Class 11 Physics - Kinetic Theory of Gases (Important Q&A)
Kinetic Theory of Gases Marks · 2M 3M
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📘 2 Mark Q&A
1+ Q&ASolution:
A real gas behaves like an ideal gas under the following conditions:
(1) Low Pressure: At low pressure, the molecules are far apart, so the intermolecular forces become negligible.
(2) High Temperature: At high temperature, the kinetic energy of the molecules is large compared to the intermolecular attraction, so the gas behaves ideally.
(3) Low Density: When the density is low, the volume occupied by the gas molecules themselves is negligible compared to the total volume of the container.
Under these conditions, the gas approximately obeys the ideal gas equation \( PV = nRT \).
📗 3 Mark Q&A
5+ Q&ASolution:
For an ideal gas, according to the law of equipartition of energy,
Internal energy per mole is \( U = \dfrac{n}{2} RT \),
where \(n\) is the number of degrees of freedom.
At constant volume,
\( C_v = \left( \dfrac{dU}{dT} \right)_V \)
\( \Rightarrow C_v = \left( \dfrac{d }{dT}\left(\dfrac{n}{2} RT \right) \right)_V \)
So, \( C_v = \dfrac{n}{2} R \)
We know for an ideal gas,
\( C_p - C_v = R \)
Therefore, \( C_p = C_v + R \)
\( \Rightarrow C_p = \dfrac{n}{2} R + R \)
\( \Rightarrow C_p = \left( \dfrac{n}{2} + 1 \right) R \)
\( \Rightarrow C_p = \dfrac{n+2}{2} R \)
Now,
\( \dfrac{C_p}{C_v} = \dfrac{ \dfrac{n+2}{2} R }{ \dfrac{n}{2} R } \)
\( \Rightarrow \dfrac{C_p}{C_v} = \dfrac{n+2}{n} \)
\( \therefore \dfrac{C_p}{C_v} = 1 + \dfrac{2}{n} \)
Hence proved.
Solution:
Degrees of Freedom: The degrees of freedom of a molecule is the number of independent ways in which the molecule can possess energy (translational, rotational, vibrational).
Principle of Equipartition of Energy: According to this principle, energy is equally distributed among all degrees of freedom of a molecule. Each degree of freedom contributes \( \dfrac{1}{2}kT \) per molecule or \( \dfrac{1}{2}RT \) per mole to the internal energy.
For a diatomic molecule at ordinary temperature:
Number of degrees of freedom \( n = 5 \) (3 translational + 2 rotational).
So, \( C_v = \dfrac{n}{2}R = \dfrac{5}{2}R \).
We know \( C_p = C_v + R \).
\( C_p = \dfrac{5}{2}R + R = \dfrac{7}{2}R \).
Therefore,
\( \gamma = \dfrac{C_p}{C_v} = \dfrac{\dfrac{7}{2}R}{\dfrac{5}{2}R} \).
\( \therefore \gamma = \dfrac{7}{5} \).
Hence, for a diatomic molecule, \( \gamma = \dfrac{7}{5} = 1.4 \).
Solution:
From kinetic theory of gases,
\( PV = \dfrac{2}{3} E \),
where \(E\) is the total kinetic energy of the gas.
Also from kinetic theory, average kinetic energy is related to temperature as
\( E = \dfrac{3}{2} nRT \) (for \(n\) moles).
\( \therefore nRT = \dfrac{2}{3}E \) (for \(n\) moles).
From ideal gas equation,
\( PV = nRT \).
Substituting \( nRT = \dfrac{2}{3}E \) into the ideal gas equation, we get
\( PV = \dfrac{2}{3} E \).
Therefore,
\( P = \dfrac{2E}{3V} \).
Hence proved.
Solution:
From kinetic theory of gases, the pressure of a gas is given by
\( P = \dfrac{1}{3} \rho \overline{c^2} \),
where \( \rho \) is the density and \( \overline{c^2} \) is the mean square velocity.
Since \( \rho = \dfrac{Nm}{V} \),
\( \therefore P = \dfrac{1}{3} \dfrac{Nm}{V} \overline{c^2} \)
Multiplying both sides by \(V\), we get
\( PV = \dfrac{1}{3} Nm \overline{c^2} \)
From ideal gas equation,
\( PV = NkT \)
On comparing the two expressions of \(PV\), we will get
\( NkT = \dfrac{1}{3} Nm \overline{c^2} \)
\( \Rightarrow kT = \dfrac{1}{3} m \overline{c^2} \)
Multiplying both sides by \( \dfrac{3}{2} \), we get
\( \dfrac{3}{2} kT = \dfrac{1}{2} m \overline{c^2} \)
But \( \dfrac{1}{2} m \overline{c^2} \) is the average kinetic energy of one molecule.
Hence, average kinetic energy \( \propto T \).
Therefore, the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature.
Solution:
Mean Free Path: The mean free path of a gas molecule is the average distance travelled by a molecule between two successive collisions with other molecules.
It is denoted by \( \lambda \).
From kinetic theory of gases,
\( \lambda = \dfrac{1}{\sqrt{2} \pi d^2 n} \),
where \(d\) is the diameter of a molecule and \(n\) is the number of molecules per unit volume.
Since \( n = \dfrac{P}{kT} \),
\( \lambda = \dfrac{kT}{\sqrt{2} \pi d^2 P} \).
Hence, mean free path depends on:
(1) Pressure \(P\): \( \lambda \propto \dfrac{1}{P} \) (decreases with increase in pressure).
(2) Temperature \(T\): \( \lambda \propto T \) (increases with increase in temperature).
(3) Molecular diameter \(d\): \( \lambda \propto \dfrac{1}{d^2} \) (larger molecules have smaller mean free path).
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