KINETIC THEORY-Exercise

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.

Solution

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.

Solution

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.

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).

Solution

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 \(1.0\ cm^3\) 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 ?

Solution

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.

Solution

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).

Solution

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 ?

Solution

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).

Solution

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 0C. 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).

Solution

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.