Kinetic Theory - Class 11 Physics - Chapter 12 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Kinetic Theory | NCERT | Physics | Class 11
For an ideal gas, the slope of the $V$-$T$ graph during an adiabatic process is $\frac{dV}{dT}=-m$ at a point where volume and temperature are $V_{0}$ and $T_{0}$. Find the value of $C_{p}$ of the gas. It is given that $\mathrm{m}$ is a positive number.
A. $\left(\frac{m T_{0}}{V_{0}}+1\right) R$
B. $\frac{m T_{0} R}{V_{0}}$
C. $\left(\frac{-m T_{0}}{V_{0}}+1\right) R$
D. $\left(\frac{-m T_{0} R}{V_{0}}\right)$
The correct option is A. $\left(\frac{m T_{0}}{V_{0}}+1\right) R$
For an adiabatic process, the equation that relates temperature $T$ and volume $V$ can be represented as: $$ TV^{\gamma-1} = \text{constant} $$ Taking the natural logarithm on both sides gives: $$ \ln T + (\gamma - 1) \ln V = \text{constant} $$ Differentiating both sides with respect to $T$, we obtain: $$ \frac{1}{T} + (\gamma - 1) \frac{dV}{V} = 0 $$ Rearranging, we have: $$ \frac{dV}{dT} = -\frac{V}{T(\gamma - 1)} $$
Given that $\left(\frac{dV}{dT}\right){V{0}, T_{0}} = -m$, equating it to the expression from the derivative we get: $$ -\frac{V_0}{T_0(\gamma - 1)} = -m $$ This implies: $$ \frac{1}{\gamma - 1} = \frac{m T_{0}}{V_{0}} $$
We know the relation $C_v = \frac{R}{\gamma-1}$. Substituting what we have found: $$ C_v = \frac{m R T_{0}}{V_{0}} $$ And the heat capacity at constant pressure, $C_p$, is given by $C_v + R$: $$ C_p = \left(\frac{m T_{0}}{V_{0}} + 1\right) R $$ Thus, the answer is option A.
Who among the following proposed the Big Bang Theory?
A. Albert Einstein
B. Georges Lemaitre
C. Charles Darwin
D. Francesco Redi
The correct option is B. Georges Lemaitre.
Georges Lemaitre proposed the Big Bang Theory in 1927. He introduced the idea that the universe originated from a singular, extremely compact point. According to his theory, the universe has been expanding since that initial moment. Lemaitre’s groundbreaking concept provides the basis for our understanding of how the universe has evolved to its current vast extent and continues to expand.
The de Broglie wavelength $\lambda$ is related to the kinetic energy (E) of the particle as
(A) $\lambda \propto E$ (B) $\lambda \propto \frac{1}{E}$ (C) $\lambda \propto \frac{1}{\sqrt{E}}$
(D) $\lambda \propto \sqrt{E}$
The correct response is (C) $\lambda \propto \frac{1}{\sqrt{E}}$. This relationship can be explained through the following derivation:
De Broglie's Hypothesis: States that any moving particle can be described by a wave. The wavelength, $\lambda$, of this wave is given by: $$ \lambda = \frac{h}{p} $$ where $h$ is Planck's constant and $p$ is the particle's momentum.
Momentum and Kinetic Energy Relationship: For a non-relativistic particle, the momentum $p$ can be expressed in terms of the mass $m$ and velocity $v$ as: $$ p = m \cdot v $$ and the kinetic energy $E$ of the particle is given by: $$ E = \frac{1}{2} m v^2 $$ Solving for $v$ in terms of $E$, we get: $$ v = \sqrt{\frac{2E}{m}} $$
Substitute for Momentum: Using the expression for $v$, the momentum $p$ can be rewritten as: $$ p = m \cdot \sqrt{\frac{2E}{m}} = \sqrt{2mE} $$
Final Expression for Wavelength $\lambda$: Substituting the expression for $p$ back into the de Broglie equation gives: $$ \lambda = \frac{h}{\sqrt{2mE}} $$ On rearranging, we observe that: $$ \lambda = \frac{h}{\sqrt{2m}} \cdot \frac{1}{\sqrt{E}} $$
Thus, the wavelength $\lambda$ is inversely proportional to the square root of the kinetic energy $E$, which confirms that the correct choice is (C) $\lambda \propto \frac{1}{\sqrt{E}}$.
