NCERT Solutions Class 11 Physics Chapter 12 — Kinetic Theory

NCERT Class 11 Physics — Kinetic Theory, Chapter 12 Exercises. All 10 questions solved.

The exercises at the end of Chapter 12 fall into two groups. Questions 12.1 to 12.6 use the ideal gas equation, to count molecules and to follow a fixed amount of gas as its pressure, volume and temperature change. Questions 12.7 to 12.10 use the kinetic picture itself: the energy and speed of individual molecules, and how far a molecule travels between collisions. They run on three results:

• The ideal gas equation — $PV = \mu RT = Nk_BT$, where $\mu$ is the number of moles, $N$ the number of molecules, $R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}$ and $k_B = R/N_A = 1.38\times10^{-23}\ \text{J K}^{-1}$.
• Temperature and molecular speed — the average translational kinetic energy of a molecule is $\tfrac32k_BT$, so $v_{\text{rms}} = \sqrt{\dfrac{3k_BT}{m}} = \sqrt{\dfrac{3RT}{M}}$, where $m$ is the mass of one molecule and $M$ the molar mass.
• Mean free path — $\bar l = \dfrac{1}{\sqrt2\,\pi n d^2}$, where $n$ is the number of molecules per unit volume and $d$ is the diameter of a molecule.

Throughout, $1\ \text{atm} = 1.013\times10^5\ \text{Pa}$ and $N_A = 6.02\times10^{23}\ \text{mol}^{-1}$. Temperatures go into every formula in kelvin, $T = t + 273$, and molar masses in $\text{kg mol}^{-1}$.

Key insight. Temperature measures the average kinetic energy of a molecule, and nothing else. At a given temperature every gas, light or heavy, has the same $\tfrac32k_BT$ per molecule (12.7). So the lighter the molecule, the faster it moves (12.8), and two gases have the same rms speed when $T/M$ is the same for both (12.9). The same counting view explains why the ideal gas equation does not care which gas it is: equal volumes at the same pressure and temperature hold equal numbers of molecules (12.2, 12.6, 12.8).

Question 12.1

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 $3\ \text{Å}$.

Solution. Compare the same amount of gas both ways. At STP one mole of gas occupies $22.4\ \text{L} = 22.4\times10^{-3}\ \text{m}^3$ (Question 12.2). The molecules themselves fill only a small part of that. Treat each as a sphere of radius $r = d/2 = 1.5\ \text{Å} = 1.5\times10^{-10}\ \text{m}$; a mole contains $N_A$ of them, with a total volume

$$V_{\text{molecules}} = N_A\cdot\tfrac43\pi r^3 = 6.02\times10^{23}\times\tfrac43\pi\,(1.5\times10^{-10})^3 = 8.5\times10^{-6}\ \text{m}^3$$

The fraction is

$$\frac{8.5\times10^{-6}}{22.4\times10^{-3}} = 3.8\times10^{-4} \approx 4\times10^{-4}$$

So less than a tenth of one per cent of the gas is molecules; the rest is empty space. That is why kinetic theory can treat the molecules of a gas as points.

Question 12.2

Molar volume is the volume occupied by $1\ \text{mol}$ of any (ideal) gas at standard temperature and pressure (STP: $1$ atmospheric pressure, $0\ ^\circ\text{C}$). Show that it is $22.4$ litres.

Solution. For $\mu = 1\ \text{mol}$ the ideal gas equation gives $V = RT/P$. At STP, $T = 273\ \text{K}$ and $P = 1.013\times10^5\ \text{Pa}$:

$$V = \frac{8.31\times273}{1.013\times10^5} = 2.24\times10^{-2}\ \text{m}^3 = 22.4\ \text{L}$$

Nothing in the working depends on which gas it is, so the result holds for any gas that behaves ideally.

Question 12.3

Figure 12.8 shows plot of $PV/T$ versus $P$ for $1.00\times10^{-3}\ \text{kg}$ of oxygen gas at two different temperatures.

(a) What does the dotted plot signify? (b) Which is true: $T_1 > T_2$ or $T_1 < T_2$? (c) What is the value of $PV/T$ where the curves meet on the $y$-axis? (d) If we obtained similar plots for $1.00\times10^{-3}\ \text{kg}$ 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 $\text{H}_2 = 2.02\ \text{u}$, of $\text{O}_2 = 32.0\ \text{u}$, $R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}$.)

