From estimating molecular volume to interpreting Maxwell–Boltzmann graphs, each NCERT solution is written to show why a method was chosen and which kinetic-theory assumption is involved.
Each answer begins with the key idea that drives the solution. This allows you to recognise which concept should trigger your approach before performing calculations.
Every final answer links back to physical meaning: how temperature affects molecular speed, why pressure alters mean free path, and why internal energy depends only on temperature.
Q1. Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
Theory: In kinetic theory, a gas is considered to be made up of a very large number of tiny molecules (treated as hard spheres) moving randomly in all directions, and the measured volume of the gas at a given temperature and pressure (for example, \(22.4 \times 10^{-3}\,\text{m}^3\) for one mole at STP) is essentially the volume of the container itself. The actual volume occupied by the molecules is much smaller because the intermolecular distances are large compared to the molecular size, so most of the container is empty space available for molecular motion. The molecular or excluded volume is obtained by summing the volumes of all molecules, each idealized as a sphere of known diameter, and the required fraction is \[ \text{Fraction} = \frac{\text{Total volume of all molecules}}{\text{Measured volume of one mole of gas at STP}}, \] which clearly demonstrates that the molecular volume is only a tiny part of the total gas volume and justifies treating ideal gas molecules as point particles to a good approximation.
At standard temperature and pressure, one mole of oxygen gas occupies a volume of \(22.4 \times 10^{-3}\,\text{m}^3\). To estimate the fraction of molecular volume, we first calculate the actual volume occupied by the oxygen molecules themselves.
The diameter of an oxygen molecule is given as \(3\,\text{Å} = 3 \times 10^{-10}\,\text{m}\). Hence, the radius of one molecule is \[ r = 1.5 \times 10^{-10}\,\text{m}. \] Assuming each molecule to be spherical, the volume of one oxygen molecule is \[ v = \frac{4}{3}\pi r^3. \]
Substituting the value of \(r\), \[ \begin{aligned} v &= \frac{4}{3}\pi (1.5 \times 10^{-10})^3 \\ &= \frac{4}{3}\pi (3.375 \times 10^{-30}) \\ &\approx 1.41 \times 10^{-29}\,\text{m}^3. \end{aligned} \]
One mole of oxygen contains Avogadro’s number of molecules, \(N_A = 6.02 \times 10^{23}\). Therefore, the total volume actually occupied by all oxygen molecules in one mole is \[ \begin{aligned} V_{\text{molecules}} &= N_A \times v \\ &= (6.02 \times 10^{23})(1.41 \times 10^{-29}) \\ &\approx 8.5 \times 10^{-6}\,\text{m}^3. \end{aligned} \]
The fraction of molecular volume to the actual volume of the gas at STP is then \[ \begin{aligned} \text{Fraction} &= \frac{V_{\text{molecules}}}{V_{\text{gas}}} \\ &= \frac{8.5 \times 10^{-6}}{22.4 \times 10^{-3}} \\ &\approx 3.8 \times 10^{-4}. \end{aligned} \]
Thus, only about \(3.8 \times 10^{-4}\), or roughly \(0.04\%\), of the volume of oxygen gas at STP is actually occupied by the molecules themselves, showing that a gas is mostly empty space.
Q2. Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
Theory: The behaviour of an ideal gas is described by the ideal gas equation \[ PV = nRT, \] which relates the pressure \(P\), volume \(V\), amount of gas in moles \(n\), absolute temperature \(T\) and the universal gas constant \(R\). For a fixed amount of gas, the volume depends only on the temperature and pressure, so at a specified temperature and pressure (such as STP) all ideal gases with \(n = 1\) mol occupy the same volume, called the molar volume. At STP, we set \(n = 1\), \(T = 273\,\text{K}\) and \(P = 1\,\text{atm}\), substitute in the ideal gas equation, and evaluate \(V = \dfrac{RT}{P}\) to obtain the numerical value of the molar volume, which comes out to approximately \(22.4\,\text{L}\) for any ideal gas.
At standard temperature and pressure, an ideal gas obeys the ideal gas equation \[ PV = nRT. \] To determine the molar volume, we consider exactly one mole of gas, so that \(n = 1\).