Describe cohesive force theory.
The cohesion-tension-transpiration pull theory explains the ascent of sap in plants. It contains two key components:
Cohesion and Adhesion: Water molecules exhibit cohesion, meaning they are naturally attracted to each other and form a continuous water column within the plant. Additionally, these water molecules adhere to the walls of the xylem elements through adhesion. These cohesive and adhesive forces together create a continuous transport channel from the roots to the leaves.
Transpiration Pull: As water evaporates from the leaf surfaces (a process known as transpiration), a negative pressure or pull is created within the xylem. This acts much like a suction force, drawing water upwards from the roots to the top of the plant.
Together, these mechanisms facilitate the upward movement of water and nutrients against gravity in plants.
Assertion (A): At constant temperature, PV vs P plot for real gases is not a straight line. Reason (R): At high pressure, all gases have $Z>1$, but at intermediate pressure, most gases have $Z<1$.
A. Both A and R are true and R is the correct explanation of A.
B. Both A and R are true but R is not the correct explanation of A.
C. A is true but R is false.
D. A is false but R is true.
SolutionThe correct option is B: Both A and R are true but R is not the correct explanation of A.
The plot of $PV$ versus $P$ for real gases deviates from a straight line due to the presence of intermolecular forces and the finite size of gas molecules, which cause deviations from ideal behavior.
For real gases, at high pressures, there is generally a positive deviation from ideal behavior represented by "$Z>1$", indicating a greater pressure than that predicted by the ideal gas law. Conversely, at intermediate pressures, most real gases exhibit a negative deviation from ideal behavior, represented by "$Z<1$", meaning the pressure is less than that predicted by the ideal gas law.
However, statement R, while true concerning the behavior of $Z$ at different pressures, does not directly explain why the $PV$ versus $P$ plot is not a straight line, which is due to the physical properties and behaviors described, rather than solely the value of $Z$ at different pressures. Thus, option B is the correct choice.
Which of the following statements are incorrect?
A. Dalton's law of partial pressures is applicable to the mixture of $\mathrm{HCl}$ and $\mathrm{NH}_{3}$.
B. According to the kinetic theory of gases, at the same temperature, equal moles of all gases have equal kinetic energy.
C. According to the kinetic theory of gases, at the same temperature, all gases have equal root mean square velocity.
D. According to the kinetic theory of gases, with an increase in temperature, pressure of the gas increases due to the increase in collision frequency.
The incorrect statements are:
A: Dalton's Law of Partial Pressures is applicable to the mixture of $\mathrm{HCl}$ and $\mathrm{NH}_{3}$.
C: According to the Kinetic Theory of Gases, at the same temperature, all gases have equal root mean square velocity.
Explanation:
A: Dalton's Law of Partial Pressures only applies to non-reacting gases. When mixing $\mathrm{HCl}$ and $\mathrm{NH}_{3}$, they react to form ammonium chloride ($\mathrm{NH}_4\mathrm{Cl}$), so Dalton's Law does not apply.
B: This statement is correct because according to the kinetic theory, the kinetic energy of a gas is determined solely by its temperature, and is independent of the gas type when moles and temperature are equal.
C: The root mean square (rms) velocity of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is represented as: $$ v_{\text{rms}} \propto \frac{1}{\sqrt{m}} $$ Therefore, not all gases have the same rms velocity at the same temperature, as it varies with molar mass.
D: This statement is correct. The pressure of a gas increases with temperature because the molecules collide more frequently and with greater force against the container's walls, enhancing the pressure.
