NCERT Solutions for Class 11 Physics Chapter 13

Q.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 3A.

Ans.

\begin{array}{l}\text{Given,}\text{diameter of the oxygen molecule,}\text{d}\text{=}\text{3A}\\ \therefore \text{Radius}\text{of the oxygen molecule,}\text{r}\text{=}\frac{\text{d}}{\text{2}}\\ \text{=}\frac{3}{2}\\ \text{=1}.\text{5}\stackrel{\text{o}}{\text{A}}\\ =\text{1}.\text{5}\times \text{1}{0}^{–\text{8}}\text{cm}\\ \text{The}\text{actual volume occupied by 1 mole of oxygen}\text{gas at}\text{STP,}{\text{V}}_{\text{A}}\text{=}{\text{22400 cm}}^{\text{3}}\\ \text{Molecular volume of oxygen gas}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ {\text{V}}_{\text{m}}=\frac{4}{3}\pi {\text{r}}^3\text{N}\\ \text{Here,}\text{N}\text{=}\text{Avogadro’s number}\text{=}\text{6}.0\text{23}\times \text{1}{0}^{\text{23}}\text{molecules}{\text{mole}}^{\text{-1}}\\ \therefore {\text{V}}_{\text{m}}=\frac{4}{3}\times 3.14\times {\left(1.5\times {10}^{-8}\right)}^3\times \text{6}.0\text{23}\times \text{1}{0}^{\text{23}}\\ \therefore {\text{V}}_{\text{m}}=8.51c{m}^3\\ \therefore \text{Ratio of molecular volume to the actual volume of}\text{oxygen}\text{gas}\text{=}\frac{{\text{V}}_{\text{m}}}{{\text{V}}_{\text{A}}}\\ =\frac{8.51}{\text{224}00}\\ \text{=3}.\text{8}\times \text{1}{0}^{–\text{4}}\end{array}

Q.2 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.

Ans.

\begin{array}{l}\text{The ideal gas equation}\text{is}\text{given}\text{as:}\\ \text{PV}=\text{nRT}\\ \text{Here,}\\ \text{R}=\text{Universal gas constant}\\ =\text{8}.{\text{314 Jmol}}^{–\text{1}}{\text{K}}^{–\text{1}}\\ \text{n}=\text{Number of moles}\\ \text{For}\text{one}\text{mole}\text{of}\text{gas,}\text{n}=1\\ \text{Standard temperature,}\text{T}=\text{273 K}\\ \text{Standard pressure,}\text{P}=\text{1 atm}\\ =\text{1}.0\text{13}\times \text{1}{0}^{\text{5}}{\text{Nm}}^{–\text{2}}\\ \therefore \text{Molar}\text{volume}\text{of}\text{an}\text{ideal}\text{gas}\text{at}\text{STP}\text{is}\text{given}\text{as:}\\ \text{V}=\frac{\text{nRT}}{\text{P}}=\frac{1\times \text{8}.\text{314}\times \text{273}}{\text{1}.0\text{13}\times \text{1}{0}^{\text{5}}}=0.0{\text{224 m}}^{\text{3}}\\ \text{V}=\text{22}.\text{4 litres}\\ \therefore \text{Molar volume of a gas at STP}\text{=}\text{22}.\text{4 litres}.\end{array}

Q.3 Figure 13.8 shows plot of PV/T versus P for 1.00×10^–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 ×10^–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 J mo1^–1K^–1.)

Ans.

