KINETIC THEORY OF GASES MCQs for JEE PHYSICS — Medium, Page 4

91 Medium

91 of 257

Remembering +4 | -1

In van der Waals’ equation, the critical pressure

\mathrm{P_c}

is given by

View answer and explanation

Correct option: B

\mathrm{a/27b}^2

Option B (

\mathrm{a/27b}^2

) is correct because the critical pressure in van der Waals' equation is given by this expression.

Solution approach

Correct Answer + Core Reason: $\mathrm{a/27b}^2$ . This is the expression for critical pressure in van der Waals' equation.

Step-by-Step Solution:

Step 1: Understand the question. The question asks for the critical pressure ( $\mathrm{P_c}$ ) in van der Waals' equation.

Step 2: Recall the van der Waals' equation and the expressions for critical constants.
- van der Waals' equation: $\left(P + \frac{a}{V^2}\right)(V - b) = RT$
- Critical volume: $V_c = 3b$
- Critical temperature: $T_c = \frac{8a}{27Rb}$
- Critical pressure: $P_c = \frac{a}{27b^2}$

Step 3: Identify the correct option. The correct option should match the expression for critical pressure.

Step 4: Final Answer: $\mathrm{a/27b}^2$

Why other options are wrong:
- Option A: $3 \mathrm{b}$ is related to critical volume, not pressure.
- Option C: $\mathrm{27a/b}^2$ is not the correct expression for critical pressure.
- Option D: $\mathrm{b^2/a}$ is not the correct expression for critical pressure.

Key Concepts:
- van der Waals' equation
- Critical constants (pressure, volume, temperature)

Memory Aids:
- $P_c = \frac{a}{27b^2}$

Quick Tips:
- Remember the expressions for critical constants in terms of van der Waals' constants.

92 Medium

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Application +4 | -1

The ratio of the molecular masses of two gases is 7/8. What is the ratio of the number of moles of two gases, if they are in two different containers and have same mass for each gas?

View answer and explanation

Correct option: A

8 / 7

Option A (

8 / 7

) is correct because the ratio of moles is inversely proportional to the ratio of molecular masses when the mass is the same.

Solution approach

Correct Answer + Core Reason: $8 / 7$ . The ratio of moles is inversely proportional to the ratio of molecular masses.

Step-by-Step Solution:

Step 1: The number of moles of a gas is given by $\mu = \frac{M}{M_0}$ , where $M$ is the total mass and $M_0$ is the molecular mass.

Step 2: Given that $\frac{(M_0)_1}{(M_0)_2} = \frac{7}{8}$ and $M$ is the same for both gases.

Step 3: The ratio of the number of moles of the two gases is $\frac{\mu_1}{\mu_2} = \frac{M / (M_0)_1}{M / (M_0)_2} = \frac{(M_0)_2}{(M_0)_1} = \frac{8}{7}$ .

Why other options are wrong:
- Option B (7 / 8): This is incorrect because it represents the ratio of molecular masses, not the ratio of moles. It's the inverse of the correct ratio.
- Option C (7 / 5): This is incorrect as it doesn't follow from the given ratio of molecular masses.
- Option D (5 / 7): This is incorrect as it doesn't follow from the given ratio of molecular masses.

Key Concepts:
- Number of moles
- Molecular mass

Memory Aids:
- Remember that moles = mass / molecular mass

Quick Tips:
- Pay attention to whether the question asks for the ratio of molecular masses or the ratio of moles.
- If the mass is constant, the ratio of moles is inversely proportional to the ratio of molecular masses.

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Understanding +4 | -1

The rms speed of a large number of molecules of a gas in a vessel is

600 \mathrm{~m} \mathrm{~s}^{-1}

. If

25 \%

of the gas leaks out at constant temperature, then rms speed of the remaining molecules will be

View answer and explanation

Correct option: D

600 \mathrm{~m} \mathrm{~s}^{-1}

Option D (

600 \mathrm{~m} \mathrm{~s}^{-1}

) is correct because the RMS speed depends only on temperature and molar mass, not the amount of gas.

Solution approach

Correct Answer + Core Reason: $600 \mathrm{~m} \mathrm{~s}^{-1}$ . The RMS speed is independent of the amount of gas at constant temperature.

Step-by-Step Solution:
Step 1: The root-mean-square (RMS) speed of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{{3RT}}{{M}}}$ , where R is the gas constant, T is the temperature, and M is the molar mass.
Step 2: The problem states that the temperature is constant.
Step 3: Since the gas is the same, the molar mass (M) is also constant.
Step 4: Therefore, the RMS speed depends only on the temperature and molar mass, which are both constant.
Step 5: Leaking out 25% of the gas does not change the temperature or the molar mass, so the RMS speed remains the same.