The ratio between the r.m.s. velocity of $\mathrm{H}_{2}$ at $50 \mathrm{~K}$ and that of $\mathrm{O}_{2}$ at $800 \mathrm{~K}$:
A) 4
B) 2
C) 1
D) $\frac{1}{4}$
The correct answer is C) 1. The ratio of the root mean square (r.m.s.) velocities is found using the formula:
$$ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} $$
where $T$ is the temperature in Kelvin, $R$ is the universal gas constant, and $M$ is the molar mass of the gas. When considering the ratio of r.m.s. velocities of hydrogen, $ \mathrm{H}_{2} $, at $ 50 , \mathrm{K} $ and oxygen, $ \mathrm{O}_{2} $, at $ 800 , \mathrm{K} $:
$$ \frac{(v_{\text{rms}})_{\mathrm{H}_2} }{(v_{\text{rms}})_{\mathrm{O}_2}} = \sqrt{\frac{3 R T_1 / M_{\mathrm{H}_2}}{3 R T_2 / M_{\mathrm{O}_2}}} = \sqrt{\frac{T_1 M_{\mathrm{O}_2}}{T_2 M_{\mathrm{H}_2}}} $$
Substituting the values, where the molar mass of $ \mathrm{H}_{2} $ (2 g/mol) and $ \mathrm{O}_{2} $ (32 g/mol), we have:
$$ \frac{(v_{\text{rms}})_{\mathrm{H}_2}}{(v_>{\text{rms}})_{\mathrm{O}_2}} = \sqrt{\frac{50 , \mathrm{K} \cdot 32 , \text{g/mol}}{800 , \mathrm{K} \cdot 2 , \text{g/mol}}} = \sqrt{\frac{1600}{1600}} = 1 $$
Thus, the ratio of the r.m.s. velocities is 1, indicating that the r.m.s. velocities of $ \mathrm{H}_{2} $ at $ 50 , \mathrm{K} $ and $ \mathrm{O}_{2} $ at $ 800 , \mathrm{K} $ are equal.
Explain the kinetic theory of matter in the case of evaporation.
Kinetic molecular theory explains that all matter is comprised of particles which are separated by spaces, are in persistent random motion, and interact through forces of attraction and repulsion. Temperature reflects the average kinetic energy of these particles.
In the context of evaporation, which is the phase change from liquid to gas, the process can be described as follows: When the particles in a liquid are heated, they gain kinetic energy, which causes them to move more vigorously and to spread farther apart. As these particles accumulate sufficient energy, they break free from the forces of attraction that keep them together in the liquid state. Consequently, they transition into the gas phase, moving very quickly and spreading widely apart.
The relation between the gas pressure $P$ and average kinetic energy per unit volume $E$ is
(A) $P=\frac{1}{2} E$
(B) $P=E$
(C) $P=\frac{3}{2} E$
(D) $P=\frac{2}{3} E$
The relationship between gas pressure $P$ and the average kinetic energy per unit volume $E$ is given by the following equation:
$$ P = \frac{2}{3} E $$
Thus, the correct option is (D) $P = \frac{2}{3} E$.
If the number of molecules of $\mathrm{H}_{2}$ are double that of $\mathrm{O}_{2}$, then the ratio of kinetic energy of hydrogen to that of oxygen at $300 \mathrm{~K}$ is
(A) $1: 1$
(B) $1: 2$
(C) $2: 1$
(D) $1: 16$
The correct answer is (A) 1:1.
Explanation:
The kinetic energy (KE) of gas molecules in an ideal gas is dependent solely on the temperature of the gas, not on the type or number of molecules present. At a given temperature, all gases have the same average kinetic energy, given by the formula:
$$
KE = \frac{3}{2} kT,
$$
where $k$ is the Boltzmann constant and $T$ is the absolute temperature.
In this case, since both $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ are at the same temperature, 300 K, their average kinetic energies are equal regardless of the difference in the number of molecules. Therefore, the ratio of kinetic energy of $\mathrm{H}_{2}$ to $\mathrm{O}_{2}$ is 1:1.
Two Formula One cars in a race have the same mass, but their kinetic energies are different. The ratio of their kinetic energies is 1:16. What would be the corresponding ratio of their velocities?
A) $4:1$ B) $1:4$ C) $2:1$ D) $1:2$
The correct answer is B) $1:4$.