Question 12.3 (a)

Solution. For an ideal gas $PV = \mu RT$, so $PV/T = \mu R$: a constant for a fixed amount of gas, whatever the pressure. Plotted against $P$ it is a horizontal straight line. The dotted line is that line, the way the oxygen would behave if it were an ideal gas.

Question 12.3 (b)

Solution. A real gas comes closest to ideal behaviour at high temperature, where the molecules move so fast that the attractions between them matter least. So of two curves for the same gas, the one that stays closer to the dotted line is at the higher temperature. The $T_1$ curve departs less from the dotted line than the $T_2$ curve does, so $T_1$ is the higher temperature.

Question 12.3 (c)

Solution. Where the curves meet the $y$-axis the pressure tends to zero. The molecules are then very far apart, every real gas behaves ideally, and $PV/T = \mu R$. For $1.00\times10^{-3}\ \text{kg}$ of oxygen, with a molar mass of $32.0\times10^{-3}\ \text{kg mol}^{-1}$:

$$\mu = \frac{1.00\times10^{-3}}{32.0\times10^{-3}} = 3.125\times10^{-2}\ \text{mol}, \qquad \frac{PV}{T} = \mu R = 3.125\times10^{-2}\times8.31 = 0.26\ \text{J K}^{-1}$$

Question 12.3 (d)

Solution. No. $PV/T = \mu R$ depends on the number of moles, not on the mass. $1.00\times10^{-3}\ \text{kg}$ of hydrogen is $1.00/2.02 = 0.495\ \text{mol}$, about sixteen times as many moles as the same mass of oxygen, and it would give $PV/T = 0.495\times8.31 = 4.1\ \text{J K}^{-1}$. For the same value as the oxygen, we need the same $3.125\times10^{-2}\ \text{mol}$ of hydrogen:

$$m = \mu M = 3.125\times10^{-2}\times2.02\times10^{-3} = 6.3\times10^{-5}\ \text{kg}$$

Question 12.4

An oxygen cylinder of volume $30$ litre has an initial gauge pressure of $15\ \text{atm}$ and a temperature of $27\ ^\circ\text{C}$. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to $11\ \text{atm}$ and its temperature drops to $17\ ^\circ\text{C}$. Estimate the mass of oxygen taken out of the cylinder ($R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}$, molecular mass of $\text{O}_2 = 32\ \text{u}$).

Solution. A pressure gauge reads the excess over atmospheric pressure. The $P$ in $PV = \mu RT$ is the absolute pressure, the gauge reading plus $1\ \text{atm}$: $16\ \text{atm}$ at the start and $12\ \text{atm}$ at the end. The gas fills the cylinder in both states, so $V = 30\ \text{L} = 3.0\times10^{-2}\ \text{m}^3$ throughout. The number of moles in each state is $\mu = PV/RT$:

$$\mu_1 = \frac{16\times1.013\times10^5\times3.0\times10^{-2}}{8.31\times300} = 19.5\ \text{mol}, \qquad \mu_2 = \frac{12\times1.013\times10^5\times3.0\times10^{-2}}{8.31\times290} = 15.1\ \text{mol}$$

The difference, $4.37\ \text{mol}$, is what was withdrawn. With a molar mass of $32\times10^{-3}\ \text{kg mol}^{-1}$:

$$m = 4.37\times32\times10^{-3} = 0.14\ \text{kg}$$

Question 12.5

An air bubble of volume $1.0\ \text{cm}^3$ rises from the bottom of a lake $40\ \text{m}$ deep at a temperature of $12\ ^\circ\text{C}$. To what volume does it grow when it reaches the surface, which is at a temperature of $35\ ^\circ\text{C}$?

Solution. The bubble holds a fixed amount of air, so $PV/T$ keeps the same value as it rises:

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

At the bottom, the pressure is the atmospheric pressure on the lake’s surface plus the pressure of $40\ \text{m}$ of water. Taking $\rho = 1000\ \text{kg m}^{-3}$ and $g = 9.8\ \text{m s}^{-2}$:

$$P_1 = 1.013\times10^5 + 1000\times9.8\times40 = 4.93\times10^5\ \text{Pa}$$

At the surface $P_2 = 1.013\times10^5\ \text{Pa}$. The temperatures are $T_1 = 285\ \text{K}$ and $T_2 = 308\ \text{K}$. So

$$V_2 = V_1\,\frac{P_1}{P_2}\,\frac{T_2}{T_1} = 1.0\times\frac{4.93\times10^5}{1.013\times10^5}\times\frac{308}{285} = 5.3\ \text{cm}^3$$

Both changes make the bubble grow: the pressure falls to about a fifth, and the air warms up.