\begin{array}{l}\left(\text{a}\right)\text{The dotted plot is a straight line parallel to pressure axis}\text{. It shows the ideal behavior of the gas}\text{.}\\ \text{It signifies that the quantity}\frac{\text{PV}}{\text{T}}\text{is constant and it does not depend upon the pressure of the gas}\text{.}\\ \text{Thus, the dotted plot is for an ideal gas}\text{.}\\ \left(\text{b}\right)\text{As the temperature increases, a real gas approaches the behavior of an ideal gas}\text{. In the given}\\ {\text{plot, the curve of the gas at temperature T}}_{\text{1}}\text{is closer to the dotted plot than the curve of the gas at}\\ {\text{temperature T}}_{\text{2}}{\text{. Thus, we conclude that T}}_{\text{1}}\text{>}{\text{T}}_{\text{2}}\text{for the given plot}\text{.}\\ \text{(c)}\text{Where the two curves}\text{meet,}\text{the value of}\frac{\text{PV}}{\text{T}}\text{is}\mu \text{R}.\\ \text{Since ideal gas equation is given as:}\\ \text{PV}=\mu \text{RT}\\ \therefore \frac{\text{PV}}{\text{T}}=\mu \text{R}\to \text{(i)}\\ \text{Here},\mu \text{and}\text{R represent}\text{number of moles}\text{and Universal}\text{gas}\text{constant}\text{respectively}\text{.}\\ \text{Since}\text{mass of oxygen}=\text{1}\times \text{1}{0}^{–\text{3}}\text{kg}=\text{1 g}\\ \text{Molecular mass of oxygen}=\text{32}.0\text{g}\\ \therefore \mu \text{=}\frac{1}{32}\to \text{(ii)}\\ \text{R}=\text{8}.{\text{314 J mole}}^{–\text{1}}{\text{K}}^{–\text{1}}\to \text{(iii)}\\ \text{From}\text{equation}\text{(i),}\text{(ii)}\text{and}\text{(iii),}\text{we}\text{obtain:}\\ \frac{\text{PV}}{\text{T}}=\frac{1}{32}\times \text{8}.\text{314}=0.26{\text{JK}}^{-1}\\ \therefore \text{The value of}\frac{\text{PV}}{\text{T}}\text{where the curves meet on}\text{the}\text{y}-\text{axis}=0.{\text{26 JK}}^{\text{-1}}\end{array} \begin{array}{l}\text{(d)}\text{Since}\text{the}\text{molecular masses of hydrogen}\left(\text{2}\text{.02 u}\right)\text{and}\text{oxygen}\left(\text{32}\text{.0 u}\right)\text{are different,}\text{therefore,}\text{if we obtained}\\ \text{similar graphs for 1}.00\times \text{1}{0}^{–\text{3}}\text{kg of}\text{hydrogen,}\text{then}\text{we}\text{will not get}\text{the same value of}\frac{\text{PV}}{\text{T}}\text{at the point where the}\\ \text{curves}\text{meet the}\text{y}-\text{axis}.\\ \text{Here,}\frac{\text{PV}}{\text{T}}=0.26{\text{JK}}^{-1}\\ \text{R}=\text{8}.{\text{314 Jmole}}^{–\text{1}}{\text{K}}^{–\text{1}}\\ \text{Molecular mass of Hydrogen,}\text{M}=\text{2}.0\text{2 u}\\ \text{At}\text{constant}\text{temperature},\frac{\text{PV}}{\text{T}}=\mu R\to \text{(i)}\\ \text{Let}\text{mass}\text{of}{\text{H}}_2=m\\ \therefore \text{Number}\text{of}\text{moles,}\mu =\frac{\text{m}}{\text{M}}\to \text{(ii)}\\ \text{From}\text{equation}\text{(i)}\text{and}\text{(ii),}\text{we}\text{have:}\\ \frac{\text{PV}}{\text{T}}=\frac{\text{m}}{\text{M}}\text{R}\\ \text{Substituting}\text{values}\text{in}\text{the}\text{above}\text{relation,}\text{we}\text{obtain:}\\ 0.26=\frac{m}{2.02}\times \text{8}.\text{314}\\ \therefore \text{m}\text{=}\frac{0.26\times 2.02}{\text{8}.\text{314}}=\text{6}.\text{3}\times \text{1}{0}^{–\text{2}}\text{g}=6.3\times {10}^{-5}\text{kg}\\ \therefore \text{Same value of}\frac{\text{PV}}{\text{T}}\text{will}\text{be}\text{obtained}\text{from}\text{6}.\text{3}\times \text{1}{0}^{–\text{5}}{\text{kg of H}}_{\text{2}}.\end{array}

Q.4 An oxygen cylinder of volume 30 litres 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 (R = 8.31J mol^–1 K^–1, molecular mass of O₂ = 32 u).

Ans.