The kinetic energy (K.E.) of an object is expressed as: $$ K.E. = \frac{1}{2} m v^2 $$ where $m$ is the mass of the object and $v$ is the velocity of the object.
Given that the ratio of the kinetic energies of two cars is $1:16$, we can write: $$ \frac{KE_1}{KE_2} = \frac{\frac{1}{2} mv_1^2}{\frac{1}{2} mv_2^2} = \frac{v_1^2}{v_2^2} = \frac{1}{16} $$ Here, $KE_1$ and $KE_2$ represent the kinetic energies of the first and second car, respectively.
Taking the square root of both sides to find the ratio of their velocities: $$ \frac{v_1}{v_2} = \sqrt{\frac{1}{16}} = \frac{1}{4} $$
Thus, the ratio of their velocities is $1:4$.
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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 Ã….
The fraction of the molecular volume to the actual volume occupied by the oxygen gas at STP is:
$$ \text{Fraction} \approx 0.000380 $$
Thus, the molecular volume is approximately (0.038%) of the total volume occupied by the gas.
12.2 Molar volume is the volume occupied by $1 \mathrm{~mol}$ of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, $0^{\circ} \mathrm{C}$ ). Show that it is 22.4 litres.
Thus, by substituting the values into the ideal gas equation, we get:
$$ V = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \cdot 273.15 \text{ K} = 22.425615 \text{~L} $$
This is very close to $22.4 \text{~L} $.
Therefore, the molar volume of an ideal gas at standard temperature and pressure (STP) is approximately 22.4 liters.
Figure 12.8 shows plot of $P V / T$ versus $P$ for $1.00 \times 10^{-3} \mathrm{~kg}$ of oxygen gas at two different temperatures.
Fig. 12.8
(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 $P V / T$ where the curves meet on the $y$-axis?
(d) If we obtained similar plots for $1.00 \times 10^{-3} \mathrm{~kg}$ of hydrogen, would we get the same value of $P V / T$ at the point where the curves meet on the $y$-axis? If not, what mass of hydrogen yields the same value of $P V / T$ (for low pressure high temperature region of the plot) ? (Molecular mass of $\mathrm{H}_{2}=2.02 \mathrm{u}$, of $\mathrm{O}_{2}=32.0 \mathrm{u}$, $R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$.)
Let's break down each question with precise answers based on the kinetic theory and properties of gases.
(a) What does the dotted plot signify?
The dotted plot signifies the ideal gas behavior. For an ideal gas, the relationship ( \frac{PV}{T} = nR ) holds constant, where ( n ) is the number of moles and ( R ) is the universal gas constant. Therefore, the dotted line represents a constant ( \frac{PV}{T} ) value, indicating ideal gas conditions where volume, pressure, and temperature follow the ideal gas law perfectly.
(b) Which is true: ( T_{1} > T_{2} ) or ( T_{1} < T_{2} ) ?
To determine the relationship between ( T_1 ) and ( T_2 ):
- At lower pressures, deviations from ideal behavior are minor, and the curves approach the ideal gas line.
- With increasing pressure, real gases exhibit deviations due to intermolecular forces. The curve with higher temperature (higher kinetic energy) shows less deviation and remains closer to the ideal gas plot over a wider range of pressures.
From the given plot:
- Since the curve labeled ( T_2 ) deviates more from the ideal behavior (greater concavity) than the curve labeled ( T_1 ), ( T_1 ) must be greater than ( T_2 ).
So, ( T_{1} > T_{2} ).
(c) What is the value of ( \frac{PV}{T} ) where the curves meet on the ( y )-axis?
At the point where the curves meet on the y-axis (low pressure and high temperature):
- The gas behaves ideally and we use the ideal gas law.
Given ( m = 1.00 \times 10^{-3} , \text{kg} ) of ( \text{O}_2 ):
-
Calculate the number of moles of ( \text{O}_2 ): [ \mu = \frac{m}{M} = \frac{1.00 \times 10^{-3} , \text{kg}}{32.0 \times 10^{-3} , \text{kg/mol}} = 0.03125 , \text{mol} ]
-
Using the ideal gas equation ( \frac{PV}{T} = \mu R ): [ \frac{PV}{T} = 0.03125 \times 8.31 , \text{J/K} \approx 0.2596875 , \text{J K}^{-1} ]
So, the value where the curves meet is approximately 0.26 J K(^{-1}).