At STP, the pressure is one atmosphere, \(P = 1.013 \times 10^{5}\,\text{Pa}\), and the temperature is \(0^\circ\text{C} = 273\,\text{K}\). The universal gas constant is \(R = 8.31\,\text{J mol}^{-1}\text{K}^{-1}\). Substituting these values into the gas equation gives \[ V = \frac{RT}{P}. \]
Thus, the molar volume at STP is obtained as \[ \begin{aligned} V &= \frac{(8.31)(273)}{1.013 \times 10^{5}} \\ &\approx 2.24 \times 10^{-2}\,\text{m}^3. \end{aligned} \]
Since \(1\,\text{m}^3 = 10^{3}\,\text{litres}\), the above volume in litres becomes \[ \begin{aligned} V &= 2.24 \times 10^{-2} \times 10^{3} \\ &= 22.4\,\text{litres}. \end{aligned} \]
Hence, it is shown that one mole of an ideal gas occupies a volume of \(22.4\,\text{litres}\) at standard temperature and pressure.
Q3. Figure 12.8 shows plot of PV/T versus P for \(1.00\times10^{–3}\) kg of oxygen gas at two
different temperatures.
(a) What does the dotted plot signify?
(b) Which is true: T1 > T2 or T1 < T2 ?
(c) What is the value of PV/T
where the curves meet on the y-axis?
(d) If we obtained similar plots for \(1.00\times10^{–3}\) kg of hydrogen, would we get the same value of
PV/T at
the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of
PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H2 = 2.02 u, of O2 =
32.0 u, R = 8.31 \(\mathrm{J\ mol^{–1}\ K^{–1}}\).)
Theory: For an ideal gas, the equation \(PV = nRT\) implies that \[ \frac{PV}{T} = nR, \] which is constant for a fixed amount of gas and does not depend on pressure. Therefore, if we plot \(\dfrac{PV}{T}\) versus pressure \(P\) for an ideal gas with fixed \(n\), we obtain a horizontal straight line, and any deviation from this line indicates non-ideal (real gas) behaviour. For real gases, intermolecular forces and finite molecular size cause \(\dfrac{PV}{T}\) to vary with pressure: at low pressure and high temperature, the gas approaches ideal behaviour and the curve tends to the horizontal ideal line, while at higher pressures and/or lower temperatures the curve bends away from it, showing greater deviation. The intercept of the \(\dfrac{PV}{T}\)–\(P\) plot on the y-axis (as \(P \to 0\)) corresponds to the ideal-gas limit and has the value \(nR\), which depends only on the number of moles present and is the same for different gases if they contain the same number of moles.
The given graph represents the variation of \(\dfrac{PV}{T}\) with pressure \(P\) for a fixed mass of oxygen gas at two different temperatures. For an ideal gas, the quantity \(\dfrac{PV}{T}\) is independent of pressure and depends only on the amount of gas present. Any deviation from constancy indicates non-ideal behavior.
The dotted horizontal line signifies the ideal gas behavior. It represents the limiting value of \(\dfrac{PV}{T}\) at very low pressure, where intermolecular forces become negligible and the gas behaves ideally irrespective of temperature.
From the graph, the curve labeled \(T_2\) lies below the curve labeled \(T_1\) in the low-pressure region before rising more sharply at higher pressures. This indicates that the deviation from ideal behavior is more pronounced at the lower temperature. Hence, it follows that \[ T_1 > T_2. \]
At the point where both curves meet the y-axis, the pressure approaches zero and the gas behaves ideally. For an ideal gas, \[ PV = nRT. \] Therefore, \[ \begin{aligned} \frac{PV}{T} &= nR. \end{aligned} \]
For oxygen, the given mass is \(1.00 \times 10^{-3}\,\text{kg} = 1.00\,\text{g}\). The molar mass of oxygen is \(32.0\,\text{g mol}^{-1}\), hence the number of moles is \[ \begin{aligned} n &= \frac{1.00}{32.0} \\ &= 3.125 \times 10^{-2}\,\text{mol}. \end{aligned} \]
Thus, the value of \(\dfrac{PV}{T}\) at the y-axis intercept is \[ \begin{aligned} \frac{PV}{T} &= nR \\ &= (3.125 \times 10^{-2})(8.31) \\ &\approx 0.26\,\text{J K}^{-1}. \end{aligned} \]
If similar plots are obtained for hydrogen, the value of \(\dfrac{PV}{T}\) at low pressure and high temperature would depend on the number of moles of hydrogen taken. For the same mass \(1.00 \times 10^{-3}\,\text{kg}\), hydrogen would have a much larger number of moles and hence a different intercept value. Therefore, the intercept would not be the same.
