{V_{\text{molecular}}}{V_{\text{actual}}} = \frac{8.51\times10^{-6}\ \text{m}^3}{2.24\times10^{-2}\ \text{m}^3} = 3.80\times10^{-4} \]Both volumes are expressed in \(\displaystyle \text{m}^3\), so the units cancel and the ratio is a pure number — it is not itself a volume.The diameter was given to two significant figures ($\displaystyle 3.0$ Å), so the ratio is kept to two significant figures: \(\displaystyle 3.8\times10^{-4}\).This means the molecules themselves occupy only about $\displaystyle 0.038$% of the space the gas fills — the remaining $\displaystyle 99.96$% is empty space between molecules. That near-total emptiness is exactly why a gas can be compressed so much more readily than a liquid or solid, where the molecules already sit close enough to touch.Answer: The molecular volume is about \(\displaystyle 3.8\times10^{-4}\) times the actual volume occupied by the gas (roughly $\displaystyle 0.04$% of it) — almost all the volume of the gas is empty space.

The ideal gas law, applied to exactly one mole at the two standard reference values of pressure and temperature, gives the molar volume directly — no measured data needed.For \(\displaystyle n \) moles of an ideal gas, the equation of state is\[PV = nRT \]where \(\displaystyle P \) is the pressure, \(\displaystyle V \) is the volume, \(\displaystyle n \) is the number of moles, \(\displaystyle T \) is the absolute temperature, and \(\displaystyle R \) is the universal gas constant. Since this holds for any ideal gas regardless of its identity, the volume of $\displaystyle 1$ mole depends only on \(\displaystyle P \) and \(\displaystyle T \), not on which gas it is — that is exactly why the question calls it "any (ideal) gas."Setting the standard values.STP (standard temperature and pressure) fixes: \[T = 0\,^\circ\text{C} = 273\ \text{K}, \qquad P = 1\ \text{atm} = 1.013 \times 10^{5}\ \text{Pa} \]A common slip here is quoting pressure as a gauge reading; STP means the absolute pressure equals $\displaystyle 1$ atmosphere, not $\displaystyle 1$ atmosphere above ambient.Take \(\displaystyle n = 1 \) mol, and \(\displaystyle R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} \) (the value of \(\displaystyle R \) is itself fixed by experiment — this is what makes the result a fact about gases and not just algebra).Solving for \(\displaystyle V \).\[V = \frac{nRT}{P} \]Substituting:\[V = \frac{(1\ \text{mol})(8.314\ \text{J mol}^{-1}\text{K}^{-1})(273\ \text{K})}{1.013 \times 10^{5}\ \text{Pa}} \]The numerator: \[8.314 \times 273 = 2269.7\ \text{J} \]So: \[V = \frac{2269.7\ \text{J}}{1.013 \times 10^{5}\ \text{Pa}} = 2.2406 \times 10^{-2}\ \text{m}^3 \](Pa = J/m³, so J/Pa comes out in m³, as it must for a volume.)Converting to litres.Since \(\displaystyle 1\ \text{m}^3 = 10^{3}\ \text{litres} \):\[V = 2.2406 \times 10^{-2} \times 10^{3}\ \text{L} = 22.406\ \text{L} \]The input data (\(\displaystyle R\) to $\displaystyle 4$ significant figures, \(\displaystyle T = 273\) to $\displaystyle 3$, \(\displaystyle P\) to $\displaystyle 4$) supports at most $\displaystyle 3$ significant figures in the result, so the mid-calculation digits are not the final word — rounding once at the end:\[V \approx 22.4\ \text{L} \]This is exactly the accepted molar volume of an ideal gas at STP, confirming the stated value of $\displaystyle 22.4$ litres.Answer: At STP (\(\displaystyle T = 273\ \text{K}\), \(\displaystyle P = 1.013\times10^{5}\ \text{Pa}\)), \(\displaystyle V = \dfrac{RT}{P} \approx 2.24 \times 10^{-2}\ \text{m}^3 = 22.4\ \text{litres}\) per mole of any ideal gas.

