NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory - Param Himalaya
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 Å.
Solution :
1. Calculate the volume of a single oxygen molecule:
Assuming an oxygen molecule is spherical, its volume ($V_{\text{molecule}}$) can be calculated using the formula for the volume of a sphere:
$$V_{\text{molecule}} = \frac{4}{3} \pi r^3$$
where $r$ is the radius of the oxygen molecule. The diameter is given as 3 Å, so the radius is
$r = \frac{3}{2} \text{ Å} = 1.5 \times 10^{-10} \text{ m}$
Plugging in the value:
$V_{\text{molecule}} = \frac{4}{3} \pi (1.5 \times 10^{-10} \text{ m})^3$
$V_{\text{molecule}} = \frac{4}{3} \pi (3.375 \times 10^{-30}) \text{ m}^3$
$V_{\text{molecule}} \approx 1.414 \times 10^{-29} \text{ m}^3$
2. Calculate the total volume occupied by one mole of oxygen molecules:
One mole of any substance contains Avogadro's number ($N_A$) of molecules, which is approximately $6.022 \times 10^{23}$. So, the total volume occupied by one mole of oxygen molecules ($V_{\text{moles}}$) is:
$V_{\text{moles}} = N_A \times V_{\text{molecule}}$
$V_{moles}=6.022 \times 10^{23} \times 1.414 \times 10^{-29}m^3$
$V_{\text{moles}} \approx 8.516 \times 10^{-6} \text{ m}^3$
To make this more intuitive, let's convert it to cubic centimeters (cm³):
$V_{\text{moles}} \approx 8.516 \times 10^{-6} \times (10^{2}cm)^{3}$
$V_{\text{moles}} \approx 8.516 \text{ cm}^3$
3. Determine the actual volume occupied by one mole of oxygen gas at STP:
At Standard Temperature and Pressure (STP), which is defined as 0 °C (273.15 K) and 1 atm pressure, one mole of any ideal gas occupies a volume of 22.4 liters. Converting this to cubic centimeters:
$V_{\text{gas}} = 22.4 \text{ L}$
$V_{\text{gas}}= 22.4 \times 1000 \text{ cm}^3$
$V_{\text{gas}} = 22400 \text{ cm}^3$
4. Calculate the fraction of molecular volume to the actual volume:
The fraction ($f$) is the ratio of the total volume of the molecules to the volume occupied by the gas:
$f = \frac{V_{\text{moles}}}{V_{\text{gas}}}$
$f = \frac{8.516 \text{ cm}^3}{22400 \text{ cm}^3}$
$f \approx 3.8 \times 10^{-4}$
Therefore, the fraction of the molecular volume to the actual volume occupied by oxygen gas at STP is approximately $\boxed{3.8 \times 10^{-4}}$. This shows that the actual volume occupied by the gas is significantly larger than the volume of the molecules themselves, indicating that gases are mostly empty space.
12 .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.
Solution:
The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as
$PV = nRT$
Where R is the universal gas constant = $8.314 J mol^{-1}K^{-1}$
n = Number of moles = 1
T = Standard temperature = 273 K
P = Standard pressure = 1 atm =
$P=1.013 \times 10^{5}Nm^{-2}$
Hence, $V = \frac{nRT}{P}$
$V= \frac{1 \times 8.314 \times 273}{1.013 \times 10^{5}}$
$V =0.0224 m^{3}$
$V= 22.4 litres$
12.3 Figure 12.8 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 $H_{2}$ = 2.02 u, of $O_{2}$ = 32.0 u,R = 8.31 J $mo1^{–1} K^{–1}$)
Solution:
(a) The dotted plot is parallel to X-axis, signifying that nR [PV/T = nR] is independent of P. Thus, it represents ideal gas behaviour.
(b) The graphs at temperature T1 are closer to ideal behaviour (because they are closer to the dotted line) hence, T1 > T2 (the higher the temperature, the ideal behaviour is higher)
(c) Use PV = nRT
PV/ T = nR
Mass of the gas = 1 x 10-3 kg = 1 g
Molecular mass of O2 = 32g/ mol
Hence, Number of mole = Given weight / Molecular weight = 1/ 32
So, nR = 1/ 32 × 8.314 = 0.26 J/ K
Hence, Value of PV / T = 0.26 J/ K
(d) 1 g of $H_{2}$ doesn’t represent the same number of mole
E.g., Molecular mass of $H_{2}$ = 2 g/mol
Hence, the number of moles of H2 requires is 1/32 (as per the question).