To obtain the same value of \(\dfrac{PV}{T}\) as oxygen, the number of moles of hydrogen must be equal to that of oxygen, namely \(3.125 \times 10^{-2}\,\text{mol}\). Since the molar mass of hydrogen is \(2.02\,\text{g mol}^{-1}\), the required mass of hydrogen is \[ \begin{aligned} m &= nM \\ &= (3.125 \times 10^{-2})(2.02) \\ &\approx 6.3 \times 10^{-2}\,\text{g}. \end{aligned} \]
Hence, a mass of approximately \(6.3 \times 10^{-2}\,\text{g}\) of hydrogen would yield the same value of \(\dfrac{PV}{T}\) in the low-pressure, high-temperature region of the plot.
Q4. An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder \(\mathrm{(R = 8.31\ J\ mol^{–1}\ K^{–1})}\), molecular mass of (\(\mathrm{O_2}\) = 32 u).
Theory: For a fixed volume container (like a gas cylinder), the amount of gas present at any state can be obtained from the ideal gas equation \[ PV = nRT, \] where \(P\) is the absolute pressure of the gas, \(V\) is the fixed volume, \(T\) is the absolute temperature, and \(n\) is the number of moles. If the cylinder volume is constant, any change in the product \(\dfrac{P}{T}\) reflects a change in the number of moles, since \[ n = \frac{PV}{RT} = \frac{V}{R}\,\frac{P}{T}. \] In practice, pressure gauges usually read gauge pressure (above atmospheric), so to apply the ideal gas law one must first convert gauge pressure to absolute pressure by adding atmospheric pressure. Once the initial and final moles are found from the two states, their difference gives the moles withdrawn, and multiplying by the molar mass yields the mass of gas taken out.
The amount of oxygen in the cylinder at any stage can be obtained using the ideal gas equation \(PV = nRT\). Since the volume of the cylinder remains constant, the number of moles of gas present initially and finally can be calculated separately and their difference will give the amount of oxygen withdrawn.
The given pressures are gauge pressures, hence the corresponding absolute pressures must be used. Taking atmospheric pressure as \(1\,\text{atm}\), the initial and final absolute pressures are \[ \begin{aligned} P_1 &= (15 + 1)\,\text{atm} \\&= 16\,\text{atm}, \\\\ P_2 &= (11 + 1)\,\text{atm} \\&= 12\,\text{atm}. \end{aligned} \] The temperatures in kelvin are \[ \begin{aligned} T_1 &= 27^\circ\text{C} \\&= 300\,\text{K}, \\ T_2 &= 17^\circ\text{C} \\&= 290\,\text{K}. \end{aligned} \] The volume of the cylinder is \(V = 30\,\text{L} = 3.0 \times 10^{-2}\,\text{m}^3\).
Using the ideal gas equation for the initial state, \[ \begin{aligned} n_1 &= \frac{P_1 V}{RT_1} \\ &= \frac{(16 \times 1.013 \times 10^{5})(3.0 \times 10^{-2})}{(8.31)(300)} \\ &\approx 19.5\,\text{mol}. \end{aligned} \]
Similarly, for the final state of the gas, \[ \begin{aligned} n_2 &= \frac{P_2 V}{RT_2} \\ &= \frac{(12 \times 1.013 \times 10^{5})(3.0 \times 10^{-2})}{(8.31)(290)} \\ &\approx 15.1\,\text{mol}. \end{aligned} \]
Hence, the number of moles of oxygen withdrawn from the cylinder is \[ \begin{aligned} \Delta n &= n_1 - n_2 \\ &\approx 19.5 - 15.1 \\ &\approx 4.4\,\text{mol}. \end{aligned} \]
The molecular mass of oxygen is \(32\,\text{g mol}^{-1}\). Therefore, the mass of oxygen taken out is \[ \begin{aligned} m &= \Delta n \times M \\ &= 4.4 \times 32 \\ &\approx 1.4 \times 10^{2}\,\text{g}. \end{aligned} \]
Thus, the estimated mass of oxygen withdrawn from the cylinder is approximately \(1.4 \times 10^{2}\,\text{g}\), or about \(0.14\,\text{kg}\).