(d) If we obtained similar plots for ( 1.00 \times 10^{-3} , \text{kg} ) of hydrogen, would we get the same value of ( \frac{PV}{T} ) at the point where the curves meet on the y-axis?
-
The value of ( \frac{PV}{T} ) depends on the number of moles (( \mu )).
-
Since the molecular mass of ( \text{H}_2 = 2.02 , \text{u} ), calculate the moles of hydrogen (\mu_{\text{H}_2}) in ( 1.00 \times 10^{-3} , \text{kg} ):
[ \mu_{\text{H}_2} = \frac{1.00 \times 10^{-3} , \text{kg}}{2.02 \times 10^{-3} , \text{kg/mol}} \approx 0.49505 , \text{mol} ]
So, the value for hydrogen would be: [ \left(\frac{PV}{T}\right)_{\text{H}_2} = 0.49505 \times 8.31 \approx 4.1139 , \text{J K}^{-1} ]
Clearly, this value is not the same as for oxygen.
To obtain the same ( \frac{PV}{T} \approx 0.26 , \text{J K}^{-1} ) for hydrogen, the mass of hydrogen should be such that the number of moles matches that of oxygen:
[ \mu_{\text{H}_2, \text{needed}} = 0.03125 , \text{mol} ]
Thus, the mass ( m_{\text{H}_2, \text{needed}} ) is: [ m_{\text{H}_2, \text{needed}} = \mu_{\text{H}_2, \text{needed}} \times M_{\text{H}_2} = 0.03125 , \text{mol} \times 2.02 , \text{g/mol} \approx 0.063125 , \text{g} = 6.31 \times 10^{-5} , \text{kg} ]
So, 6.31 mg of hydrogen would yield the same ( PV/T ) value.
An oxygen cylinder of volume 30 litre has an initial gauge pressure of $15 \mathrm{~atm}$ and a temperature of $27^{\circ} \mathrm{C}$. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to $11 \mathrm{~atm}$ and its temperature drops to $17^{\circ} \mathrm{C}$. Estimate the mass of oxygen taken out of the cylinder ($R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ molecular mass of $\mathrm{O}_{2}=32 \mathrm{u}$ ).
From the computations, we have:
Initial number of moles $n_1 = 25.657 $
Final number of moles $ n_2 = 19.906 $
The difference in the number of moles:
[ \Delta n = n_1 - n_2 = 25.657 - 19.906 = 5.751 ]
Now, we calculate the mass of oxygen taken out. The molar mass of $O_2$ is $32 \ \text{g/mol}$.
[ \text{Mass} = \Delta n \times \text{molar mass of } O_2 ] [ \text{Mass} = 5.751 \times 32 = 184.032 \ \text{g} ]
So, the mass of oxygen taken out of the cylinder is 184.032 g.
An air bubble of volume $1.0 \mathrm{~cm}^{3}$ rises from the bottom of a lake $40 \mathrm{~m}$ deep at a temperature of $12{ }^{\circ} \mathrm{C}$. To what volume does it grow when it reaches the surface, which is at a temperature of $35^{\circ} \mathrm{C}$ ?
The final volume $ V_2 $ of the air bubble when it reaches the surface is approximately: [ V_2 \approx 5.39 \times 10^{-6} , \text{m}^3 ]
To convert this back to cubic centimeters: [ V_2 \approx 5.39 \times 10^{-6} , \text{m}^3 \times 10^6 ,\text{cm}^3/\text{m}^3 \approx 5.39 , \text{cm}^3 ]
So the air bubble grows to a volume of approximately 5.39 cm³ when it reaches the surface at a temperature of $35^{\circ}C$.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity $25.0 \mathrm{~m}^{3}$ at a temperature of $27^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure.