Therefore, Mass of $H_{2}$ required = No. of mole of $H_{2}$ × Molecular mass of $H_{2}$= 1/ 32 × 2 = 1 / 16 g = 0.0625 g = 6.3 × $10^{-5}$ kg
Hence, 6.3 × $10^{-5}$ kg of $H_{2}$ would yield the same value.
12.4 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 ($R = 8.31 Jmol^{–1} K^{–1}$, molecular mass of $O_{2}$= 32 u).
Solution:
Volume of gas :
$V_{1} = 30 litres = 30 \times 10^{-3} m^{3}$
Gauge pressure,
$P_{1} = 15 atm = 15 \times 1.013 \times 10^{5}$ Pa
Temperature,
$T_{1}= 27^{0}C = 273+27= 300 K$
Universal gas constant,
$R = 8.314 J mol^{-1} K^{-1}$
Let the initial number of moles of oxygen gas in the cylinder be $n_{1}$
The gas equation is given as follows:
$P_{1}V_{1} = n_{1}RT_{1}$
Hence,
$n_{1} = \frac{P_{1}V_{1}}{ RT_{1}}$
$n_{1} = \frac{(15.195 \times 10^{5} \times 30 \times 10^{-3}}{(8.314 \times 300)}$
$n_{1} = 18.276$
But $n_{1} = \frac{m_{1}}{M}$
Where,
$m_{1}$ = Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
Thus, $m_{1} = n_{1}M$
$m_{1}= 18.276 \times 32 = 584.84 g$
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduce.
Volume,
$V_{2}= 30 litres = 30 \times 10^{-3} m^{3}$
Gauge pressure,
$P_{2} = 11 atm = 11 \times 1.013 \times 10^{5} Pa$
Temperature,
$T_{2} = 17^{0}C = 273+17 = 290 K$
Let $n_{2}$ be the number of moles of oxygen left in the cylinder.
The gas equation is given as
$P_{2}V_{2}= n_{2}RT_{2}$
Hence, $n_{2} = \frac{P_{2}V_{2}}{ RT_{2}}$
$n_{2}= \frac{(11.143 \times 10^{5} \times 30 \times 10^{-30}$)} {(8.314 \times 290)}$
$n_{2}= 13.86$
But $n_{2} = \frac{m_{2}}{M}$
Where, $m_{2}$ is the mass of oxygen remaining in the cylinder.
Therefore, $m_{2} = n_{2} \times M$
$m_{2}= 13.86 \times 32$
$m_{2}= 453.1 g$
The mass of oxygen taken out of the cylinder is given by the relation,
Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder
$m= m_{1} – m_{2} = 584.84 g – 453.1 g$
$m= 131.74 g = 0.131 kg
Hence, 0.131 kg of oxygen is taken out of the cylinder.
12.5 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 :
1. Define initial and final states:
Initial state (bottom of the lake):
Volume, $V_1 = 1.0 \text{ cm}^3$
Temperature, $T_1 = 12 \text{ °C} = 12 + 273.15 = 285.15 \text{ K}$
Depth, $h = 40 \text{ m}$
Final state (surface of the lake):
Volume, $V_2 = ?$
Temperature, $T_2 = 35 \text{ °C} = 35 + 273.15 = 308.15 \text{ K}$
2. Calculate the pressure at the bottom of the lake ($P_1$):
The pressure at the bottom of the lake is the sum of the atmospheric pressure at the surface and the pressure due to the column of water.
Atmospheric pressure, $P_{atm} = 1 \text{ atm} = 1.01325 \times 10^5 \text{ Pa}$
Density of water, $\rho = 1000 \text{ kg/m}^3$
Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$
Pressure due to water column:
$P_h = \rho g h$
$P_h= 1000 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 40 \text{ m} = 392000 \text{ Pa}$
Total pressure at the bottom:
$P_1 = P_{atm} + P_h$
$P_1= 1.01325 \times 10^5 \text{ Pa} + 392000 \text{ Pa} = 493325 \text{ Pa}$
$P_1= 101325\text{ Pa} + 392000 \text{ Pa} = 493325 \text{ Pa}$
3. Determine the pressure at the surface of the lake ($P_2$):
At the surface, the pressure is simply the atmospheric pressure.