Question 12.6

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity $25.0\ \text{m}^3$ at a temperature of $27\ ^\circ\text{C}$ and $1\ \text{atm}$ pressure.

Solution. The ideal gas equation in the form $PV = Nk_BT$ counts molecules directly. It does not matter that air is a mixture: every molecule, of whatever kind, counts once.

$$N = \frac{PV}{k_BT} = \frac{1.013\times10^5\times25.0}{1.38\times10^{-23}\times300} = 6.1\times10^{26}$$

Note on the book’s answer. The key prints $6.10\times10^{26}$; the working above gives $6.12\times10^{26}$. The difference comes only from the value taken for atmospheric pressure ($1.01\times10^5\ \text{Pa}$ gives $6.10\times10^{26}$), so it is rounding, not an error.

Question 12.7

Estimate the average thermal energy of a helium atom at (i) room temperature ($27\ ^\circ\text{C}$), (ii) the temperature on the surface of the Sun ($6000\ \text{K}$), (iii) the temperature of $10$ million kelvin (the typical core temperature in the case of a star).

Solution. Helium is monatomic, so its thermal energy is entirely the kinetic energy of the atom’s motion from place to place, which on average is $\tfrac32k_BT$. The energy is simply proportional to the absolute temperature:

$$\text{(i)}\quad \tfrac32\times1.38\times10^{-23}\times300 = 6.2\times10^{-21}\ \text{J}$$

$$\text{(ii)}\quad \tfrac32\times1.38\times10^{-23}\times6000 = 1.24\times10^{-19}\ \text{J}$$

$$\text{(iii)}\quad \tfrac32\times1.38\times10^{-23}\times10^7 = 2.1\times10^{-16}\ \text{J}$$

In more familiar units, that is about $0.04\ \text{eV}$ at room temperature and about $1.3\ \text{keV}$ in the core of a star.

Question 12.8

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 $v_{\text{rms}}$ the largest?

Solution. Numbers of molecules. Yes. Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules; that is Avogadro’s law, and it is just $PV = Nk_BT$ solved for $N$. Whether a molecule has one atom or seven makes no difference to the count.

Speeds. No. At the same temperature the molecules of all three gases have the same average kinetic energy, $\tfrac12m\overline{v^2} = \tfrac32k_BT$, so $v_{\text{rms}} = \sqrt{3k_BT/m}$ is largest for the lightest molecule. The molar masses are about $20\ \text{g mol}^{-1}$ for neon, $71\ \text{g mol}^{-1}$ for chlorine and $352\ \text{g mol}^{-1}$ for uranium hexafluoride. Neon is therefore the fastest: its rms speed is about $1.9$ times that of chlorine and $4.2$ times that of uranium hexafluoride.

Question 12.9

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 $-20\ ^\circ\text{C}$? (atomic mass of Ar $= 39.9\ \text{u}$, of He $= 4.0\ \text{u}$).

Solution. Since $v_{\text{rms}} = \sqrt{3RT/M}$, two gases have the same rms speed when $T/M$ is the same for both. The helium is at $-20\ ^\circ\text{C} = 253\ \text{K}$:

$$\frac{T_{\text{Ar}}}{39.9} = \frac{253}{4.0} \quad\Rightarrow\quad T_{\text{Ar}} = 253\times\frac{39.9}{4.0} = 2.52\times10^3\ \text{K}$$

An argon atom is ten times as heavy as a helium atom, so the argon has to be ten times as hot to move as fast.

Question 12.10

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0\ \text{atm}$ and temperature $17\ ^\circ\text{C}$. Take the radius of a nitrogen molecule to be roughly $1.0\ \text{Å}$. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of $\text{N}_2 = 28.0\ \text{u}$).