Why other options are wrong:
- Option A: This is incorrect because the RMS speed does not change with the amount of gas.
- Option B: This is incorrect because the RMS speed does not change with the amount of gas.
- Option C: This is incorrect because the RMS speed does not change with the amount of gas.

Key Concepts:
- Root-mean-square (RMS) speed
- Kinetic theory of gases

Memory Aids:
- Remember the formula for RMS speed: $v_{rms} = \sqrt{\frac{{3RT}}{{M}}}$

Quick Tips:
- Understand the factors that affect RMS speed.

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Application +4 | -1

Let

A

and

B

the two gases and given

\frac{{{T_B}}}{{{M_A}}} = 4.\frac{{{T_B}}}{{{M_B}}};

where

T

is the temperature and

M

is molecular mass. If

{C_A}

and

{C_B}

are the

r.m.s.

speed, then the ratio

\frac{{{C_A}}}{{{C_B}}}

will be equal to

View answer and explanation

Correct option: A

2

Option A (2) is correct because the ratio of RMS speeds is proportional to the square root of the ratio of (T/M).

Solution approach

Correct Answer + Core Reason: 2. The ratio of RMS speeds is proportional to the square root of the ratio of (T/M).

Step-by-Step Solution:

Step 1: Given: $\frac{{{T_A}}}{{{M_A}}} = 4\frac{{{T_B}}}{{{M_B}}}$ .

Step 2: The RMS speed is given by $C = \sqrt {\frac{{3RT}}{M}}$ .

Step 3: Therefore, $\frac{{{C_A}}}{{{C_B}}} = \frac{{\sqrt {\frac{{3R{T_A}}}{{{M_A}}}} }}{{\sqrt {\frac{{3R{T_B}}}{{{M_B}}}} }} = \sqrt {\frac{{{T_A}}}{{{M_A}}} \div \frac{{{T_B}}}{{{M_B}}}} = \sqrt {\frac{{{T_A}}}{{{M_A}}} \times \frac{{{M_B}}}{{{T_B}}}}$ .

Step 4: Substituting the given condition, $\frac{{{C_A}}}{{{C_B}}} = \sqrt {4\frac{{{T_B}}}{{{M_B}}} \times \frac{{{M_B}}}{{{T_B}}}} = \sqrt 4 = 2$ .

Why other options are wrong:
- Option B (4): This is incorrect. It is the ratio of $\frac{{{T_A}}}{{{M_A}}}$ to $\frac{{{T_B}}}{{{M_B}}}$ , not the ratio of RMS speeds.
- Option C (1): This is incorrect. It implies that $\frac{{{T_A}}}{{{M_A}}} = \frac{{{T_B}}}{{{M_B}}}$ , which is not the given condition.
- Option D (0.5): This is incorrect. It is the inverse of the correct ratio.

Key Concepts:
- Root mean square speed

Memory Aids:
- $C = \sqrt {\frac{{3RT}}{M}}$

Quick Tips:
- Remember the formula for RMS speed and how it relates to temperature and molecular mass.

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Understanding +4 | -1

Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will

View answer and explanation

Correct option: C

Remain same

Option C (Remain same) is correct because the temperature of the gas molecules depends on their average kinetic energy, which is not affected by the uniform motion of the lorry.

Solution approach

Correct Answer + Core Reason: Remain same. The temperature depends on the average kinetic energy of the molecules, which is unaffected by the lorry's motion.

Step-by-Step Solution:

Step 1: Recall the relationship between temperature and kinetic energy: Temperature is directly proportional to the average kinetic energy of the gas molecules.

Step 2: Uniform motion of the lorry does not change the relative motion of the gas molecules within the container. The average kinetic energy of the molecules remains the same.

Step 3: Since the average kinetic energy remains the same, the temperature remains the same.

Why other options are wrong:
- Option A: Increase: This would imply an increase in the average kinetic energy, which is not caused by uniform motion.
- Option B: Decrease: This would imply a decrease in the average kinetic energy, which is not caused by uniform motion.
- Option D: Decrease for some, while increase for others: The uniform motion affects all molecules equally, so there's no reason for some to decrease and others to increase.

Key Concepts:
- Temperature
- Kinetic energy
- Ideal gas law

Memory Aids:
- Temperature is a measure of average kinetic energy.

