Class 11 Physics (Physics Part-II) Chapter 12 Kinetic Theory

Key Concept: Temperature Effect

c) Number density decreases, mean free path increases

[Solution Description]

As the temperature increases, according to the ideal gas law, the volume of the gas tends to increase if pressure remains constant. This results in a decrease in number density $(n)$, as $n$ is defined as the number of molecules per unit volume. With the decrease in number density, the mean free path $(l = \frac{1}{n\pi d^2})$ increases because of the inverse relationship.

Key Concept: Conceptual Reasoning

d) It reduces to one fourth

[Solution Description] The mean free path is given by the formula $l = \frac{1}{\sqrt{2} \pi n d^2}$. If the molecular diameter $d$ is increased by a factor of 2, then $l’ = \frac{1}{\sqrt{2} \pi n (2d)^2} = \frac{1}{4}\left(\frac{1}{\sqrt{2} \pi n d^2}\right)$. Hence, the mean free path decreases to one fourth of its original value.

Key Concept: Advanced Formula Application

b) $45R$

[Solution Description] The heat required can be calculated using $Q = nC\Delta T$. Here, $n = 3$ moles, $C = 3R$, and $\Delta T = 10$ K:

$Q = 3 \times 3R \times 10 = 90R.$

Key Concept: Molecular Speed

a) Neon

[Solution Description] To determine the root mean square (rms) speed, we use the formula $v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}}$. The rms speed is inversely proportional to the square root of the molar mass ($m$), meaning lighter gases move faster at a given temperature. Let’s calculate the relative rms speeds:

Given options are:

– Neon ($M = 20.18 \, \text{g/mol}$)

– Chlorine ($M = 70.90 \, \text{g/mol}$)

– Uranium hexafluoride ($M = 352.02 \, \text{g/mol}$)

– Argon ($M = 39.95 \, \text{g/mol}$)

Since the gas with the lowest molar mass will have the highest rms speed, we compare these values:

$v_{\text{Neon}} > v_{\text{Argon}} > v_{\text{Chlorine}} > v_{\text{Uranium hexafluoride}}$

Thus, Neon has the highest rms speed.

Key Concept: Specific Heat Ratio

a) $\frac{5}{3}$

[Solution Description] For a monoatomic gas, the degrees of freedom are different from a diatomic gas. Typically, a monoatomic gas has 3 translational degrees of freedom, which results in $\gamma = \frac{5}{3}$. This is because $C_v = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$, leading to:
$\gamma = \frac{5}{3}$

Key Concept: Intermediate Calculation

b) $45R \text{ J}$

[Solution Description] For a monatomic ideal gas, we use the formula $Q = nC_v\Delta T$. Given that $C_v = \frac{3}{2}R$ for a monatomic gas, $n = 3$ moles, and $\Delta T = 10 \text{ K}$, the heat required is $Q = 3 \times \frac{3}{2}R \times 10 = 45R \text{ J}$.

Key Concept: Formula Application

c) 15 liters

[Solution Description] Charles’ Law states that $V \propto T$, or $V_1/T_1 = V_2/T_2$. Given $V_1 = 10$ liters, $T_1 = 300$ K, and $T_2 = 450$ K, we solve for $V_2$:
$V_2 = \frac{V_1 \cdot T_2}{T_1} = \frac{10 \text{ L} \times 450 \text{ K}}{300 \text{ K}} = 15 \text{ L}$

Thus, the new volume is 15 liters.

Key Concept: Basic Definitions

b) $\frac{5}{2} R$

[Solution Description] A diatomic gas has $C_v = \frac{5}{2} R$. This is derived from the fact that diatomic gases have both translational and rotational degrees of freedom contributing to the specific heat.

Key Concept: Equipartition Law

b) $\frac{1}{2}k_BT$

[Solution Description] The law of equipartition of energy states that each degree of freedom contributes $\frac{1}{2}k_BT$ to the total energy in thermal equilibrium. Therefore, the average energy per degree of freedom is $\frac{1}{2}k_BT$.

Key Concept: Translational Degrees of Freedom

c) 3

[Solution Description] A molecule that can move freely in space has three translational degrees of freedom because it can translate in the x, y, and z directions.

Key Concept: Degrees of Freedom

c) 3

[Solution Description] Monatomic gas molecules move freely in three-dimensional space. Each independent movement direction represents a degree of freedom. Therefore, a monatomic gas molecule has three translational degrees of freedom.

Key Concept: Pressure Derivation

b) Molecular collisions with the walls

[Solution Description]

The pressure exerted by an ideal gas is due to molecular collisions with the walls of the container. As molecules collide with the walls, they exert force per unit area, which results in measurable pressure.

Key Concept: Basic Formula Application

a) $E = N \times \left(\frac{1}{2}\right) m \overline{v^2}$

[Solution Description]

The average translational kinetic energy $E$ can be represented using the formula:
$E = N \times \left(\frac{1}{2}\right) m \overline{v^2}$
where $N$ is the number of molecules, $m$ is the mass of each molecule, and $\overline{v^2}$ is the mean square speed.
This equation provides the total translational kinetic energy of all molecules in the gas.

Key Concept: Basic Concept

d) Molecules are in random motion and collisions are elastic

[Solution Description] The molecules in a gas are assumed to be in random motion, and their collisions are elastic according to the kinetic theory of gases.