\begin{array}{l}\text{Here,}\text{initial}\text{volume of oxygen,}{\text{V}}_{\text{i}}\text{=}\text{30 litres}\\ =\text{3}0\times \text{1}{0}^{–\text{3}}{\text{m}}^{\text{3}}\\ \text{Initial}\text{gauge pressure,}{\text{P}}_{\text{i}}\text{=}\text{15 atm}\\ =\text{15}\times \text{1}.0\text{13}\times \text{1}{0}^{\text{5}}\text{Pa}\\ \text{=}\text{15}\text{.195}\times \text{1}{0}^{\text{5}}\text{Pa}\\ \text{Initial}\text{temperature},{\text{T}}_i=\text{27}°\text{C}\\ =\text{3}00\text{K}\\ \text{Universal gas constant,}\text{R}=\text{8}.{\text{314 Jmole}}^{–\text{1}}{\text{K}}^{–\text{1}}\\ \text{Molecular}\text{mass}\text{of}{\text{O}}_{\text{2}}\text{,}\text{M}=32\text{g}\\ \text{Let}\text{initial mass of oxygen}\text{=}{\text{m}}_{\text{i}}\\ \text{Let the initial number of moles of oxygen gas in the}\text{cylinder}\text{=}{\text{n}}_{\text{i}}\\ \text{The gas equation is given by:}\\ {\text{P}}_i{\text{V}}_i={\text{n}}_i{\text{RT}}_i\\ \therefore {\text{n}}_i=\frac{{\text{P}}_i{\text{V}}_i}{{\text{RT}}_i}\\ =\frac{\text{15}\text{.195}\times \text{1}{0}^{\text{5}}\times \text{3}0\times \text{1}{0}^{–\text{3}}}{\text{8}.\text{314}\times \text{3}00}\\ =\text{18}.\text{276}\\ \text{Since,}{\text{n}}_i=\frac{m_i}{M}\\ \therefore {\text{m}}_i={\text{n}}_i\text{M}\\ =\text{18}.\text{276}\times \text{32}\\ =\text{584}.\text{84 g}\\ \text{Since}\text{initial}\text{values}\text{of}\text{pressure}\text{and}\text{temperature}\text{decrease,}\\ \text{after withdrawing}\text{some oxygen from the cylinder,}\\ \text{Given,}\text{final}\text{volume}\text{of}\text{oxygen,}{\text{V}}_{\text{f}}\text{=}\text{30 litres}\\ =\text{3}0\times \text{1}{0}^{–\text{3}}{\text{m}}^{\text{3}}\\ \text{Final}\text{gauge pressure,}{\text{P}}_{\text{f}}\text{=}\text{11 atm}\\ =\text{11}\times \text{1}.0\text{13}\times \text{1}{0}^{\text{5}}\text{Pa}\\ \text{=}\text{11}\text{.143}\times \text{1}{0}^{\text{5}}\text{Pa}\\ \text{Final}{\text{temperature,T}}_f=\text{17}°\text{C}=\text{29}0\text{K}\\ \text{Let the number of moles of oxygen left in the cylinder}\text{=}{\text{n}}_{\text{f}}\\ \text{The gas equation is given as:}\\ {\text{P}}_f{\text{V}}_f={\text{n}}_f{\text{RT}}_f\\ \therefore {\text{n}}_f=\frac{{\text{P}}_f{\text{V}}_f}{{\text{RT}}_f}\\ =\frac{\text{11}\text{.143}\times \text{1}{0}^{\text{5}}\times \text{3}0\times \text{1}{0}^{–\text{3}}}{\text{8}.\text{314}\times 290}\\ =\text{13}.\text{86}\\ \text{But,}\text{number}\text{of}\text{moles}\text{left},{\text{n}}_f=\frac{m_f}{M}\\ \text{Here,}{\text{m}}_{\text{f}}=\text{Mass of oxygen left in the cylinder}\\ \therefore {\text{m}}_{\text{f}}={\text{n}}_f\text{M}\\ =\text{13}.\text{86}\times \text{32}\\ =\text{453}.\text{1 g}\\ \text{Mass of oxygen withdrawn from the cylinder,}\\ \Delta \text{M}\text{=}\text{Initial mass of oxygen–Final mass of oxygen}\\ \Delta \text{M}={\text{m}}_i–{\text{m}}_f\\ =\text{584}.\text{84 g}–\text{453}.\text{1 g}\\ \text{=131}\text{.74}\text{g}\\ \therefore \Delta \text{M=0}\text{.131}\text{kg}\\ \therefore \text{0}\text{.131}\text{kg}\text{of}\text{oxygen is}\text{withdrawn}\text{from the cylinder}\text{.}\end{array}

Q.5 An air bubble of volume 1.0 cm³ 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?

Ans.

\begin{array}{l}\text{Here,}\text{initial}\text{volume of the air bubble,}\text{V}=\text{1}.0{\text{cm}}^{\text{3}}\\ =\text{1}.0\times \text{1}{0}^{–\text{6}}{\text{m}}^{\text{3}}\\ \text{Height}\text{upto}\text{which}\text{the}\text{air}\text{bubble}\text{rises}\text{,}\text{d}\text{=}\text{40 m}\\ \text{Temperature at the}\text{bottom}\text{of}\text{the}{\text{lake,T}}_{\text{1}}\text{=}\text{12°C}\\ =\text{285 K}\\ {\text{Temperature at the surface of the lake,T}}_{\text{s}}=\text{35}°\text{C}\\ =\text{3}0\text{8 K}\\ \text{Pressure on the surface of the lake,}{\text{P}}_s=\text{1 atm}\\ =\text{1}\times \text{1}.0\text{13}\times \text{1}{0}^{\text{5}}\text{Pa}\\ \text{Pressure at the bottom}\text{of}\text{the}\text{lake,}{\text{P}}_{\text{b}}=\text{1 atm}+\text{d}\rho \text{g}\\ \text{Here,}\\ \rho =\text{Density of water}\\ =\text{1}{0}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ \text{g =Acceleration due to gravity}\\ =\text{9}.{\text{8 ms}}^{\text{-2}}\\ \therefore {\text{P}}_b=\text{1}.0\text{13}\times \text{1}{0}^{\text{5}}+\text{4}0\times \text{1}{0}^{\text{3}}\times \text{9}.\text{8}\\ =\text{4933}00\text{Pa}\\ \text{From}\text{the}\text{relation,}\\ \frac{P_b{V}_b}{T_b}=\frac{P_s{V}_s}{T_s}\\ \text{Here,}{\text{V}}_{\text{s}}\text{=}\text{Volume of the air bubble at the surface}\text{of}\text{lake}\\ {\text{V}}_s\text{=}\frac{P_b{V}_b{T}_s}{T_b{P}_s}=\frac{\text{4933}00\times \text{1}.0\times \text{1}{0}^{–\text{6}}\times \text{3}0\text{8}}{\text{285}\times \text{1}.0\text{13}\times \text{1}{0}^{\text{5}}}\\ =\text{5}.\text{263}\times \text{1}{0}^{–\text{6}}{\text{m}}^{\text{3}}\text{=}\text{5}.{\text{263 cm}}^{\text{3}}\\ \therefore \text{When the air bubble reaches the surface,}\text{its volume = 5}.{\text{263 cm}}^{\text{3}}\end{array}

Q.6 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.