The point where both curves meet on the \(\displaystyle y\)-axis is where the real gas behaves ideally — that single fact answers three of the four parts.(a) What the dotted line meansFor an ideal gas, \(\displaystyle PV = nRT\), so \[\frac{PV}{T} = nR \] which is a constant — it does not change as \(\displaystyle P\) changes. A horizontal dotted line on this graph is exactly that: a plot of \(\displaystyle PV/T\) that stays fixed no matter what the pressure is. So the dotted line is the ideal-gas prediction. The two solid curves for oxygen bend away from it as \(\displaystyle P\) grows, because a real gas only obeys \(\displaystyle PV=nRT\) exactly in the limit of low pressure (weak intermolecular forces, molecules acting almost independently). The dotted plot signifies ideal-gas behaviour — the value \(\displaystyle PV/T\) would have at every pressure if oxygen were a perfect ideal gas.(b) Which temperature is higherA real gas departs from ideal behaviour because of the forces between its molecules. Raising the temperature gives molecules more kinetic energy, so those intermolecular forces matter relatively less, and the gas tracks the ideal-gas line more closely over a wider range of pressure. The curve that stays nearer to the dotted (ideal) line for longer as \(\displaystyle P\) increases is therefore the higher-temperature curve. In the figure that is the \(\displaystyle T_1\) curve, so \[T_1 > T_2 . \] (This is the general rule for real gases: they approach ideal-gas behaviour at high temperature and low pressure — never assume the lower curve is hotter just because it looks "smaller".)(c) The value of \(\displaystyle PV/T\) at the meeting pointThe two curves meet each other — and the dotted line — as \(\displaystyle P \to 0\), the regime where oxygen genuinely behaves like an ideal gas. There, \(\displaystyle PV/T = nR\), with \(\displaystyle n\) the number of moles of the actual sample, not its mass. This is the step people skip: you must convert mass to moles using the molar mass before you can use \(\displaystyle R\). \[n = \frac{m}{M} \] Here \(\displaystyle m = 1.00\times10^{-3}\text{ kg} = 1.00\text{ g}\) of oxygen, and \(\displaystyle M(\mathrm{O_2}) = 32.0\text{ g mol}^{-1}\): \[n = \frac{1.00\text{ g}}{32.0\text{ g mol}^{-1}} = 0.03125\text{ mol} \] So \[\frac{PV}{T} = nR = (0.03125\text{ mol})\times\left(8.31\ \mathrm{J\ mol^{-1}\ K^{-1}}\right) = 0.2597\ \mathrm{J\ K^{-1}} \] All the given data (\(\displaystyle 1.00\times10^{-3}\), \(\displaystyle 32.0\), \(\displaystyle 8.31\)) carry three significant figures, so the answer is rounded, once, to three: \[\frac{PV}{T} = 0.260\ \mathrm{J\ K^{-1}} \](d) Would hydrogen give the same intercept, and if not, what mass would?The intercept is \(\displaystyle nR\) — it depends on how many moles are present, not on which gas it is and not on the mass by itself. \(\displaystyle 1.00\times10^{-3}\) kg of hydrogen is not the same number of moles as \(\displaystyle 1.00\times10^{-3}\) kg of oxygen, because the molecules have different masses: \[n(\mathrm{H_2}) = \frac{1.00\text{ g}}{2.02\text{ g mol}^{-1}} = 0.495\text{ mol} \neq n(\mathrm{O_2}) = 0.03125\text{ mol} \] So no — equal masses of hydrogen and oxygen do not give the same \(\displaystyle PV/T\) intercept; hydrogen's would be about nineteen times larger.To make hydrogen's curve meet the \(\displaystyle y\)-axis at the same value \(\displaystyle 0.260\ \mathrm{J\ K^{-1}}\), it must supply the same number of moles as the oxygen sample did, \(\displaystyle n = 0.03125\) mol — because \(\displaystyle PV/T = nR\) is fixed once \(\displaystyle n\) is fixed, whatever the gas. The required mass is \[m(\mathrm{H_2}) = n \times M(\mathrm{H_2}) = (0.03125\text{ mol})\times\left(2.02\text{ g mol}^{-1}\right) = 0.0631\text{ g} \] Converting to kilograms and keeping three significant figures: \[m(\mathrm{H_2}) = 6.31\times10^{-5}\text{ kg} \]Answer: (a) The dotted line is the ideal-gas value of \(\displaystyle PV/T\) (constant, independent of \(\displaystyle P\)); the real oxygen curves approach it only as \(\displaystyle P\to0\). (b) \(\displaystyle T_1 > T_2\), since the curve closer to the dotted line over more of its range is the higher-temperature one. (c) \(\displaystyle PV/T = 0.260\ \mathrm{J\ K^{-1}}\) at the meeting point. (d) No — equal mass does not give equal \(\displaystyle PV/T\); hydrogen would need a mass of \(\displaystyle 6.31\times10^{-5}\) kg (about $\displaystyle 63.1$ mg) to meet the axis at the same \(\displaystyle 0.260\ \mathrm{J\ K^{-1}}\).