Quick Tips:
- Uniform motion does not affect the internal energy or temperature of a system.

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Understanding +4 | -1

Given below are two statements:

Statement (I): The mean free path of gas molecules is inversely proportional to square of molecular diameter.

Statement (II): Average kinetic energy of gas molecules is directly proportional to absolute temperature of gas.

In the light of the above statements, choose the correct answer from the option given below:

View answer and explanation

Correct option: D

Both Statement I and Statement II are true

Option D is correct because the mean free path is inversely proportional to the square of the molecular diameter, and the average kinetic energy is directly proportional to the absolute temperature.

Solution approach

Correct Answer + Core Reason: Both Statement I and Statement II are true. Statement I is based on the mean free path formula, and Statement II is based on the kinetic theory of gases.

Step-by-Step Solution:

Step 1: Analyze Statement I: The mean free path ( $\lambda$ ) is given by $\lambda=\frac{{\mathrm{{RT}}}}{{\sqrt{{2}} \pi \mathrm{{d}}^2 \mathrm{{N}}_{{\mathrm{{A}}}} \mathrm{{P}}}}$ , where d is the molecular diameter.

Step 2: Analyze Statement II: The average kinetic energy (KE) is given by $\mathrm{{KE}}=\frac{{\mathrm{{f}}}}{{2}} \mathrm{{nRT}}$ , where T is the absolute temperature.

Step 3: From the formula in Step 1, it's clear that $\lambda$ is inversely proportional to $d^2$ .

Step 4: From the formula in Step 2, it's clear that KE is directly proportional to T.

Why other options are wrong:
- Option A: Statement I is true, so this option is incorrect.
- Option B: Statement II is true, so this option is incorrect.
- Option C: Both statements are true, so this option is incorrect.

Key Concepts:
- Mean Free Path
- Kinetic Theory of Gases

Memory Aids:
- Remember the formula for mean free path and kinetic energy.

Quick Tips:
- Pay attention to the relationships between variables in the formulas.

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Understanding +4 | -1

For a gas at a temperature

T

the root-mean-square velocity

{v_{rms}},

the most probable speed

{v_{mp}},

and the average speed

{v_{av}}

obey the relationship

View answer and explanation

Correct option: B

{v_{rms}} > {v_{av}} > {v_{mp}}

Option B (

{v_{rms}} > {v_{av}} > {v_{mp}}

) is correct because the root-mean-square velocity is always greater than the average velocity, which is greater than the most probable speed.

Solution approach

Correct Answer + Core Reason: ${v_{rms}} > {v_{av}} > {v_{mp}}$ . This is the correct relationship between the three speeds.

Step-by-Step Solution:

Step 1: Define root-mean-square velocity: $v_{rms} = \sqrt{\frac{3kT}{m}}$

Step 2: Define average speed: $v_{av} = \sqrt{\frac{8kT}{\pi m}}$

Step 3: Define most probable speed: $v_{mp} = \sqrt{\frac{2kT}{m}}$

Step 4: Comparing the coefficients, we have $\sqrt{3} > \sqrt{\frac{8}{\pi}} > \sqrt{2}$

Step 5: Therefore, ${v_{rms}} > {v_{av}} > {v_{mp}}$

Why other options are wrong:
- Option A: Incorrect order of velocities.
- Option C: Incorrect order of velocities.
- Option D: Incorrect order of velocities.

Key Concepts:
- Root-mean-square velocity
- Average speed
- Most probable speed

Memory Aids:
- RMS is the fastest, Most Probable is the slowest.

Quick Tips:
- Remember the order of the three speeds.

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Application +4 | -1

If

\gamma

be the ratio of specific heats of a perfect gas, the number of degrees of freedom of a molecule of the gas is

View answer and explanation

Correct option: C

\frac{2}{\gamma-1}

Option C (

\frac{2}{\gamma-1}

) is correct because the number of degrees of freedom is related to the ratio of specific heats by the formula

n = \frac{2}{\gamma-1}

.

Solution approach

Correct Answer + Core Reason: $\frac{2}{\gamma-1}$ . The number of degrees of freedom is related to the ratio of specific heats.