Ans.

\begin{array}{l}\text{Here,}\text{volume of room,V}\text{=}\text{25}{\text{.0 m}}^{\text{3}}\\ \text{Temperature of room,T}\text{=}\text{27}°\text{C}=\text{300 K}\\ \text{Pressure inside the room,}\text{P}\text{=}\text{1 atm}\text{=}\text{1}\text{×}\text{1}\text{.013}\text{×}{\text{10}}^{\text{5}}\text{Pa}\\ \text{The}\text{ideal}\text{gas}\text{equation}\text{is}\text{given}\text{as:}\\ \text{PV}={\text{k}}_{\text{B}}\text{NT}\\ \text{Here},{\text{K}}_{\text{B}}\text{=}\text{Boltzmann constant}\\ =\text{1}.\text{38}\times \text{1}{0}^{–\text{23}}{\text{m}}^{\text{2}}\text{kg}{\text{s}}^{\text{–2}}{\text{K}}^{\text{–1}}\\ \text{N}\text{=}\text{Number}\text{of}\text{air}\text{molecules}\text{inside}\text{the}\text{room}\\ \therefore \text{N}\text{=}\frac{\text{PV}}{{\text{k}}_{\text{B}}\text{T}}=\frac{\text{1}.0\text{13}\times \text{1}{0}^{\text{5}}\times \text{25}}{\text{1}.\text{38}\times \text{1}{0}^{–\text{23}}\times \text{3}00}\\ =\text{6}.\text{11}\times \text{1}{0}^{\text{26}}\text{molecules}\\ \therefore \text{Total number of air molecules in the given room}\\ \text{= 6}.\text{11}\times \text{1}{0}^{\text{26}}\end{array}

Q.7 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).

Ans.

\begin{array}{l}\left(\text{i}\right)\text{At room temperature,}\\ \text{Temperature,}\text{T}=\text{27}°\text{C}=\text{3}00\text{K}\\ \text{Average thermal energy}\text{=}\frac{3}{2}\text{kT}\\ \text{Where}\text{k=Boltzmann constant}\\ =\text{1}.\text{38}\times \text{1}{0}^{–\text{23}}{\text{m}}^{\text{2}}\text{kg}{\text{s}}^{–\text{2}}{\text{K}}^{–\text{1}}\\ \therefore \text{Average}\text{thermal}\text{energy}=\frac{3}{2}\text{kT}\\ =\frac{3}{2}\times \text{1}.\text{38}\times \text{1}{0}^{–\text{23}}\times \text{3}00\\ =\text{6}.\text{21}\times \text{1}{0}^{–\text{21}}\text{J}\\ \therefore \text{Average thermal energy of a helium atom at room temperature}\left(\text{27}°\text{C}\right)\text{is}\text{6}.\text{21}\times \text{1}{0}^{–\text{21}}\text{J}.\\ \left(\text{ii}\right)\text{On the surface of the sun,}\\ \text{Temperature,}\text{T}\text{=}\text{6000 K}\\ \text{Average thermal energy}\text{=}\frac{\text{3}}{\text{2}}\text{kT}\\ =\frac{3}{2}\times \text{1}.\text{38}\times \text{1}{0}^{–\text{23}}\times 6000\\ \text{=}\text{1}.\text{241}\times \text{1}{0}^{–\text{19}}\text{J}\\ \therefore \text{Average thermal energy of a helium atom on the}\text{surface of the sun is}\text{1}.\text{241}\times \text{1}{0}^{–\text{19}}\text{J}.\\ \left(\text{iii}\right)\text{At temperature,T}=\text{1}{0}^{\text{7}}\text{K}\\ \text{Average thermal energy}\text{=}\frac{3}{2}\text{kT}\\ =\frac{3}{2}\times \text{1}.\text{38}\times \text{1}{0}^{–\text{23}}\times \text{1}{0}^{\text{7}}\\ =\text{2}.0\text{7}\times \text{1}{0}^{–\text{16}}\text{J}\\ \therefore \text{Average thermal energy of a helium atom at the core of a star}\text{=}\text{2}.0\text{7}\times \text{1}{0}^{–\text{16}}\text{J}\end{array}

Q.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_rms the largest?

Ans.

\begin{array}{l}\text{Since all the three vessels have same volume at same temperature and pressure,}\\ \text{Therefore, according to Avogadro’s law, all the three vessels contain the same number of the respective molecules}\text{.}\\ {\text{The root mean square speed (v}}_{\text{rms}}\text{) of a gas of mass}\text{m,}\text{and temperature}\text{T, is given as:}\\ {\text{v}}_{\text{rms}}=\sqrt{\frac{3kT}{m}}\\ \text{Where,}\text{k}\text{=}\text{Boltzmann constant}\\ \text{Since}\text{for}\text{the}\text{given}\text{gases,}\text{k}\text{and}\text{T}\text{are}\text{constants,}\\ \therefore {\text{v}}_{\text{rms}}\propto \sqrt{\frac{1}{m}}\\ \therefore \text{The}\text{root}\text{mean}\text{square}\text{speed}\text{of}\text{the}\text{molecules}\text{is}\text{not}\text{the}\text{same}\text{in}\text{the}\text{three}\text{cases}\text{.}\\ \text{Out of the given gases neon has the smallest mass, therefore, root mean square}\text{speed of neon is the largest}\text{.}\end{array}

Q.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°C? (Atomic mass of Ar = 39.9 u, of He = 4.0 u).