To estimate the total number of air molecules in a room with a capacity of $25.0 \ \text{m}^3$ at a temperature of $27^{\circ} \ \text{C}$ (which is $300 \ \text{K}$) and $1 \ \text{atm}$ pressure, we can use the ideal gas law:
$$ PV = Nk_BT $$
Where:
$P$ is the pressure
$V$ is the volume
$N$ is the number of molecules
$k_B$ is the Boltzmann constant
$T$ is the absolute temperature in Kelvin
Given:
$P = 1 \ \text{atm} = 1.01 \times 10^5 \ \text{Pa}$
$V = 25 \ \text{m}^3$
$T = 27^{\circ} \ \text{C} = 300 \ \text{K}$
$k_B = 1.38 \times 10^{-23} \ \text{J K}^{-1}$
Re-arranging the equation to solve for $N$:
$$ N = \frac{PV}{k_{B}T} $$
Plugging in the values:
$$ N = \frac{(1.01 \times 10^5 \ \text{Pa}) \times (25 \ \text{m}^3)}{(1.38 \times 10^{-23} \ \text{J K}^{-1}) \times (300 \ \text{K})} $$
Using the result from the computation: $$ N \approx 6.63 \times 10^{26} $$
So, the total number of air molecules in the room is approximately $6.63 \times 10^{26}$ molecules.
Estimate the average thermal energy of a helium atom at (i) room temperature $\left(27^{\circ} \mathrm{C}\right.$ ), (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).
The average thermal energy of a helium atom at various temperatures is:
(i) Room Temperature ($27^{\circ} \mathrm{C}$ or 300.15 K)
The average thermal energy is: $$ \approx 6.21 \times 10^{-21} \text{ Joules} $$
(ii) Temperature on the Surface of the Sun (6000 K)
The average thermal energy is: $$ \approx 1.24 \times 10^{-19} \text{ Joules} $$
(iii) Temperature of the Core of a Star (10 million K)
The average thermal energy is: $$ \approx 2.07 \times 10^{-16} \text{ Joules} $$
These values reflect the increase in average thermal energy with rising temperatures.
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 ?
Number of Molecules
Since the three vessels have gases at the same temperature, pressure, and equal volume, according to the ideal gas law: $$ PV = nRT $$ where:
$P$ is the pressure,
$V$ is the volume,
$n$ is the number of moles,
$R$ is the universal gas constant,
$T$ is the temperature.
Given that $P$, $V$, and $T$ are the same for all three vessels, the number of moles $n$ will also be the same for each gas. Since the number of molecules ($N$) is proportional to the number of moles ($n$), we have: $$ N = n \cdot N_A $$ where $N_A$ is Avogadro's number. Consequently, the number of molecules in each vessel will be the same.
Root Mean Square Speed ($V_{\text {rms }}$)
The root mean square speed of molecules in a gas is given by: $$ V_{\text{rms}} = \sqrt{\frac{3k_B T}{m}} $$ where:
$k_B$ is the Boltzmann constant,
$T$ is the absolute temperature,
$m$ is the mass of one molecule of the gas.
Even though the temperature $T$ is the same for all gases, the molecular masses ($m$) of the three gases are different. Thus, the $V_{\text{rms}}$ will differ for each gas:
Neon (monatomic): lower molecular mass.
Chlorine (diatomic): intermediate molecular mass.
Uranium hexafluoride (polyatomic): highest molecular mass.
Since $V_{\text{rms}}$ is inversely proportional to the square root of the molecular mass $V_{\text{rms}} \propto \frac{1}{\sqrt{m}}$, the gas with the lowest molecular mass will have the highest $V_{\text{rms}}$.
Conclusion
The vessels do contain an equal number of molecules.
The root mean square speed ($V_{\text{rms}}$) is not the same for the three gases.
The largest $V_{\text{rms}}$ is for the gas with the smallest molecular mass, which is neon (monatomic).
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 $-20^{\circ} \mathrm{C}$ ? (atomic mass of $\mathrm{Ar}$ $=39.9 \mathrm{u}$, of $\mathrm{He}=4.0 \mathrm{u}$ ).