Step-by-Step Solution:

Step 1: Recall the relationship between $C_v$ and degrees of freedom ( $n$ ): ${C_v} = n\frac{R}{2}$

Step 2: Recall the relationship between $C_p$ and $C_v$ : ${C_p} = {C_v} + R$

Step 3: Substitute ${C_v} = n\frac{R}{2}$ into the equation for $C_p$ : ${C_p} = n\frac{R}{2} + R = \left( {\frac{{n + 2}}{2}} \right)R$

Step 4: Define the ratio of specific heats: $\gamma = \frac{{{C_p}}}{{{C_v}}}$

Step 5: Substitute the expressions for $C_p$ and $C_v$ : $\gamma = \frac{{\left( {\frac{{n + 2}}{2}} \right)R}}{{n\frac{R}{2}}} = \frac{{n + 2}}{n}$

Step 6: Solve for $n$ : $n\gamma = n + 2 \Rightarrow n\gamma - n = 2 \Rightarrow n(\gamma - 1) = 2 \Rightarrow n = \frac{2}{{\gamma - 1}}$

Why other options are wrong:
- Option A: $\frac{25}{2}(\gamma-1)$ : This is incorrect because it does not represent the correct relationship between degrees of freedom and the ratio of specific heats.
- Option B: $\frac{3 \gamma-1}{2 \gamma-1}$ : This is incorrect because it does not represent the correct relationship between degrees of freedom and the ratio of specific heats.
- Option D: $\frac{9}{2}(\gamma-1)$ : This is incorrect because it does not represent the correct relationship between degrees of freedom and the ratio of specific heats.

Key Concepts:
- Degrees of freedom
- Specific heat at constant volume ( $C_v$ )
- Specific heat at constant pressure ( $C_p$ )
- Ratio of specific heats ( $\gamma$ )

Memory Aids:
- Remember the relationship: $\gamma = \frac{{{C_p}}}{{{C_v}}}$
- Remember that $C_p > C_v$ for gases.

Quick Tips:
- Understand the relationship between degrees of freedom and the type of gas (monoatomic, diatomic, etc.).

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Understanding +4 | -1

The average kinetic energy of a gas molecules is

View answer and explanation

Correct option: D

Directly proportional to absolute temperature of gas

Option D (Directly proportional to absolute temperature of gas) is correct because the average kinetic energy of a gas molecule is directly proportional to the absolute temperature, as described by the equation E = (3/2)kT.

Solution approach

Correct Answer + Core Reason: Directly proportional to absolute temperature of gas. The average kinetic energy of a gas molecule is directly proportional to the absolute temperature.

Step-by-Step Solution:

Step 1: Recall the formula for the average kinetic energy of a gas molecule: $E = \frac{{3}}{{2}}kT$ , where E is the average kinetic energy, k is the Boltzmann constant, and T is the absolute temperature.

Step 2: Analyze the relationship between E and T: From the formula, it is clear that the average kinetic energy (E) is directly proportional to the absolute temperature (T).

Why other options are wrong:
- Option A: The average kinetic energy is not directly proportional to the pressure of the gas. While pressure is related to kinetic energy through the ideal gas law, the direct proportionality is with temperature.
- Option B: The average kinetic energy is not inversely proportional to the volume of the gas. Volume is related to kinetic energy through the ideal gas law, but the direct proportionality is with temperature.
- Option C: The average kinetic energy is not inversely proportional to the absolute temperature of the gas; it is directly proportional.

Key Concepts:
- Average kinetic energy of a gas molecule
- Boltzmann constant
- Absolute temperature
- Kinetic theory of gases

Memory Aids:
- Remember the formula: $E = \frac{{3}}{{2}}kT$

Quick Tips:
- The average kinetic energy of a gas molecule depends only on the absolute temperature.

100 Medium

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Application +4 | -1

A gas cylinder of 50 litres capacity contained helium at 80 atmospheric pressure. Due to slow leakage it was found after a while that the pressure had dropped to 40 atmosphere. Find the proportion of the gas that has escaped and also the volume the escaped gas would occupy at atmospheric pressure.

View answer and explanation

Correct option: A

2000 litre

Option A (2000 litre) is correct because it is calculated using Boyle's law to determine the volume of the escaped gas at atmospheric pressure.

Solution approach

Correct Answer + Core Reason: 2000 litre. This is the volume of escaped gas at atmospheric pressure.