Ans.

\begin{array}{l}\text{Here,}\text{Temperature of helium atom,}{\text{T}}_{\text{He}}=–\text{2}0°\text{C}\\ =\text{253 K}\\ \text{Atomic mass of the}\text{argon}\text{gas,}{\text{M}}_{\text{Ar}}=\text{39}.\text{9 u}\\ \text{Atomic mass of the}\text{helium}\text{gas},{\text{M}}_{\text{He}}=\text{4}.0\text{u}\\ \text{Let}\text{rms speed of argon}\text{gas}\text{=}{\left({\text{v}}_{\text{rms}}\right)}_{\text{Ar}}\\ \text{Let rms speed of helium}\text{gas}\text{=}{\left({\text{v}}_{\text{rms}}\right)}_{\text{He}}\\ \text{Rms speed of the}\text{argon gas}\text{is given as:}\\ {\left({\text{v}}_{\text{rms}}\right)}_{\text{Ar}}=\sqrt{\frac{3R{T}_{Ar}}{M_{Ar}}}\to \left(\text{i}\right)\\ \text{Where,}\text{R}\text{=}\text{Universal gas constant}\\ {\text{T}}_{\text{Ar}}\text{=}\text{Temperature of argon gas}\\ \text{Rms speed of the}\text{helium gas}\text{is given as:}\\ {\left({\text{v}}_{\text{rms}}\right)}_{\text{He}}=\sqrt{\frac{3R{T}_{He}}{M_{He}}}\to \left(\text{ii}\right)\\ \text{As}\text{per}\text{the}\text{question:}\\ {\left({\text{v}}_{\text{rms}}\right)}_{\text{Ar}}={\left({\text{v}}_{\text{rms}}\right)}_{\text{He}}\\ \sqrt{\frac{3R{T}_{Ar}}{M_{Ar}}}=\sqrt{\frac{3R{T}_{He}}{M_{He}}}\\ \frac{T_{Ar}}{M_{Ar}}=\frac{T_{He}}{M_{He}}\\ T_{Ar}=\frac{T_{He}}{M_{He}}\times M_{Ar}=\frac{253}{4}\times 39.9\\ =\text{2523}.\text{675}=\text{2}.\text{52}\times \text{1}{0}^{\text{3}}\text{K}\\ \therefore \text{Temperature of the argon gas}\text{atom}\text{=}\text{2}.\text{52}\times \text{1}{0}^{\text{3}}\text{K}\end{array}

Q.10 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 N₂ = 28.0 u).

Ans.

\begin{array}{l}\text{Here,}\text{pressure inside the cylinder,}\text{P = 2}\text{.0 atm = 2}{\text{.026×10}}^{\text{5}}\text{Pa}\\ \text{Temperature inside the cylinder,}\text{T = 17°C = 290 K}\\ \text{Radius of a nitrogen molecule,}\text{r = 1}{\text{.0 Å = 1×10}}^{\text{10}}\text{m}\\ \text{Diameter,}{\text{d = 2r = 2×1×10}}^{\text{10}}{\text{m = 2×10}}^{\text{10}}\text{m}\\ \text{Molecular mass of nitrogen,}{\text{M}}_{\text{N}}\text{= 28}{\text{.0 g = 28×10}}^{\text{–3}}\text{kg}\\ \text{The root mean square speed of nitrogen is given as:}\\ {\text{v}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3RT}}{\text{M}}}\\ \text{Where,R = Universal gas constant = 8}{\text{.314 Jmol}}^{\text{–1}}{\text{K}}^{\text{–1}}\\ \therefore {\text{v}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3×8}\text{.314×290}}{{\text{28×10}}^{\text{-3}}}}\text{= 508}{\text{.26 ms}}^{\text{-1}}\\ \text{Mean free path, l =}\frac{\text{kT}}{\sqrt{\text{2}}{\text{×d}}^{\text{2}}\text{×P}}\\ \text{Here,}\text{k=Boltzmann constant = 1}{\text{.38×10}}^{\text{–23}}{\text{kgm}}^{\text{2}}{\text{s}}^{\text{–2}}{\text{K}}^{\text{–1}}\\ \therefore \text{l =}\frac{\text{1}{\text{.38×10}}^{\text{–23}}{\text{kgm}}^{\text{2}}{\text{s}}^{\text{–2}}{\text{K}}^{\text{–1}}\text{×290 K}}{\sqrt{\text{2}}\text{×3}\text{.14×}{\left({\text{2×10}}^{\text{-10}}\text{m}\right)}^{\text{2}}\text{×2}{\text{.026×10}}^{\text{5}}Pa}\text{= 1}{\text{.11×10}}^{\text{–7}}\text{m}\\ \text{Collision frequency}\text{=}\frac{{\text{v}}_{\text{rms}}}{\text{l}}\text{=}\frac{\text{508}\text{.26}}{\text{1}{\text{.11×10}}^{\text{–7}}}\text{= 4}{\text{.58×10}}^{\text{9}}{\text{s}}^{\text{–1}}\\ \text{Collision time is given by}\text{the}\text{relation:}\\ \text{T =}\frac{\text{d}}{{\text{v}}_{\text{rms}}}\\ \text{T =}\frac{{\text{2×10}}^{\text{-10}}}{\text{508}\text{.26}}\text{= 3}{\text{.93×10}}^{\text{-13}}\text{s}\\ \text{Time taken between successive collisions}\text{is}\text{given}\text{as:}\\ \text{T’ =}\frac{\text{l}}{{\text{v}}_{\text{rms}}}\text{=}\frac{\text{1}{\text{.11×10}}^{\text{–7}}}{\text{508}\text{.26}}\text{= 2}{\text{.18×10}}^{\text{–10}}\text{s}\\ \therefore \frac{\text{T’}}{\text{T}}\text{=}\frac{\text{2}{\text{.18×10}}^{\text{–10}}}{\text{3}{\text{.93×10}}^{\text{-13}}}\approx \text{500}\\ \therefore \text{Time taken between successive collisions = 500 times the}\text{time taken for}\text{a collision}\text{.}\end{array}