The temperature at which the root mean square speed of an argon gas atom equals the rms speed of a helium gas atom at $-20^{\circ} \mathrm{C}$ is 2525 K.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0 \mathrm{~atm}$ and temperature $17^{\circ} \mathrm{C}$. Take the radius of a nitrogen molecule to be roughly 1.0 Ã…. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of $\left.\mathrm{N}_{2}=28.0 \mathrm{u}\right)$.
Number density ($n$):[ n \approx 5.06 \times 10^{25} , \text{m}^{-3} ]
Mean free path ($\lambda$):[ \lambda = \frac{1}{\sqrt{2} \pi d^2 n} \approx 3.50 \times 10^{-8} , \text{m} ]
Average speed ($\overline{v}$):[ \overline{v} \approx 468.27 , \text{m/s} ]
Collision frequency ($f$):[ f = \frac{\overline{v}}{\lambda} \approx \frac{468.27}{3.50 \times 10^{-8}} , \text{s}^{-1} ] [ f \approx 1.34 \times 10^{10} , \text{s}^{-1} ]
Finally, the collision time ($\tau$) is the inverse of the collision frequency: [ \tau = \frac{1}{f} \approx 7.46 \times 10^{-11} , \text{s} ]
Summary
Mean free path ($\lambda$): $3.50 \times 10^{-8}$ m
Collision frequency ($f$): $1.34 \times 10^{10}$ s$^{-1}$
Collision time ($\tau$): $7.46 \times 10^{-11}$ s
Comparison
You can now compare the collision time $\tau$ with the time the molecule moves freely between collisions:
Free time between collisions ($\tau_{free}$): $\tau_{free} \approx 7.46 \times 10^{-11}$ s. This represents the average time a nitrogen molecule moves before colliding with another molecule in the given conditions.
These values demonstrate the rapid and frequent collisions nitrogen molecules undergo in the given conditions.
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Comprehensive Kinetic Theory Class 11 Notes: Understanding Gases and Molecular Motion
Understanding the kinetic theory of gases is crucial for class 11 students preparing for their physics curriculum. This article will help you grasp the fundamental concepts of kinetic theory, the historical development, the behaviour of gases, and important laws, equations, and applications.
Introduction to Kinetic Theory
What is the Kinetic Theory of Gases?
The kinetic theory of gases explains the behaviour of gases by considering them as composed of many small particles, primarily atoms or molecules, which are in constant random motion. It provides a molecular-level explanation for the macroscopic properties of gases, such as pressure, temperature, and volume.
Historical Background
Development of Kinetic Theory
The development of kinetic theory is attributed to the works of several scientists over centuries.
Early Discoveries by Boyle and Newton
In 1661, Boyle discovered Boyle's Law, and later, Newton also contributed by proposing that gases comprise tiny atomic particles.
Contributions of Maxwell and Boltzmann
In the 19th century, Maxwell and Boltzmann made significant advancements, giving a molecular interpretation of gas behaviour, which is consistent with gas laws and Avogadro’s hypothesis.
Molecular Nature of Matter
The Atomic Hypothesis
Richard Feynman highlighted the importance of the atomic hypothesis, which states that all matter is made up of small particles in constant motion.
Contributions by Ancient Scholars
Ancient scholars like Kanada in India and Democritus in Greece speculated about the existence of indivisible particles (atoms), laying the foundation for modern atomic theory.
Modern Acceptance by Dalton and Avogadro
John Dalton’s atomic theory explained the laws of definite and multiple proportions, while Avogadro proposed that equal volumes of all gases contain the same number of molecules under the same conditions of temperature and pressure.
Behaviour of Gases
Explanation of Gas Laws
Boyle’s Law
Boyle's Law states that the pressure of a given mass of gas is inversely proportional to its volume at a constant temperature.
Charles’ Law
Charles' Law states that the volume of a gas is directly proportional to its absolute temperature at constant pressure.
Kinetic Theory of an Ideal Gas
Understanding Ideal Gas
An ideal gas perfectly follows the gas laws under all conditions of pressure and temperature.
Characteristics of an Ideal Gas
In an ideal gas, the molecules are assumed to occupy negligible space and have no interactions except during collisions.