Q5. An air bubble of volume \(1.0\ \text{cm}^3\) rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ?
Theory: For a given fixed amount of gas, the ideal gas equation \(PV = nRT\) can be written in the form \[ \frac{PV}{T} = \text{constant}, \] which is known as the combined gas law and relates the pressure, volume and temperature of the gas between two states. When a gas bubble rises slowly through a liquid, we assume that the amount of gas inside the bubble remains constant and that the gas approximately obeys the ideal gas law, so the initial and final states are connected by \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}. \] In this problem, the pressure at the bottom is higher due to the hydrostatic pressure of water in addition to atmospheric pressure, while at the surface the pressure is just atmospheric, and the temperatures at the two levels are different; using the combined gas law with these pressures and temperatures allows us to find the new volume of the bubble at the surface.
As the air bubble rises slowly through water, its amount of gas remains constant. Hence, the change in its volume can be analyzed using the ideal gas relation in the combined form \[ \frac{PV}{T} = \text{constant}. \] This relation is applied between the bottom of the lake and the surface.
At the bottom of the lake, the pressure on the bubble is the sum of atmospheric pressure and hydrostatic pressure due to water. Taking atmospheric pressure as \(P_0 = 1.013 \times 10^{5}\,\text{Pa}\), water density as \(\rho = 1000\,\text{kg m}^{-3}\), and acceleration due to gravity as \(g = 9.8\,\text{m s}^{-2}\), the pressure at depth \(h = 40\,\text{m}\) is \[ \begin{aligned} P_1 &= P_0 + \rho g h \\ &= 1.013 \times 10^{5} + (1000)(9.8)(40) \\ &= 4.93 \times 10^{5}\,\text{Pa}. \end{aligned} \]
The initial volume of the bubble is \(V_1 = 1.0\,\text{cm}^3\). The initial temperature is \(12^\circ\text{C} = 285\,\text{K}\). At the surface, the pressure is atmospheric, \(P_2 = 1.013 \times 10^{5}\,\text{Pa}\), and the temperature is \(35^\circ\text{C} = 308\,\text{K}\).
Using the combined gas equation, \[ \begin{aligned} \frac{P_1 V_1}{T_1} &= \frac{P_2 V_2}{T_2}, \end{aligned} \] the final volume \(V_2\) of the bubble at the surface is \[ \begin{aligned} V_2 &= V_1 \frac{P_1}{P_2} \frac{T_2}{T_1}. \end{aligned} \]
Substituting the numerical values, \[ \begin{aligned} V_2 &= (1.0)\,\frac{4.93 \times 10^{5}}{1.013 \times 10^{5}} \times \frac{308}{285} \\ &\approx 5.3\,\text{cm}^3. \end{aligned} \]
Therefore, when the air bubble reaches the surface of the lake, its volume increases to approximately \(5.3\,\text{cm}^3\).
Q6. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.
Theory: Air is a mixture of several gases (mainly nitrogen, oxygen, argon, carbon dioxide and water vapour), but if the mixture behaves ideally, the whole sample can be treated as a single ideal gas characterized by its total pressure, volume and temperature. For any ideal gas, the equation \[ PV = nRT \] relates the macroscopic variables pressure \(P\), volume \(V\), temperature \(T\) and the amount of gas \(n\) in moles. Once the number of moles \(n\) is known, the total number of molecules in the sample is obtained using Avogadro’s relation \[ N = n N_A, \] where \(N_A\) is Avogadro’s constant; this result is independent of the composition of the air, since all ideal gases contain the same number of molecules per mole at a given temperature and pressure.