Q.11 A meter long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Ans.

\begin{array}{l}\text{Here,}\text{length of narrow bore,}\text{L}\text{=}\text{1 m}=\text{1}00\text{cm}\\ \text{Length of mercury thread,}\text{l}\text{=}\text{76 cm}\\ \text{Length of air column between mercury and the}\text{closed end,}{\text{l}}_{\text{a}}\text{=}\text{15 cm}\\ \text{As the bore is held vertically in air with the open}\text{end}\text{at the bottom,}\\ \text{The air space}\text{occupied}\text{by}\text{the mercury}\text{length}\\ \text{=1}00–\left(\text{76}+\text{15}\right)\\ =\text{9 cm}\\ \therefore \text{Total length of the air column}\text{=}\text{15}\text{+}\text{9}\text{=}\text{24 cm}\\ \text{Let mercury flow out due}\text{to}\text{atmospheric pressure}\text{=}\text{h}\text{cm}\\ \therefore \text{Length of air column inside the bore}=\text{24}+\text{h}\text{cm}\\ \text{Length of mercury column}\text{=}\text{76}–\text{h}\text{cm}\\ \text{Initial pressure,}{\text{P}}_{\text{i}}\text{=}\text{76 cm of mercury}\\ \text{Initial volume,}{\text{V}}_{\text{i}}\text{=}{\text{15 cm}}^{\text{3}}\\ \text{Final pressure,}{\text{P}}_{\text{f}}=\text{76}–\left(\text{76}–\text{h}\right)=\text{h}\text{cm of mercury}\\ \text{Final volume,}{\text{V}}_{\text{f}}=\left(\text{24}+\text{h}\right){\text{cm}}^{\text{3}}\\ \text{As}\text{temperature remains constant,}\\ \therefore \text{From}\text{Boyle’s}\text{law}\\ {\text{P}}_{\text{i}}{\text{V}}_{\text{i}}\text{=}{\text{P}}_{\text{f}}{\text{V}}_{\text{f}}\\ \text{76}\times \text{15}=\text{h}\left(\text{24}+\text{h}\right)\\ {\text{h}}^{\text{2}}+\text{24h}–\text{114}0=0\\ \therefore \text{h}=\frac{-24\pm \sqrt{{\left(\text{24}\right)}^2+4\times 1\times 1140}}{2\times 1}\\ =\text{23}.\text{8 cm or}–\text{47}.\text{8 cm}\\ \text{Since}\text{height cannot be negative,}\\ \therefore \text{23}.\text{8 cm of mercury will flow out from the bore}\text{and}\\ \left(76\text{cm}-\text{23}.\text{8}\text{cm}\right)\text{52}.\text{2 cm of mercury will be}\text{left in it}\text{.}\\ \therefore \text{Length of the air column}\text{=}\text{24}\text{+}\text{23}\text{.8}\text{=}\text{47}\text{.8 cm}\text{.}\end{array}

Q.12 From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^–1. Identify the gas.
[Hint: Use Graham’s law of diffusion: R₁/R₂ = (M₂/M₁)^1/2, where R₁, R₂ are diffusion rates of gases 1 and 2, and M₁ and M₂ their respective molecular masses. The law is a simple consequence of kinetic theory.]