Derivation of Pressure Equation
Through kinetic theory, the pressure of an ideal gas can be derived using the relation: [ P = \frac{1}{3} n m \overline{v^2} ] where ( n ) is the number density of molecules, ( m ) is the mass of a molecule, and ( \overline{v^2} ) is the mean square speed.
graph TB
a[Ideal Gas Equation: PV = nRT]
b[Where,]
c[P = Pressure]
d[V = Volume]
e[n = Number of moles]
f[R = Universal gas constant]
g[T = Temperature]
a --> b
b --> c
b --> d
b --> e
b --> f
b --> g
Equipartition of Energy
Law of Equipartition of Energy
The law states that, at thermal equilibrium, the total energy is equally distributed among all available degrees of freedom, contributing ( k_B T / 2 ) to the energy.
Application to Different Types of Gases
Monatomic Gases: Only translational degrees of freedom.
Diatomic Gases: Translational and rotational degrees of freedom.
Polyatomic Gases: Translational, rotational, and vibrational degrees of freedom.
Vibrational, Rotational, and Translational Modes
Each mode contributes to the total energy. For example, a diatomic molecule has 5 degrees of freedom due to translational and rotational movement.
Specific Heat Capacity
Specific Heat for Monatomic, Diatomic, and Polyatomic Gases
The specific heat capacity at constant volume ( C_v ) depends on the number of degrees of freedom ( f ): [ C_v = ( f / 2 ) R ] where ( R ) is the universal gas constant.
Calculations and Equations
Monatomic Gases: ( C_v = \frac{3}{2} R )
Diatomic Gases: ( C_v = \frac{5}{2} R ) (without vibrational modes)
Polyatomic Gases: Depends on the number of vibrational modes ( f )
Experimental versus Theoretical Values
While the theoretical values derived through kinetic theory provide a good approximation, discrepancies may always exist due to real gas behaviour.
Mean Free Path
Calculating Mean Free Path
The mean free path (( l )) is the average distance a molecule travels between collisions: [ l = \frac{1}{\sqrt{2} n \pi d^2} ] where ( n ) is the number density and ( d ) is the diameter of the molecule.
Importance in Gas Behaviour
The mean free path explains why gases can diffuse quickly despite high molecular speeds because they undergo frequent collisions.
Summary of Key Equations
Comprehensive List of Kinetic Theory Equations
Ideal Gas Law: ( PV = nRT )
Pressure Equation: ( P = \frac{1}{3} n m \overline{v^2} )
Mean Free Path: ( l = \frac{1}{\sqrt{2} n \pi d^2} )
graph TB
A[Ideal Gas Law: PV = nRT]
B["Pressure Equation: P = (1/3)nm(v^2)"]
C["Mean Free Path: l = 1/(sqrt(2) n pi d^2)"]
Applications and Implications
Real versus Ideal Gases
While the kinetic theory provides excellent insights, real gases deviate from ideal behaviour at high pressures and low temperatures due to intermolecular forces and finite molecular size.
Observations and Limitations
Kinetic theory's assumptions simplify the understanding but may not fully capture the complexity of real gases.
Practical Applications in Science and Engineering
Understanding gas behaviour is crucial in various fields, including meteorology, aerospace engineering, and chemical engineering.
Common Misconceptions and Points to Ponder
Clarifying Common Misunderstandings
Importance of Molecular Collisions
Random molecular collisions are crucial for understanding gas pressure and temperature.
Relationship Between Pressure, Temperature, and Volume
Derived gas laws explain this relationship comprehensively.
Exercises and Practice Problems
Practice Questions with Solutions
Conceptual Questions:
Explain the kinetic theory of gases.
What is the law of equipartition of energy?
Numerical Problems:
Calculate the mean free path for air molecules at STP.
Derive the pressure equation for an ideal gas using kinetic theory principles.
Understanding the kinetic theory of gases is pivotal for comprehending many macroscopic properties of gases from a molecular perspective. This article provides a detailed overview that should equip class 11 students with essential concepts and knowledge in this domain.
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