The air inside the room can be treated as an ideal gas mixture. Since all gases in the mixture obey the ideal gas law, the total number of molecules present depends only on the total pressure, volume, and temperature, irrespective of the composition of air.
For an ideal gas, the equation \[ PV = nRT \] relates the pressure \(P\), volume \(V\), temperature \(T\), and number of moles \(n\). The given pressure is \(1\,\text{atm} = 1.013 \times 10^{5}\,\text{Pa}\), the volume of the room is \(V = 25.0\,\text{m}^3\), and the temperature is \(27^\circ\text{C} = 300\,\text{K}\).
Substituting these values into the ideal gas equation gives \[ \begin{aligned} n &= \frac{PV}{RT} \\ &= \frac{(1.013 \times 10^{5})(25.0)}{(8.31)(300)} \\ &\approx 1.02 \times 10^{3}\,\text{mol}. \end{aligned} \]
The total number of molecules \(N\) is related to the number of moles by Avogadro’s constant \(N_A = 6.02 \times 10^{23}\,\text{mol}^{-1}\). Hence, \[ \begin{aligned} N &= n N_A \\ &= (1.02 \times 10^{3})(6.02 \times 10^{23}) \\ &\approx 6.1 \times 10^{26}. \end{aligned} \]
Therefore, the total number of air molecules present in the room is of the order of \(6 \times 10^{26}\), including all constituents such as oxygen, nitrogen, water vapour, and trace gases.
Q7. Estimate the average thermal energy of a helium atom at
(i) room temperature (27 °C),
(ii) the temperature on the surface of the Sun (6000 K),
(iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Theory: In kinetic theory, the random motion of gas atoms or molecules is associated with thermal energy, and for an ideal gas this microscopic energy is purely kinetic. For a monoatomic ideal gas such as helium, each atom has three translational degrees of freedom (motion along the x, y and z directions), and the equipartition theorem states that each degree of freedom contributes an average energy of \(\tfrac{1}{2}kT\), where \(k\) is the Boltzmann constant and \(T\) is the absolute temperature. Hence, the average thermal (kinetic) energy per atom of a monoatomic ideal gas is \[ \langle E \rangle = \frac{3}{2}kT, \] which shows that the mean energy of an individual atom is directly proportional to the absolute temperature and increases linearly as the temperature rises.
A helium atom is a monoatomic gas particle. For a monoatomic ideal gas, the average thermal energy per atom is purely translational and is given by \[ \langle E \rangle = \frac{3}{2}kT, \] where \(k = 1.38 \times 10^{-23}\,\text{J K}^{-1}\) is the Boltzmann constant and \(T\) is the absolute temperature.
At room temperature, \(27^\circ\text{C} = 300\,\text{K}\). Substituting this value, \[ \begin{aligned} \langle E \rangle &= \frac{3}{2}(1.38 \times 10^{-23})(300) \\ &\approx 6.2 \times 10^{-21}\,\text{J}. \end{aligned} \] Thus, the average thermal energy of a helium atom at room temperature is of the order of \(10^{-21}\,\text{J}\).
On the surface of the Sun, the temperature is approximately \(6000\,\text{K}\). The average thermal energy then becomes \[ \begin{aligned} \langle E \rangle &= \frac{3}{2}(1.38 \times 10^{-23})(6000) \\ &\approx 1.24 \times 10^{-19}\,\text{J}. \end{aligned} \] This shows a substantial increase in thermal energy due to the much higher temperature.
For a temperature of \(10^{7}\,\text{K}\), typical of stellar cores, the average thermal energy of a helium atom is \[ \begin{aligned} \langle E \rangle &= \frac{3}{2}(1.38 \times 10^{-23})(10^{7}) \\ &\approx 2.1 \times 10^{-16}\,\text{J}. \end{aligned} \]
Hence, the average thermal energy of a helium atom rises dramatically with temperature, ranging from about \(10^{-21}\,\text{J}\) at room temperature to about \(10^{-16}\,\text{J}\) in the extremely hot interiors of stars.