Ans.

\begin{array}{l}\text{Here,}\text{average rate of diffusion of hydrogen},{\text{R}}_{\text{1}}=\text{28}.{\text{7 cm}}^{\text{3}}{\text{s}}^{–\text{1}}\\ \text{Average}\text{Rate of diffusion of another gas,}{\text{R}}_{\text{2}}=\text{7}.{\text{2 cm}}^{\text{3}}{\text{s}}^{–\text{1}}\\ \text{Molecular mass of hydrogen,}{\text{M}}_{\text{1}}\text{=}\text{2}.0\text{2}0\text{g}\\ \text{Let}\text{molecular mass of the unknown gas}\text{=}{\text{M}}_{\text{2}}\\ \text{As}\text{per Graham’s Law of diffusion,}\\ \frac{{\text{R}}_{\text{1}}}{{\text{R}}_{\text{2}}}=\sqrt{\frac{{\text{M}}_{\text{2}}}{{\text{M}}_{\text{1}}}}\\ \therefore {\text{M}}_{\text{2}}\text{=}{\text{M}}_{\text{1}}{\left(\frac{{\text{R}}_{\text{1}}}{{\text{R}}_{\text{2}}}\right)}^2\\ =\text{2}.0\text{2}0{\left(\frac{28.7}{7.2}\right)}^2\\ =32.09\text{g}\\ \text{Since molecular mass of oxygen is 32, therefore, the unknown gas is oxygen}\text{.}\end{array}

Q.13 A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have
uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
n₂ = n₁ exp [-mg (h₂ – h₁)/ k_BT]
Where n₂, n₁ refer to number density at heights h₂ and h₁ respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n₂ = n₁ exp [-mg N_A(ρ – P′) (h₂ –h₁)/ (ρRT)]
Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium.
[N_A is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

Ans.

\begin{array}{l}\text{Given}\text{that,}\text{as}\text{per}\text{the law of atmospheres,}\\ {\text{n}}_{\text{2}}={\text{n}}_{\text{1}}\text{exp}\left[-\text{mg}\left({\text{h}}_{\text{2}}–{\text{h}}_{\text{1}}\right)/{\text{k}}_{\text{B}}\text{T}\right]\to \left(\text{i}\right)\\ {\text{Here,n}}_{\text{1}}\text{and}{\text{n}}_{\text{2}}\text{is}\text{the number density of}\text{particles}\text{at height}{\text{h}}_{\text{1}}\text{and}{\text{h}}_{\text{2}}\text{respectively}\text{.}\\ \text{mg}\text{=}\text{Weight of the particle hanged in the gas}\text{column}\\ \text{Let}{\rho }_m\text{and}{\rho }_{s}bethed\text{ensity of medium}\text{and}d\text{ensity of}\text{the}\text{suspended particle}\text{respectively}\text{.}\\ \text{Let}\text{mass of one suspended particle}={\text{m}}_{\text{s}}\\ \text{Let}\text{mass of the medium displaced}={\text{m}}_{\text{m}}\\ \text{Let}\text{volume of the suspended particle}=\text{V}\\ \text{As}\text{per Archimedes’ principle for a particle suspended}\text{in a liquid column,}\text{we}\text{have:}\\ \text{Effective weight of the suspended particle}\text{=}\text{Weight of the medium displaced}–\text{Weight of the}\text{suspended particle}\\ =\text{mg}–\text{m}‘\text{g}\\ \text{=}\text{mg}–\text{V}\rho ‘\text{g}\\ =\text{mg}–\left(\frac{\text{m}}{\rho }\right)\rho ‘\text{g}\\ =\text{mg}\left(1-\frac{\rho ‘}{\rho }\right)\to \text{(ii)}\\ \text{Gas constant}\text{is}\text{given}\text{as:}\\ \text{R}={\text{k}}_{\text{B}}\text{N}\\ {\text{k}}_{\text{B}}=\frac{\text{R}}{\text{N}}\to \left(\text{iii}\right)\\ \text{Putting equation}\left(\text{ii}\right)\text{in place of}\text{mg in equation}\left(\text{i}\right)\text{and}\text{then using equation}\left(\text{iii}\right)\text{,we obtain:}\\ {\text{n}}_{\text{2}}={\text{n}}_{\text{1}}\text{exp}\left[\frac{-\text{mg}\left({\text{h}}_{\text{2}}–{\text{h}}_{\text{1}}\right)}{{\text{k}}_{\text{B}}\text{T}}\right]\\ ={\text{n}}_{\text{1}}\text{exp}\left[-\text{mg}\left(1-\frac{\rho ‘}{\rho }\right)\left({\text{h}}_{\text{2}}–{\text{h}}_{\text{1}}\right)\frac{\text{N}}{\text{RT}}\right]\\ ={\text{n}}_{\text{1}}\text{exp}\left[-\text{mg}\left(\rho -\rho ‘\right)\left({\text{h}}_{\text{2}}–{\text{h}}_{\text{1}}\right)\frac{\text{N}}{\text{RT}\rho }\right]\end{array}

Q.14 Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few A].