A pressure gauge reads only the excess over the atmosphere around it — to use the ideal gas law you must add that atmosphere back on to get the absolute pressure. The cylinder itself does not change volume; what changes between the two readings is how much gas is inside it. So the way to find the mass removed is to find the number of moles present before and after, using the ideal gas equation each time, and subtract.Step $\displaystyle 1$ — Convert every quantity to SI units, and gauge pressure to absolute pressure.Volume of the cylinder (fixed throughout): \[V = 30\ \text{litre} = 30\times10^{-3}\ \text{m}^3 = 3.0\times10^{-2}\ \text{m}^3\]Temperatures, in kelvin — the gas law needs absolute temperature, not Celsius: \[T_1 = 27\,^\circ\text{C} = 300\ \text{K}, \qquad T_2 = 17\,^\circ\text{C} = 290\ \text{K}\]Absolute pressure = gauge pressure + atmospheric pressure (here, \(\displaystyle 1\ \text{atm} = 1.013\times10^{5}\ \text{Pa}\)): \[P_1 = (15+1)\ \text{atm} = 16\ \text{atm} = 16\times1.013\times10^{5}\ \text{Pa} = 1.621\times10^{6}\ \text{Pa}\] \[P_2 = (11+1)\ \text{atm} = 12\ \text{atm} = 12\times1.013\times10^{5}\ \text{Pa} = 1.216\times10^{6}\ \text{Pa}\]Step $\displaystyle 2$ — Use the ideal gas law, \(\displaystyle PV = nRT\), to find the amount of gas present at each stage.Here \(\displaystyle P\) is absolute pressure, \(\displaystyle V\) is the cylinder's volume, \(\displaystyle n\) is the number of moles of gas, \(\displaystyle R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}\) is the universal gas constant, and \(\displaystyle T\) is the absolute temperature. Solving for \(\displaystyle n\): \[n = \frac{PV}{RT}\]Before withdrawal: \[n_1 = \frac{P_1V}{RT_1} = \frac{(1.621\times10^{6}\ \text{Pa})(3.0\times10^{-2}\ \text{m}^3)}{(8.31\ \text{J mol}^{-1}\text{K}^{-1})(300\ \text{K})} = \frac{4.863\times10^{4}}{2.493\times10^{3}} = 19.51\ \text{mol}\]After withdrawal: \[n_2 = \frac{P_2V}{RT_2} = \frac{(1.216\times10^{6}\ \text{Pa})(3.0\times10^{-2}\ \text{m}^3)}{(8.31\ \text{J mol}^{-1}\text{K}^{-1})(290\ \text{K})} = \frac{3.648\times10^{4}}{2.410\times10^{3}} = 15.14\ \text{mol}\]Step $\displaystyle 3$ — Turn the drop in mole number into a drop in mass.The number of moles that left the cylinder is \[\Delta n = n_1 - n_2 = 19.51\ \text{mol} - 15.14\ \text{mol} = 4.37\ \text{mol}\]A molecular mass of $\displaystyle 32$ u for \(\displaystyle O_2\) means the molar mass is \(\displaystyle M = 32\ \text{g mol}^{-1}\) — the mistake to avoid here is treating $\displaystyle 32$ as a mass rather than a mass per mole. The mass withdrawn is \[\Delta m = \Delta n \times M = 4.37\ \text{mol} \times 32\ \text{g mol}^{-1} = 139.8\ \text{g}\]The given data ($\displaystyle 15$ atm, $\displaystyle 11$ atm, $\displaystyle 30$ litre, $\displaystyle 27$ °C, $\displaystyle 17$ °C) all carry two significant figures, so the result is rounded to two significant figures at this final step, not before: \[\Delta m \approx 1.4\times10^{2}\ \text{g} = 0.14\ \text{kg}\]Answer: About $\displaystyle 140$ g ($\displaystyle 0.14$ kg) of oxygen was withdrawn from the cylinder.