Solution. Number density. From $P = nk_BT$, with $P = 2.0\ \text{atm} = 2.026\times10^5\ \text{Pa}$ and $T = 290\ \text{K}$:

$$n = \frac{P}{k_BT} = \frac{2.026\times10^5}{1.38\times10^{-23}\times290} = 5.1\times10^{25}\ \text{m}^{-3}$$

Mean free path. The formula needs the diameter, not the radius: $d = 2\times1.0\ \text{Å} = 2.0\times10^{-10}\ \text{m}$.

$$\bar l = \frac{1}{\sqrt2\,\pi n d^2} = \frac{1}{1.414\times3.14\times5.1\times10^{25}\times(2.0\times10^{-10})^2} = 1.1\times10^{-7}\ \text{m}$$

Speed. $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3\times8.31\times290}{28.0\times10^{-3}}} = 5.1\times10^2\ \text{m s}^{-1}$.

Collision frequency. The molecule covers a distance $\bar l$ between one collision and the next, so it collides $v_{\text{rms}}/\bar l$ times each second:

$$\nu = \frac{v_{\text{rms}}}{\bar l} = \frac{5.1\times10^2}{1.1\times10^{-7}} = 4.6\times10^9\ \text{s}^{-1}$$

Comparison. A collision lasts roughly as long as the molecule takes to travel its own diameter, $d/v_{\text{rms}} = 2.0\times10^{-10}/(5.1\times10^2) = 3.9\times10^{-13}\ \text{s}$. The time between collisions is $\bar l/v_{\text{rms}} = 1/\nu = 2.2\times10^{-10}\ \text{s}$. Their ratio is $\bar l/d \approx 550$. A molecule spends several hundred times as long moving freely as it does colliding, which is why kinetic theory can treat the molecules as free particles that only occasionally collide.

Note on the book’s answer. The key rounds the mean free path to $1.0\times10^{-7}\ \text{m}$ before dividing, which gives a collision frequency of $5.1\times10^9\ \text{s}^{-1}$ and a ratio of $500$. Unrounded, the figures are $1.1\times10^{-7}\ \text{m}$, $4.6\times10^9\ \text{s}^{-1}$ and about $550$. For an estimate they agree; the difference is rounding, not an error.

Common mistakes

• Question 12.1: using the diameter as the radius. The volume of a sphere is $\tfrac43\pi r^3$ with $r = d/2 = 1.5\ \text{Å}$. Putting in $3\ \text{Å}$ makes the molecular volume eight times too large.
• Question 12.3(b): picking the curve that dips further as the hotter one. A larger dip looks like “more happening”. But heating a real gas brings it closer to ideal behaviour, so the hotter curve is the one nearer the dotted line, $T_1$.
• Question 12.3(d): expecting the same value for the same mass. $PV/T = \mu R$ counts moles, not kilograms. A gram of hydrogen is about sixteen times as many moles as a gram of oxygen.
• Question 12.4: putting the gauge pressure into $PV = \mu RT$. The gas equation needs the absolute pressure, gauge plus atmospheric. Here the slip happens to leave the mass near $0.14\ \text{kg}$, which makes it easy to miss, but the number of moles in each state comes out wrong.
• Question 12.5: forgetting the atmosphere above the lake. The pressure at the bottom is $P_0 + \rho gh$, not $\rho gh$ alone. Leaving out $P_0$ gives $4.2\ \text{cm}^3$ instead of $5.3\ \text{cm}^3$.
• Questions 12.5 and 12.9: leaving temperatures in degrees Celsius. The gas laws need kelvin. In 12.5 the ratio $35/12$ is nearly $3$, while the correct $308/285$ is under $1.1$; in 12.9, $-20$ would give a negative temperature.
• Question 12.8: expecting fewer molecules of the heavier or more complex gas. The gas equation counts molecules, whatever they are made of, so equal volumes at equal $P$ and $T$ hold equal numbers. It is the speed that depends on the mass.
• Question 12.10: using the radius in the mean free path formula. The formula has the diameter $d$. With $1.0\ \text{Å}$ in place of $2.0\ \text{Å}$, the mean free path comes out four times too large.

Practise next

• Chapter 13, Oscillations — the next chapter; its simple harmonic motion is also the model for the vibrations of the molecules met here.
• Chapter 11, Thermodynamics — the pressure, volume, temperature and internal energy that kinetic theory explains molecule by molecule.