$P_2 = P_{atm} = 1.01325 \times 10^5 \text{ Pa}$
4.Apply the Combined Gas Law:}
The combined gas law states:
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
We want to find $V_2$, so rearrange the formula:
$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$
5. Substitute the values and calculate $V_2$:
$V_2 = \frac{(493325 \text{ Pa}) \times (1.0 \text{ cm}^3) \times (308.15 \text{ K})}{(1.01325 \times 10^5 \text{ Pa}) \times (285.15 \text{ K})}$
$V_2 = \frac{151978258.75}{28880628.75} \text{ cm}^3$
$V_2 \approx 5.26 \text{ cm}^3$
$V_2 \approx 5.26 \times 10^{-6} \text{m}^3$
Answer:
The air bubble grows to approximately
$5.26 \text{ cm}^3$ when it reaches the surface.
12.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity $25 m^{3}$ at a temperature of 27°C and 1 atm pressure.
Solution:
We use the Ideal Gas Equation:
$PV = nRT$
Where:
$P = Pressure = 1 atm$
$P= 1.013 \times 10^5 \text{Pa}$
$V = Volume = 25 \text{m}^3$
$T = Temperature = 27^\circ\text{C}$
$T = 27+273=300 K$
$R = 8.314 \text{J mol}^{-1} \text{K}^{-1}$
Step 1: Calculate number of moles of air, $n$
$n = \frac{PV}{RT} = \frac{(1.013 \times 10^5) \times 25}{8.314 \times 300}$
$n = \frac{2.5325 \times 10^6}{2494.2} \approx 1015.2 mol$
Step 2: Convert moles to number of molecules
Number of molecules = $n \times N_A$
Where $N_A = 6.022 \times 10^{23} \text{mol}^{-1}$
Number of molecules = $1015.2 \times 6.022 \times 10^{23} \approx 6.11 \times 10^{26}$
Final Answer:
Number of Molecules: $6.11 \times 10^{26}molecules$
12.7 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).
Solution :
The average thermal energy of a helium atom is given by the formula:
$E = \frac{3}{2} k_B T$
where:
$k_B = 1.38 \times 10^{-23} \text{J/K}$
(i) At room temperature,
$T= 27^\circ \text{C} = 300 \text{K}$
$E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300$
$E = \frac{3}{2} \times 4.14 \times 10^{-21} = 6.21 \times 10^{-21} \text{J}$
(ii) At the surface of the Sun,
$T = 6000 \text{K}$
$E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 6000$
$E = \frac{3}{2} \times 8.28 \times 10^{-20} = 1.242 \times 10^{-19} \text{J}$
(iii) At core temperature of a star,
$T= 10^7 \text{K}$
$E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 10^7$
$E = \frac{3}{2} \times 1.38 \times 10^{-16} = 2.07 \times 10^{-16} \text{J}$
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 vrms the largest ?
Solution :
(i) Equal number of molecules:
Using the ideal gas law:
$PV = nRT$
Given that pressure ($P$), volume ($V$), and temperature ($T$) are the same for all three vessels, it follows that the number of moles ($n$) is the same for each gas:
$n = \frac{PV}{RT} = \text{constant}$
Since the number of molecules is:
$N = n \times N_A$
$\Rightarrow N = \text{constant}$
Therefore, each vessel contains the same number of molecules.
(ii) Root mean square speed ($v_{\text{rms}}$):
The formula for root mean square speed is:
$v_{\text{rms}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$
Where:
$R$ is the universal gas constant,
$T$ is the absolute temperature (same for all),
$M$ is the molar mass of the gas.
Since $v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$, the gas with the lowest molar mass will have the highest $v_{\text{rms}}$.
Molar masses:
| Gas | Molar Mass (g/mol) |
|---|---|
| Neon (Ne) | 20.18 |
| Chlorine (Cl₂) | 70.90 |
| Uranium hexafluoride (UF₆) | 352.02 |
Comparison of $v_{rms}$:
$M_{\text{Ne}} < M_{\text{Cl}_2} < M_{\text{UF}_6}$
$v_{\text{rms, Ne}} > v_{\text{rms, Cl}_2} > v_{\text{rms, UF}_6}$
Final Answer:
Yes, all three vessels contain the same number of molecules.
No, the root mean square speeds are not the same.