The bubble obeys the ideal gas law as both its pressure and temperature change, so use \(\displaystyle \dfrac{PV}{T} = \text{constant} \) between the bottom and the surface — not just Boyle's Law or just Charles's Law alone.The air trapped in the bubble does not escape or mix with more air as it rises, so the number of moles \(\displaystyle n\) stays fixed. For a fixed amount of gas, the ideal gas equation \(\displaystyle PV = nRT\) gives\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]where the subscript $\displaystyle 1$ is the bottom of the lake and $\displaystyle 2$ is the surface.Step $\displaystyle 1$: Convert the temperatures to kelvin.Gas laws need absolute temperature, not the Celsius reading, because the relations above come from \(\displaystyle PV \propto T\) measured from absolute zero.\[T_1 = 12\,^\circ\text{C} = 12 + 273 = 285\ \text{K}, \qquad T_2 = 35\,^\circ\text{C} = 35 + 273 = 308\ \text{K} \]Step $\displaystyle 2$: Find the pressure at the bottom of the lake.The bubble is surrounded by water, so the pressure inside it equals the local pressure of the water pushing on it — atmospheric pressure at the surface plus the weight of the water column above, using \(\displaystyle P = P_{\text{atm}} + \rho g h\), where \(\displaystyle \rho\) is the density of water, \(\displaystyle g\) is the acceleration due to gravity, and \(\displaystyle h\) is the depth. This is the step people get wrong: you cannot use atmospheric pressure alone at the bottom — the water column adds to it.Taking the standard values \(\displaystyle P_{\text{atm}} = 1.01 \times 10^{5}\ \text{Pa}\), \(\displaystyle \rho = 1.0 \times 10^{3}\ \text{kg/m}^3\), and \(\displaystyle g = 9.8\ \text{m/s}^2\), with \(\displaystyle h = 40\ \text{m}\):\[P_1 = P_{\text{atm}} + \rho g h = 1.01\times10^{5}\ \text{Pa} + (1.0\times10^{3}\ \text{kg/m}^3)(9.8\ \text{m/s}^2)(40\ \text{m}) \]\[P_1 = 1.01\times10^{5}\ \text{Pa} + 3.92\times10^{5}\ \text{Pa} = 4.93\times10^{5}\ \text{Pa} \]At the surface, the bubble feels only atmospheric pressure:\[P_2 = P_{\text{atm}} = 1.01\times10^{5}\ \text{Pa} \]Step $\displaystyle 3$: Solve for the new volume.Rearranging \(\displaystyle \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\) for \(\displaystyle V_2\):\[V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} \]Substituting \(\displaystyle V_1 = 1.0\ \text{cm}^3\):\[V_2 = \frac{(4.93\times10^{5}\ \text{Pa})(1.0\ \text{cm}^3)(308\ \text{K})}{(285\ \text{K})(1.01\times10^{5}\ \text{Pa})} \]\[V_2 = \frac{1.518\times10^{8}}{2.879\times10^{7}}\ \text{cm}^3 = 5.275\ \text{cm}^3 \]Step $\displaystyle 4$: Round to the correct number of significant figures.The given data — the volume \(\displaystyle 1.0\ \text{cm}^3\), the depth \(\displaystyle 40\ \text{m}\), and the temperatures \(\displaystyle 12\,^\circ\text{C}\) and \(\displaystyle 35\,^\circ\text{C}\) — each carry two significant figures, so the answer should be stated to two significant figures as well:\[V_2 \approx 5.3\ \text{cm}^3 \]The bubble grows because the pressure squeezing it drops sharply as it rises (from \(\displaystyle 4.93\times10^{5}\ \text{Pa}\) down to \(\displaystyle 1.01\times10^{5}\ \text{Pa}\)) while the temperature rises only slightly (\(\displaystyle 285\ \text{K}\) to \(\displaystyle 308\ \text{K}\)) — the pressure drop, not the warming, is what does almost all the expanding.Answer: The bubble's volume grows to about \(\displaystyle 5.3\ \text{cm}^3\) (from \(\displaystyle 1.0\ \text{cm}^3\)).