Q8. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest ?
Theory: According to Avogadro’s law, equal volumes of all gases at the same temperature and pressure contain the same number of molecules, regardless of the nature (monatomic, diatomic, or polyatomic) or molar mass of the gas. For an ideal gas, the state equation \[ PV = nRT \] shows that, for fixed \(P\), \(V\) and \(T\), the amount of gas \(n\) (and hence the number of molecules) must be the same in each vessel. On the other hand, the root mean square speed of gas molecules depends on both the temperature and the molar mass and is given by \[ v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}, \] so at a common temperature, lighter molecules (smaller \(M\)) have larger \(v_{\mathrm{rms}}\), and heavier molecules (larger \(M\)) move more slowly on average.
All three vessels have the same volume and contain gases at the same temperature and pressure. Under these conditions, each gas obeys the ideal gas equation \[ PV = nRT. \] Since \(P\), \(V\), and \(T\) are identical for all three vessels, the number of moles \(n\) of gas in each vessel must be the same.
The number of molecules present in a gas is directly proportional to the number of moles. Therefore, each vessel contains an equal number of molecules, irrespective of whether the gas is monatomic, diatomic, or polyatomic. The internal structure of the molecules does not affect the count of molecules present at given pressure, volume, and temperature.
The root mean square speed of gas molecules depends on the temperature and the molecular mass of the gas and is given by \[ v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}} \] where \(M\) is the molar mass of the gas. Although the temperature is the same for all three gases, their molar masses are different.
Neon, being monatomic, has a molar mass of about \(20\,\text{g mol}^{-1}\). Chlorine, which exists as a diatomic molecule, has a molar mass of about \(71\,\text{g mol}^{-1}\). Uranium hexafluoride is a heavy polyatomic molecule with a much larger molar mass. Since \(v_{\mathrm{rms}}\) varies inversely as the square root of the molar mass, the speeds of the molecules cannot be the same in the three vessels.
Because neon has the smallest molar mass among the three gases, its molecules have the highest root mean square speed. Chlorine molecules have a lower \(v_{\mathrm{rms}}\) than neon, while uranium hexafluoride molecules, being extremely heavy, have the smallest root mean square speed.
Q9. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Theory: The random thermal motion of gas molecules is quantified, in kinetic theory, by the root mean square (rms) speed, which for an ideal gas is given by \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}}, \] where \(R\) is the universal gas constant, \(T\) is the absolute temperature, and \(M\) is the molar mass of the gas. This expression shows that, at a fixed temperature, lighter molecules (smaller \(M\)) move faster on average, while heavier molecules (larger \(M\)) move more slowly. If two different gases are required to have the same rms speed, the condition \[ \sqrt{\frac{3RT_1}{M_1}} = \sqrt{\frac{3RT_2}{M_2}} \] leads to the proportionality \(\dfrac{T_1}{M_1} = \dfrac{T_2}{M_2}\); thus, the heavier gas must be at a proportionately higher temperature to match the rms speed of a lighter gas.
The root mean square speed of gas atoms depends on the absolute temperature and the atomic mass of the gas. It is given by \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where \(T\) is the temperature in kelvin and \(M\) is the molar mass of the gas.
The rms speed of argon atoms is stated to be equal to that of helium atoms at \(-20^\circ\text{C}\). Converting this temperature into kelvin gives \[ T_{\text{He}} = -20^\circ\text{C} = 253\,\text{K}. \] Let the required temperature of argon be \(T_{\text{Ar}}\).
Equating the rms speeds for argon and helium, \[ \begin{aligned} \sqrt{\frac{3RT_{\text{Ar}}}{M_{\text{Ar}}}} &= \sqrt{\frac{3RT_{\text{He}}}{M_{\text{He}}}}. \end{aligned} \] Squaring both sides and cancelling the common factor \(3R\), we obtain \[ \begin{aligned} \frac{T_{\text{Ar}}}{M_{\text{Ar}}} &= \frac{T_{\text{He}}}{M_{\text{He}}}. \end{aligned} \]
Substituting the given molar masses \(M_{\text{Ar}} = 39.9\,\text{u}\) and \(M_{\text{He}} = 4.0\,\text{u}\), \[ \begin{aligned} T_{\text{Ar}} &= T_{\text{He}} \frac{M_{\text{Ar}}}{M_{\text{He}}} \\ &= 253 \times \frac{39.9}{4.0} \\ &\approx 2.52 \times 10^{3}\,\text{K}. \end{aligned} \]
Thus, the temperature at which argon atoms have the same root mean square speed as helium atoms at \(-20^\circ\text{C}\) is approximately \(2.5 \times 10^{3}\,\text{K}\).