Step-by-Step Solution:

Step 1: Initial state: $P_1 = 80 \text{ atm}$ , $V_1 = 50 \text{ litres}$

Step 2: Final state: $P_2 = 40 \text{ atm}$ , $V_2 = 50 \text{ litres}$ (volume of the cylinder remains the same)

Step 3: Calculate the amount of gas remaining at 80 atm:\nUsing Boyle's law, $P_1V_1 = P_2V_2$ , where $V_2$ is the volume the remaining gas would occupy at 80 atm.
$80 \times V = 40 \times 50$ \nV = $\frac{40 \times 50}{80} = 25 \text{ litres}$

Step 4: Calculate the volume of gas that escaped at 80 atm:
$V_{\text{escaped}} = 50 - 25 = 25 \text{ litres}$

Step 5: Calculate the volume of the escaped gas at atmospheric pressure (1 atm):\nUsing Boyle's law again, $P_1V_1 = P_3V_3$ , where $P_3 = 1 \text{ atm}$ and $V_3$ is the volume of the escaped gas at 1 atm.
$80 \times 25 = 1 \times V_3$ \nV_3 = 2000 litres $\n\n**Why other options are wrong:**\n- **Option B:** Incorrect. This is likely a calculation error.\n- **Option C:** Incorrect. This is likely a calculation error.\n- **Option D:** Incorrect. This is likely a calculation error.\n\n**Key Concepts:**\n- Boyle's Law\n- Gas Laws\n\n**Memory Aids:**\n- Remember Boyle's Law:$ P_1V_1 = P_2V_2 $\n\n**Quick Tips:**\n- Always identify the initial and final states clearly.$

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Understanding +4 | -1

Root mean square velocity of a particle is

v

at pressure

P.

If pressure is increased two times, then the

r.m.s.

velocity becomes

View answer and explanation

Correct option: D

v

Option D (v) is correct because the root mean square velocity depends on temperature and molar mass, not pressure.

Solution approach

Correct Answer + Core Reason: v. RMS velocity is independent of pressure.

Step-by-Step Solution:

Step 1: Recall the formula for RMS velocity: $v_{rms} = \sqrt{{\frac{{3RT}}{M}}}$ , where $R$ is the gas constant, $T$ is the temperature, and $M$ is the molar mass.

Step 2: Observe that the formula does not include pressure.

Step 3: Conclude that changing the pressure does not affect the RMS velocity if the temperature remains constant.

Why other options are wrong:
- Option A: Incorrect; RMS velocity doesn't depend on pressure.
- Option B: Incorrect; RMS velocity doesn't depend on pressure.
- Option C: Incorrect; RMS velocity doesn't depend on pressure.

Key Concepts:
- Root mean square velocity
- Kinetic theory of gases

Memory Aids:
- RMS velocity depends on temperature and molar mass.

Quick Tips:
- Remember the RMS velocity formula.

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Analyzing +4 | -1

Imagine a planet whose diameter and mass are both one half of those of Earth. The day's temperature of this planet's surface reaches up to

800 \mathrm{~K}

. Find whether oxygen molecules are possible in the atmosphere of this planet.

View answer and explanation

Correct option: A

v_{\operatorname{rms}}

is very small compared to escape velocity on the planet, molecules cannot escape from the surface of the planet's atmosphere

Option A (

v_{\operatorname{rms}}

is very small compared to escape velocity on the planet, molecules cannot escape from the surface of the planet's atmosphere) is correct because the calculated RMS velocity is much less than the escape velocity.

Solution approach

Correct Answer + Core Reason: $v_{\operatorname{rms}}$ is very small compared to escape velocity on the planet, molecules cannot escape from the surface of the planet's atmosphere. The RMS velocity of oxygen molecules is significantly less than the escape velocity of the planet.

Step-by-Step Solution:

Step 1: Calculate the escape velocity of the planet. The escape velocity is given by $v_e = \sqrt{\frac{{2GM}}{R}}$ . Since the planet's mass and radius are half of Earth's, the escape velocity is the same as Earth's.

Step 2: Calculate the escape velocity of Earth: $v_E = \sqrt{\frac{{2GM_E}}{R_E}} = 11.2 \text{{ km/s}}$

Step 3: Calculate the RMS velocity of oxygen molecules at 800 K. The RMS velocity is given by $v_{\text{{rms}}} = \sqrt{\frac{{3kT}}{m}}$ , where k is Boltzmann's constant, T is the temperature, and m is the mass of an oxygen molecule.

Step 4: Substitute the values: $v_{\text{{rms}}} = \sqrt{\frac{{3 \times 1.38 \times 10^{{-23}} \times 800}}{{5.3 \times 10^{{-26}}}}} = 0.79 \text{{ km/s}}$

Step 5: Compare the RMS velocity and escape velocity. Since $0.79 \text{{ km/s}} << 11.2 \text{{ km/s}}$ , the oxygen molecules cannot escape the planet's atmosphere.