Air is a mixture of gases, but you don't need to know the mixture to count molecules — the ideal gas law gives the total number directly from just \(\displaystyle P\), \(\displaystyle V\), and \(\displaystyle T\).For an ideal gas containing \(\displaystyle N\) molecules (of any species, or mixed species — it doesn't matter which), \[PV = Nk_BT \] where \(\displaystyle P\) is the absolute pressure, \(\displaystyle V\) is the volume, \(\displaystyle T\) is the absolute temperature, and \(\displaystyle k_B = 1.38 \times 10^{-23}\ \text{J K}^{-1}\) is Boltzmann's constant. This form doesn't reference moles or molar mass at all, which is exactly why it works for a mixture like air (oxygen, nitrogen, water vapour, argon, …) without needing the composition. Rearranging, \[N = \frac{PV}{k_BT} \]Get every quantity into SI units first. \[T = 27\,^\circ\text{C} + 273 = 300\ \text{K} \]Substitute.\[N = \frac{PV}{k_BT} = \frac{(1.013 \times 10^{5}\ \text{Pa})(25.0\ \text{m}^3)}{(1.38 \times 10^{-23}\ \text{J K}^{-1})(300\ \text{K})} \]Numerator (using \(\displaystyle 1\ \text{Pa}\cdot\text{m}^3 = 1\ \text{J}\)): \[PV = 1.013 \times 10^{5} \times 25.0 = 2.5325 \times 10^{6}\ \text{J} \]Denominator: \[k_BT = 1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-21}\ \text{J} \]Divide: \[N = \frac{2.5325 \times 10^{6}}{4.14 \times 10^{-21}} = 6.118 \times 10^{26} \]The pressure and Boltzmann's constant are each known to $\displaystyle 3$ significant figures, so the answer is rounded to $\displaystyle 3$ significant figures — this is not a quantity you can pin down more finely than the least-precise input, so a longer decimal string would be false precision.\[N \approx 6.12 \times 10^{26}\ \text{molecules} \]This is a huge number, which makes sense — a room-sized volume of air at everyday pressure holds on the order of \(\displaystyle 10^{26}\) molecules, roughly the same order as a mole (\(\displaystyle 6.02 \times 10^{23}\)) times a few hundred.Answer: The room contains approximately \(\displaystyle 6.12 \times 10^{26}\) air molecules in total.

Average thermal energy depends only on temperature, not on which gas it is — so "helium" here is really just a label.The kinetic theory of gases says every molecule of an ideal gas, treated as a point particle with only translational motion, shares its energy equally among its three directions of motion (the x, y and z axes) — this is the law of equipartition of energy. Each of the three translational degrees of freedom carries an average energy \(\displaystyle \tfrac{1}{2}k_BT \), so the total average translational kinetic energy per molecule is\[E = \frac{3}{2}k_BT \]where \(\displaystyle k_B = 1.38 \times 10^{-23}\ \text{J K}^{-1} \) is Boltzmann's constant and \(\displaystyle T \) is the absolute temperature in kelvin. Helium is monatomic, so this translational energy is its entire thermal energy — there is no rotational or vibrational energy to add, unlike for a diatomic or polyatomic molecule. Notice that mass does not appear anywhere in this formula: at a given temperature, every ideal-gas molecule — light or heavy — carries the same average thermal energy. That is the point of picking helium: the number you get would be identical for any monatomic gas at that temperature.The one step people skip is converting to kelvin before substituting — this law only holds for the absolute temperature, not the Celsius value.(i) Room temperature, \(\displaystyle 27\,^\circ\text{C}\)Converting to kelvin: \(\displaystyle T = 27 + 273 = 300\ \text{K} \).\[E = \frac{3}{2}(1.38\times10^{-23}\ \text{J K}^{-1})(300\ \text{K}) = 6.21\times10^{-21}\ \text{J} \](ii) Surface of the Sun, \(\displaystyle T = 6000\ \text{K}\)This temperature is already in kelvin.\[E = \frac{3}{2}(1.38\times10^{-23}\ \text{J K}^{-1})(6000\ \text{K}) = 1.242\times10^{-19}\ \text{J} \](iii) Stellar core, \(\displaystyle T = 10^7\ \text{K}\)\[E = \frac{3}{2}(1.38\times10^{-23}\ \text{J K}^{-1})(1\times10^{7}\ \text{K}) = 2.07\times10^{-16}\ \text{J} \]In each case \(\displaystyle k_B \) is given to $\displaystyle 3$ significant figures, so the results are rounded to $\displaystyle 3$ significant figures ($\displaystyle 300$ K and $\displaystyle 6000$ K are here taken as exact conversions/given values, so they don't further restrict the precision).Answer: (i) \(\displaystyle 6.21 \times 10^{-21}\ \text{J}\) (ii) \(\displaystyle 1.242 \times 10^{-19}\ \text{J}\) (iii) \(\displaystyle 2.07 \times 10^{-16}\ \text{J}\)