The root mean square speed is largest for neon (Ne), because it has the smallest molar mass.
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 °C ? (atomic mass of Ar= 39.9 u, of He = 4.0 u).
Given :
1. Formula for RMS Speed :
The root mean square (rms) speed ($v_{rms}$) of a gas is given by the formula:
$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$
Where:
$R$ is the ideal gas constant.
$T$ is the temperature in Kelvin (K).
$M$ is the molar mass of the gas in kg/mol.
2. Problem Setup :
Atomic mass of Argon $M_{\text{Ar}} = 39.9u$
Atomic mass of Helium $M_{\text{He}} =4.0u$
Temperature of Helium $T_{\text{He}}$= 20°C
First, we convert the temperature of Helium from Celsius to Kelvin:
$T_{\text{He}} = -20 + 273.15 = 253.15 K$
The problem states that the rms speed of Argon is equal to the rms speed of Helium:
$v_{rms-AR}= v_{rms-He}$
3. Equating the RMS Speeds
Using the formula, we can set up the equation:
$\sqrt{\frac{3RT_{\text{Ar}}}{M_{\text{Ar}}}} = \sqrt{\frac{3RT_{\text{He}}}{M_{\text{He}}}}$
By squaring both sides and canceling the common term $3R$, we simplify the equation to:
$\frac{T_{\text{Ar}}}{M_{\text{Ar}}} = \frac{T_{\text{He}}}{M_{\text{He}}}$
4. Solving for the Temperature of Argon
Now, we rearrange the formula to solve for the temperature of Argon ($T_{\text{Ar}}$):
$T_{\text{Ar}} = T_{\text{He}} \times \frac{M_{\text{Ar}}}{M_{\text{He}}}$
Substitute the known values into the equation. We can use the atomic masses directly since the units (u) will cancel out.
$T_{\text{Ar}} = 253.15 \times \frac{39.9}{4.0}$
$T_{\text{Ar}}= 253.15 \times 9.975$
$T_{\text{Ar}} \approx 2525.2 kelvin$
To express this temperature in degrees Celsius we subtract 273.15:
$T_{\text{Ar}} = 2525.2 - 273.15$
$T_{\text{Ar}}\approx$ 2252.1°C
12.10 Estimate the mean free path and collision frequency f 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 $N_{2}$ = 28.0 u.
Solution :
Mean free path = 1.11 × $10^{-7}$ m
Collision frequency = 4.58 × $10^{9} s^{-1}$
Successive collision time ≅ 500 × (Collision time)
The pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × $10^{5}$ Pa
The temperature inside the cylinder, T = 17°C = 290 K
The radius of a nitrogen molecule, r = 1.0 Å = 1 × $10^{10}$ m
Diameter, d = 2 × 1 × $10^{10}$ = 2 × $10^{10}$ m
The molecular mass of nitrogen, M = 28.0 g = 28 × $10^{-3}$ kg
The root mean square speed of nitrogen is given by the relation $V_{rms}$ = √3RT / M
Where, R is the universal gas constant = 8.314 J $mol^{-1} K^{-1}$
Hence, $V_{rms}$ = 3 × 8.314 × 290 / 28 × $10^{-3}$
On calculation,
we get = 508.26 m/s
The mean free path (l) is given by the relation l = KT / √2 × π × $d^{2}$ × P
Where, k is the Boltzmann constant = 1.38 × $10^{-23} kg m^{2} s^{-2} K^{-1}$
Hence, l = (1.38 × $10^{-23}$ × 290) / (√2 × 3.14 × (2 × $10^{-10})^{2}$ × 2.026 × $10^{5}$
We get, = 1.11 × $10^{-7}$ m
Collision frequency = $V_{rms}$ / l = 508.26 / 1.11 × $10^{-7}$
On calculation, we get = 4.58 × $10^{9} s^{-1}$
The collision time is given as
T = d / Vrms = 2 × $10^{-10}$ / 508.26
On further calculation, we get = 3.93 × $10^{-13}$ s
Time taken between successive collisions T’ = l / $V_{rms}$ = 1.11 × $10^{-7}$ / 508.26
We get, = 2.18 × $10^{-10}$
Hence,
T’ / T = 2.18 × $10^{-10}$ / 3.93 × $10^{-13}$
On calculation, we get
= 500
Therefore, the time taken between successive collisions is 500 times the time taken for a collision.