Q10. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of \(\mathrm{N_2}\) = 28.0 u).
Theory: In the kinetic theory of gases, molecules move in straight lines between brief collisions with other molecules, and the average distance travelled between two successive collisions is called the mean free path \(\lambda\). For a gas of hard spherical molecules, the mean free path is approximately \[ \lambda = \frac{kT}{\sqrt{2}\,\pi d^{2}P}, \] where \(k\) is the Boltzmann constant, \(T\) the absolute temperature, \(d\) the molecular diameter, and \(P\) the gas pressure. Once \(\lambda\) and the typical molecular speed (e.g. the root mean square speed \(v_{\text{rms}}\)) are known, the collision frequency \[ z \approx \frac{v_{\text{rms}}}{\lambda} \] gives the average number of collisions per second suffered by a molecule, and the time between successive collisions is \(\tau_{\text{free}} = \lambda / v_{\text{rms}}\), which is usually much larger than the very short duration of the collision itself.
The motion of a nitrogen molecule in the gas can be described using the kinetic theory of gases. The mean free path is the average distance travelled by a molecule between two successive collisions and is given by \[ \lambda = \frac{kT}{\sqrt{2}\,\pi d^{2}P}, \] where \(k\) is the Boltzmann constant, \(T\) the absolute temperature, \(d\) the molecular diameter, and \(P\) the pressure.
The temperature of the gas is \(17^\circ\text{C} = 290\,\text{K}\) and the pressure is \(2.0\,\text{atm} = 2.03 \times 10^{5}\,\text{Pa}\). The given radius of a nitrogen molecule is \(1.0\,\text{Å}\), so the molecular diameter is \[ d = 2.0\,\text{Å} = 2.0 \times 10^{-10}\,\text{m}. \]
Substituting the values, \[ \begin{aligned} \lambda &= \frac{(1.38 \times 10^{-23})(290)}{\sqrt{2}\,\pi (2.0 \times 10^{-10})^{2}(2.03 \times 10^{5})} \\ &\approx 1.1 \times 10^{-7}\,\text{m}. \end{aligned} \] Thus, the mean free path of a nitrogen molecule under the given conditions is of the order of \(10^{-7}\,\text{m}\).
To find the collision frequency, the root mean square speed of nitrogen molecules is required. It is given by \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where \(M = 28.0 \times 10^{-3}\,\text{kg mol}^{-1}\) is the molar mass of nitrogen.
Substituting the values, \[ \begin{aligned} v_{\text{rms}} &= \sqrt{\frac{3(8.31)(290)}{28.0 \times 10^{-3}}} \\ &\approx 5.1 \times 10^{2}\,\text{m s}^{-1}. \end{aligned} \]
The collision frequency \(z\), defined as the number of collisions suffered by a molecule per second, is approximately \[ \begin{aligned} z &= \frac{v_{\text{rms}}}{\lambda} \\ &= \frac{5.1 \times 10^{2}}{1.1 \times 10^{-7}} \\ &\approx 4.6 \times 10^{9}\,\text{s}^{-1}. \end{aligned} \]
The time for which a molecule moves freely between two successive collisions is \[ \begin{aligned} \tau_{\text{free}} &= \frac{\lambda}{v_{\text{rms}}} \\ &\approx 2.2 \times 10^{-10}\,\text{s}. \end{aligned} \] The actual collision time, which is the duration of interaction during a collision, is much smaller, typically of the order of \(10^{-12}\,\text{s}\).
Hence, a nitrogen molecule spends most of its time moving freely between collisions, while the collision itself occupies only a very small fraction of the total time.
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