Equal volume, same T and P means equal molecule count — the ideal gas law does not care what the gas is made of.Part $\displaystyle 1$ — number of moleculesFor an ideal gas, \[PV = Nk_BT \] where \(\displaystyle N\) is the number of molecules, \(\displaystyle k_B\) is Boltzmann's constant, \(\displaystyle P\) the pressure, \(\displaystyle V\) the volume and \(\displaystyle T\) the absolute temperature. Solving for \(\displaystyle N\), \[N = \frac{PV}{k_BT} \]This expression has no term for molar mass, atomicity, or molecular structure — it depends only on \(\displaystyle P\), \(\displaystyle V\), and \(\displaystyle T\). Since the three vessels have equal capacity (same \(\displaystyle V\)) and are stated to be at the same pressure and the same temperature, all three give the same value of \(\displaystyle N\).So yes, all three vessels — neon, chlorine, and uranium hexafluoride — contain the same number of molecules. (This is exactly Avogadro's law: equal volumes of any gas at the same temperature and pressure hold equal numbers of molecules, whether the gas is monatomic, diatomic, or a heavy polyatomic molecule.)Part $\displaystyle 2$ — root mean square speedThe rms speed of a gas's molecules comes from kinetic theory: \[v_{rms} = \sqrt{\dfrac{3RT}{M}} \] where \(\displaystyle R\) is the universal gas constant, \(\displaystyle T\) the absolute temperature, and \(\displaystyle M\) the molar mass of that gas.Here is the step people trip on: the average translational kinetic energy per molecule, \(\displaystyle \tfrac{1}{2}mv_{rms}^2 = \tfrac{3}{2}k_BT\), depends only on \(\displaystyle T\) and is exactly the same for a neon atom, a chlorine molecule, and a UF\(\displaystyle _6\) molecule at a given temperature — kinetic theory shares that energy equally regardless of mass or complexity. But equal energy with unequal mass \(\displaystyle m\) forces unequal speed: a lighter molecule must move faster to carry the same kinetic energy. That is why \(\displaystyle v_{rms}\) is not the same across the three vessels even though \(\displaystyle T\) is.Since \(\displaystyle T\) is identical in all three vessels, \(\displaystyle v_{rms} \propto 1/\sqrt{M}\): the lightest molecule has the largest rms speed.Molar masses: \[M(\text{Ne}) \approx 20.2\ \text{g mol}^{-1} \] \[M(\text{Cl}_2) = 2 \times 35.5 \approx 70.9\ \text{g mol}^{-1} \] \[M(\text{UF}_6) = 238.0 + 6(19.0) \approx 352\ \text{g mol}^{-1} \]Comparing \(\displaystyle 1/\sqrt{M}\) for the three (same \(\displaystyle T\), same \(\displaystyle R\), so this ratio alone fixes the ordering): \[\frac{1}{\sqrt{M(\text{Ne})}} : \frac{1}{\sqrt{M(\text{Cl}_2)}} : \frac{1}{\sqrt{M(\text{UF}_6)}} \approx \frac{1}{4.49} : \frac{1}{8.42} : \frac{1}{18.8} \approx 0.223 : 0.119 : 0.0533 \]Neon's value is roughly \(\displaystyle 1.9\times\) chlorine's and roughly \(\displaystyle 4.2\times\) uranium hexafluoride's. Neon is by far the lightest of the three, so its molecules must move fastest to hold the same average kinetic energy — the rms speed is largest in the neon vessel.Answer: Yes — all three vessels contain the same number of molecules, since \(\displaystyle N = PV/(k_BT)\) depends only on \(\displaystyle P\), \(\displaystyle V\), \(\displaystyle T\), which are equal for all three. No — \(\displaystyle v_{rms}\) is not the same, because \(\displaystyle v_{rms} = \sqrt{3RT/M}\) depends on molar mass \(\displaystyle M\) even though \(\displaystyle T\) is common; \(\displaystyle v_{rms}\) is largest for neon, since it has by far the smallest molar mass (\(\displaystyle \approx 20.2\ \text{g mol}^{-1}\), versus \(\displaystyle \approx 70.9\ \text{g mol}^{-1}\) for Cl\(\displaystyle _2\) and \(\displaystyle \approx 352\ \text{g mol}^{-1}\) for UF\(\displaystyle _6\)).