Key Concept: Pressure Calculation

c) $P = \frac{1}{3} nm \overline{v^2}$

[Solution Description] The pressure exerted by an ideal gas can be derived using the formula:

$P = \frac{1}{3} nm \overline{v^2}$

where $m$ is the mass of a single molecule, $n$ is the number density, and $\overline{v^2}$ is the mean squared speed of the gas molecules.

Key Concept: Basic Equation

c) $PV = nRT$

[Solution Description] The ideal gas equation relates the pressure, volume, temperature, and amount of substance of an ideal gas. The general formula for the ideal gas law is $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the universal gas constant, and $T$ is the temperature in Kelvin.

Key Concept: Basic Concept

c) $PV = \mu RT$

[Solution Description]

The perfect gas equation describes the state of an ideal gas using pressure $P$, volume $V$, amount of substance $\mu$, and temperature $T$. It is given by the formula $PV = \mu RT$, where $R$ is the universal gas constant.

Key Concept: Charles’ Law

d) The volume doubles

[Solution Description] Charles’ Law states that the volume of an ideal gas is directly proportional to its absolute temperature at constant pressure, i.e., $\frac{V_1}{T_1} = \frac{V_2}{T_2}$. Here, $T_2 = 2 T_1$, so $V_2$ will be $2 V_1$. Therefore, the volume doubles.

Key Concept: Boyle’s Law

c) It halves

[Solution Description] Boyle’s Law states that $PV = \text{constant}$ when temperature is constant. This means $P_1V_1 = P_2V_2$. Given $P_1 = 2$ atm, $P_2 = 4$ atm. If we take $V_1 = V$, then $V_2 = \frac{P_1}{P_2} \cdot V_1 = \frac{2}{4} \cdot V = \frac{1}{2} V$. Thus, the volume halves.

Key Concept: Atomic Size

c) Atoms are spaced tens of angstroms apart allowing freedom of motion.

[Solution Description]

In gases, atoms or molecules are spaced relatively far apart compared to solids and liquids. This spacing allows them to move freely, which is characteristic of gases.

Key Concept: Mole Concept

b) 32 grams

[Solution Description]

At standard temperature and pressure (S.T.P), the volume occupied by one mole of any ideal gas is 22.4 litres. The molecular weight of oxygen (O$$_2$$) is approximately 32 grams per mole.

Therefore, the mass of 22.4 litres of oxygen gas is equal to its molecular weight:

$$\text{Mass} = 32 \text{ grams}$$

Key Concept: Basic Concepts

c) Identical

[Solution Description] According to Dalton’s Atomic Theory, atoms of the same element are identical in mass and properties.

Key Concept: Atomic Size

d) 1 angstrom

[Solution Description] The size of a typical atom is roughly about $1$ angstrom $1 \times 10^{-10}$ meters. This measurement indicates the approximate distance across the diameter of an atom.

Key Concept: Basic Concept

c) John Dalton

[Solution Description] The atomic hypothesis was most famously put forward by John Dalton in the 19th century, who proposed that matter is made up of small indivisible particles called atoms.

Key Concept: Pressure Derivation

b) $P = \frac{1}{3} n m \overline{v^2}$

[Solution Description] Kinetic theory describes gases as consisting of rapidly moving molecules, and pressure arises from collisions with container walls. The derived formula for pressure $P$ in terms of number density $n$, molecular mass $m$, and average squared velocity $\overline{v^2}$ is:

$P = \frac{1}{3} n m \overline{v^2}$

This equation shows that pressure depends on these factors in a specific ratio.

Key Concept: Molecular Speed

b) 482 m/s

[Solution Description] The root mean square speed ($v_{rms}$) of a gas molecule is given by the formula:

$v_{rms} = \sqrt{\frac{3k_B T}{m}}$

where $k_B$ is the Boltzmann constant ($1.38 \times 10^{-23}\ \text{J/K}$), $T$ is the temperature in Kelvin, and $m$ is the mass of one molecule in kg. First, we need to convert the molar mass from g/mol to kg:

$m = \frac{32\ \text{g/mol}}{1000} \times \frac{1}{6.022 \times 10^{23}\ \text{mol}^{-1}} = 5.31 \times 10^{-26}\ \text{kg}$

Now substitute into the formula for $v_{rms}$:

$v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23}\ \text{J/K} \times 400\ \text{K}}{5.31 \times 10^{-26}\ \text{kg}}}$

$v_{rms} = \sqrt{\frac{1.656 \times 10^{-20}}{5.31 \times 10^{-26}}}$

$v_{rms} \approx 482\ \text{m/s}$

Key Concept: Theory Development

c) Maxwell and Boltzmann

[Solution Description] The development of the kinetic theory was an important achievement in the nineteenth century, with significant contributions by James Clerk Maxwell and Ludwig Boltzmann. They advanced the understanding of molecular motion and how it relates to macroscopic properties.

Key Concept: Gas Laws

b) Charles’s Law

[Solution Description] Charles’s Law states that the volume of a gas is directly proportional to its temperature at constant pressure. This is justified by the kinetic theory as increasing temperature increases molecular motion which leads to greater volume.

Key Concept: Law Identification

c) Avogadro’s Law

[Solution Description]

Avogadro’s Law states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. This principle helps in understanding the behavior of gases under similar conditions.

Therefore, this description matches Avogadro’s Law.

Key Concept: Ideal Gas Equation

c) Universal gas constant

[Solution Description]

In the ideal gas equation $PV = \mu RT$, $R$ represents the universal gas constant.