Ans.

\begin{array}{l}\text{Let}\text{radius}\text{of}\text{an}\text{atom}\text{=}\text{r}\\ \therefore \text{Volume of an atom}\text{=}\frac{\text{4}}{\text{3}}\pi {\text{r}}^{\text{3}}\\ \text{Avogadro’s number,}\text{N}=\text{6}\text{.023}\text{×}{\text{10}}^{\text{23}}\\ \text{Volume of}\text{N}\text{number of molecules}\text{=}\frac{\text{4}}{\text{3}}\pi {\text{r}}^{\text{3}}\text{N}\to \text{(i)}\\ \text{Volume of one mole of the substance}\text{=}\frac{\text{M}}{\rho }\to \text{(ii)}\\ \frac{\text{4}}{\text{3}}\pi {\text{r}}^{\text{3}}\text{N}=\frac{\text{M}}{\rho }\\ \therefore \text{Radius}\text{of}\text{an}\text{atom,}\text{r}=\sqrt[3]{\frac{3M}{4\pi \rho N}}\\ \text{For carbon}\text{(diamond)}:\\ \text{Atomic}\text{mass,}\text{M}=\text{12}.0\text{1}\times \text{1}{0}^{–\text{3}}\text{kg},\\ \text{Density,}\rho =\text{2}.\text{22}\times \text{1}{0}^{\text{3}}{\text{kgm}}^{–\text{3}}\\ \therefore \text{Radius},r={\left(\frac{3\times \text{12}.0\text{1}\times \text{1}{0}^{–\text{3}}}{4\pi \times \text{2}.\text{22}\times \text{1}{0}^{\text{3}}\times \text{6}.0\text{23}\times \text{1}{0}^{\text{23}}}\right)}^{\frac{1}{3}}\\ =\text{1}\text{.77}\stackrel{\text{o}}{\text{A}}\\ =\text{1}.\text{29 A}\\ \therefore \text{The radius of a carbon atom is 1}\text{.29}\stackrel{\text{o}}{\text{A}}\text{.}\\ \text{For gold:}\\ \text{M}=\text{197}.00\times \text{1}{0}^{–\text{3}}\text{kg}\\ \text{Density},\rho =\text{19}.\text{32}\times \text{1}{0}^{\text{3}}{\text{kg m}}^{\text{–3}}\\ =\text{1}\text{.59}\stackrel{\text{o}}{\text{A}}\\ \therefore \text{Radius,}\text{r}={\left(\frac{3\times \text{197}\times \text{1}{0}^{–\text{3}}}{4\pi \times 19.32\times \text{1}{0}^{\text{3}}\times \text{6}.0\text{23}\times \text{1}{0}^{\text{23}}}\right)}^{\frac{1}{3}}\\ =\text{1}\text{.59}\stackrel{\text{o}}{\text{A}}\\ \therefore \text{The radius of a gold atom is 1}\text{.59 A}\text{.}\\ \text{For liquid nitrogen:}\\ \text{Atomic}\text{mass,}\text{M}=\text{14}.0\text{1}\times \text{1}{0}^{–\text{3}}\text{kg}\\ \text{Density,}\rho =\text{1}.00\times \text{1}{0}^{\text{3}}{\text{kg m}}^{–\text{3}}\\ \therefore \text{Radius,}\text{r}={\left(\frac{3\times \text{14}\text{.01}\times \text{1}{0}^{–\text{3}}}{4\pi \times 1.00\times \text{1}{0}^{\text{3}}\times \text{6}.0\text{23}\times \text{1}{0}^{\text{23}}}\right)}^{\frac{1}{3}}\\ =1.77\stackrel{\text{o}}{\text{A}}\\ \therefore \text{The radius of a liquid nitrogen atom is 1}\text{.77}\stackrel{\text{o}}{\text{A}}\text{.}\\ \text{For lithium:}\\ \text{Atomic}\text{mass,}\text{M}=\text{6}.\text{94}\times \text{1}{0}^{–\text{3}}\text{kg}\\ \text{Density},\rho =0.\text{53}\times \text{1}{0}^{\text{3}}{\text{kgm}}^{–\text{3}}\\ \therefore \text{Radius,}\text{r}={\left(\frac{3\times 6.\text{94}\times \text{1}{0}^{–\text{3}}}{4\pi \times 0.53\times \text{1}{0}^{\text{3}}\times \text{6}.0\text{23}\times \text{1}{0}^{\text{23}}}\right)}^{\frac{1}{3}}\\ =1.73\stackrel{\text{o}}{\text{A}}\\ \therefore \text{The radius of a lithium atom is 1}\text{.73}\stackrel{\text{o}}{\text{A}}\text{.}\\ \text{For}\text{fluorine}\text{(liquid):}\\ \text{Atomic}\text{mass,}\text{M}=\text{19}\text{.00}\times \text{1}{0}^{–\text{3}}\text{kg}\end{array} \begin{array}{l}\text{Density,}\rho =\text{1}.\text{14}\times \text{1}{0}^{\text{3}}{\text{kg m}}^{\text{–3}}\\ \therefore \text{Radius,}\text{r}={\left(\frac{3\times 19\times \text{1}{0}^{–\text{3}}}{4\pi \times 1.14\times \text{1}{0}^{\text{3}}\times \text{6}.0\text{23}\times \text{1}{0}^{\text{23}}}\right)}^{\frac{1}{3}}\\ =\text{1}\text{.88}\stackrel{\text{o}}{\text{A}}\\ \therefore \text{Radius of a liquid fluorine atom}\text{=}\text{1}\text{.88}\stackrel{\text{o}}{\text{A}}\end{array}