The trick is that the molar mass in the rms-speed formula cancels when you compare two gases at different temperatures — you never have to convert atomic mass units to kilograms.The root-mean-square speed of a gas, from kinetic theory, is \[v_{rms} = \sqrt{\dfrac{3RT}{M}} \] where \(\displaystyle R\) is the universal gas constant, \(\displaystyle T\) is the absolute temperature, and \(\displaystyle M\) is the molar (or atomic) mass of the gas.Convert the given temperature to kelvin first — this formula only works with absolute temperature, never Celsius: \[T_{He} = -20\,^\circ\text{C} = -20 + 273 = 253\ \text{K} \]We want the argon temperature \(\displaystyle T_{Ar}\) at which its rms speed equals helium's rms speed at $\displaystyle 253$ K: \[\sqrt{\dfrac{3RT_{Ar}}{M_{Ar}}} = \sqrt{\dfrac{3RT_{He}}{M_{He}}} \]Squaring both sides, the \(\displaystyle 3R\) common to both gases cancels: \[\dfrac{T_{Ar}}{M_{Ar}} = \dfrac{T_{He}}{M_{He}} \]\[T_{Ar} = T_{He}\times\dfrac{M_{Ar}}{M_{He}} \]Because only the ratio of the two atomic masses appears, they can be left in atomic mass units (u) — the unit cancels along with everything else, so there is no need to convert to kg.Substituting \(\displaystyle T_{He} = 253\ \text{K}\), \(\displaystyle M_{Ar} = 39.9\ \text{u}\), \(\displaystyle M_{He} = 4.0\ \text{u}\): \[T_{Ar} = 253\ \text{K}\times\dfrac{39.9}{4.0} = 253\ \text{K}\times 9.975 \] \[T_{Ar} = 2523.675\ \text{K} \]Argon atoms are almost ten times as massive as helium atoms, so they need a temperature almost ten times as high to reach the same rms speed — this same fact (heavier molecules move slower at a given temperature) is why light gases like hydrogen and helium escape a planet's atmosphere far more easily than heavier gases do.Rounding to match the precision of the given atomic masses ($\displaystyle 39.9$ u, three significant figures):Answer: \(\displaystyle T_{Ar} \approx 2523.7\ \text{K}\) (about \(\displaystyle 2251\,^\circ\text{C}\))

A molecule's mean free path depends on how "big" it looks to its neighbours — and "big" means diameter, not radius. The mean free path is the average distance a molecule travels before it collides with another molecule. Kinetic theory gives it as\[l = \frac{1}{\sqrt{2}\,\pi d^{2} n} \]where \(\displaystyle d\) is the molecular diameter and \(\displaystyle n\) is the number density (molecules per unit volume). The radius given here is $\displaystyle 1.0$ Å, so the diameter — the quantity that actually enters the collision cross-section — is\[d = 2r = 2.0\times10^{-10}\ \text{m} \]Using the radius by mistake here would shrink \(\displaystyle d^2\) by a factor of $\displaystyle 4$ and inflate \(\displaystyle l\) by the same factor, so this substitution is worth double-checking.Step $\displaystyle 1$: Convert the given data to SI units.\[T = 17\,^{\circ}\text{C} = 17 + 273 = 290\ \text{K} \] \[P = 2.0\ \text{atm} = 2.0\times(1.013\times10^{5}\ \text{Pa}) = 2.026\times10^{5}\ \text{Pa} \]Step $\displaystyle 2$: Find the number density \(\displaystyle n\) from the ideal gas law. For an ideal gas, \(\displaystyle PV = Nk_BT\), so the number of molecules per unit volume is\[n = \frac{N}{V} = \frac{P}{k_B T} \]with Boltzmann's constant \(\displaystyle k_B = 1.38\times10^{-23}\ \text{J K}^{-1}\). Substituting,\[n = \frac{2.026\times10^{5}\ \text{Pa}}{(1.38\times10^{-23}\ \text{J K}^{-1})(290\ \text{K})} = \frac{2.026\times10^{5}}{4.00\times10^{-21}}\ \text{m}^{-3} = 5.06\times10^{25}\ \text{m}^{-3} \]Step $\displaystyle 3$: Compute the mean free path.\[l = \frac{1}{\sqrt{2}\,\pi d^{2} n} = \frac{1}{\sqrt{2}\,\pi (2.0\times10^{-10}\ \text{m})^{2}(5.06\times10^{25}\ \text{m}^{-3})} \]The denominator: \(\displaystyle d^2 = 4.0\times10^{-20}\ \text{m}^2\), and \(\displaystyle \sqrt2\,\pi d^2 n = (1.414)(3.1416)(4.0\times10^{-20})(5.06\times10^{25}) = 8.99\times10^{6}\ \text{m}^{-1}\). So\[l = \frac{1}{8.99\times10^{6}\ \text{m}^{-1}} = 1.11\times10^{-7}\ \text{m} \]Rounding to two significant figures (matching the $\displaystyle 2.0$ atm and $\displaystyle 1.0$ Å given), \(\displaystyle l \approx 1.1\times10^{-7}\ \text{m}\) — about $\displaystyle 550$ times the molecule's own diameter, which is why treating a gas molecule as travelling in straight lines between rare collisions is a reasonable picture.Step $\displaystyle 4$: Find the molecule's rms speed, since collision frequency needs a speed to turn a distance between collisions into a time between collisions. From kinetic theory,\[v_{rms} = \sqrt{\frac{3k_BT}{m}} \]where \(\displaystyle m\) is the mass of one \(\displaystyle N_2\) molecule, not the molar mass. Converting: \(\displaystyle m = 28.0\ \text{u} = 28.0\times(1.66\times10^{-27}\ \text{kg}) = 4.65\times10^{-26}\ \text{kg}\). Then\[v_{rms} = \sqrt{\frac{3(1.38\times10^{-23}\ \text{J K}^{-1})(290\ \text{K})}{4.65\times10^{-26}\ \text{kg}}} = \sqrt{2.58\times10^{5}\ \text{m}^2\text{s}^{-2}} = 5.1\times10^{2}\ \text{m/s} \]Step $\displaystyle 5$: Collision frequency and the time between collisions. The collision frequency is how many collisions the molecule undergoes per second, which is the speed divided by the distance it covers between collisions:\[\nu = \frac{v_{rms}}{l} = \frac{5.08\times10^{2}\ \text{m/s}}{1.11\times10^{-7}\ \text{m}} = 4.6\times10^{9}\ \text{collisions/s} \]The average time the molecule moves freely between two successive collisions is the reciprocal,\[\tau = \frac{1}{\nu} = \frac{l}{v_{rms}} = \frac{1.11\times10^{-7}\ \text{m}}{5.08\times10^{2}\ \text{m/s}} = 2.2\times10^{-10}\ \text{s} \]Step $\displaystyle 6$: The collision time itself. This is the (much shorter) time the molecule spends actually overlapping with its collision partner — roughly the time to cross one molecular diameter at its own speed:\[t_{c} = \frac{d}{v_{rms}} = \frac{2.0\times10^{-10}\ \text{m}}{5.08\times10^{2}\ \text{m/s}} = 3.9\times10^{-13}\ \text{s} \]Comparing the two times. Because both times share the same speed in the denominator, their ratio reduces to a ratio of lengths — the mean free path to the molecular diameter — so it does not depend on which "average speed" convention is used:\[\frac{\tau}{t_c} = \frac{l/v_{rms}}{d/v_{rms}} = \frac{l}{d} = \frac{1.11\times10^{-7}\ \text{m}}{2.0\times10^{-10}\ \text{m}} \approx 5.6\times10^{2} \]So the molecule spends free time between collisions that is on the order of $\displaystyle 500$ times longer than the collision itself takes — collisions are brief, violent interruptions in an otherwise long, straight-line free flight.Answer: Mean free path \(\displaystyle l \approx 1.1\times10^{-7}\ \text{m}\); collision frequency \(\displaystyle \nu \approx 4.6\times10^{9}\ \text{s}^{-1}\), giving a free-flight time of \(\displaystyle \tau \approx 2.2\times10^{-10}\ \text{s}\) between collisions; the collision itself lasts only \(\displaystyle t_c \approx 3.9\times10^{-13}\ \text{s}\), so the free-flight time is roughly $\displaystyle 500$ times longer than the collision time.