Why other options are wrong:
- Option B: This option incorrectly states that molecules can escape when their RMS velocity is small compared to the escape velocity.
- Option C: This option is similar to Option B and is incorrect for the same reason.
- Option D: The correct answer is one of the provided options.

Key Concepts:
- Escape Velocity: $v_e = \sqrt{\frac{{2GM}}{R}}$
- RMS Velocity: $v_{\text{{rms}}} = \sqrt{\frac{{3kT}}{m}}$

Memory Aids:
- Remember the formulas for escape velocity and RMS velocity.

Quick Tips:
- Always convert units to SI units before calculations.
- Compare the RMS velocity with the escape velocity to determine if molecules can escape the atmosphere.

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Understanding +4 | -1

Kinetic theory of gases provide a base for

View answer and explanation

Correct option: C

Charle’s law and Boyle’s law

Option C (Charle’s law and Boyle’s law) is correct because the kinetic theory of gases provides the theoretical foundation for both Boyle's and Charles's laws.

Solution approach

Correct Answer + Core Reason: Charle’s law and Boyle’s law. Kinetic theory provides the base for both laws.

Step-by-Step Solution:

Step 1: Recall Boyle's Law.

Step 2: Boyle's Law states that for a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume: $P \propto \frac{1}{V}$ .

Step 3: Recall Charles's Law.

Step 4: Charles's Law states that for a fixed mass of gas at constant pressure, the volume is directly proportional to the absolute temperature: $V \propto T$ .

Step 5: The kinetic theory of gases provides a theoretical basis for both of these laws by relating pressure and temperature to the average kinetic energy of the gas molecules.

Why other options are wrong:
- Option A: Charle’s law - While kinetic theory supports Charles's law, it also supports Boyle's law.
- Option B: Boyle’s law - While kinetic theory supports Boyle's law, it also supports Charles's law.
- Option D: None of these - Kinetic theory provides a base for both Charles's and Boyle's laws.

Key Concepts:
- Kinetic theory of gases
- Boyle's Law
- Charles's Law

Memory Aids:
- Remember that Boyle's Law relates pressure and volume, while Charles's Law relates volume and temperature.

Quick Tips:
- Understand the assumptions of the kinetic theory of gases.

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Understanding +4 | -1

Mean free path of

{O}_2

gas molecules in rigid container is

\lambda

. The temperature of gas is doubled, then new mean free path of

{O}_2

gas molecules is

View answer and explanation

Correct option: A

\lambda

Option A is correct. The mean free path of a gas molecule in a rigid container is independent of temperature because the number density remains constant.

Solution approach

Correct Answer + Core Reason: $\lambda$ . For rigid container $V=$ constant $\Rightarrow n$ is fixed \nHence $\lambda=\frac{1}{\sqrt{2} \pi n d^2} \Rightarrow$ is independent of temperature

Step-by-Step Solution:

Step 1: The mean free path ( $\lambda$ ) is given by the formula: $\lambda = \frac{1}{\sqrt{2} \pi n d^2}$ , where n is the number density (number of molecules per unit volume) and d is the diameter of the molecule.

Step 2: In a rigid container, the volume (V) is constant. Since the number of molecules is also constant, the number density (n = N/V) remains constant.

Step 3: Because the number density (n) and the diameter (d) of the molecules are constant, the mean free path ( $\lambda$ ) is independent of temperature.

Step 4: Therefore, doubling the temperature does not change the mean free path.

Why other options are wrong:
- Option B: This is incorrect because the mean free path does not double with temperature in a rigid container.
- Option C: This is incorrect because the mean free path does not halve with temperature in a rigid container.
- Option D: This is incorrect because the mean free path does not decrease by a factor of 4 with temperature in a rigid container.

Key Concepts:
- Mean free path
- Number density
- Rigid container

Memory Aids:
- Mean free path formula: $\lambda = \frac{1}{\sqrt{2} \pi n d^2}$

Quick Tips:
- Remember that in a rigid container, the number density remains constant.

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Application +4 | -1

Let

A

and

B

the two gases and given

\frac{{{T_B}}}{{{M_A}}} = 4.\frac{{{T_B}}}{{{M_B}}};

where

T

is the temperature and

M

is molecular mass. If

{C_A}

and

{C_B}

are the

r.m.s.

speed, then the ratio

\frac{{{C_A}}}{{{C_B}}}

will be equal to

View answer and explanation

Correct option: A

2

Option A (2) is correct because the ratio of RMS speeds is proportional to the square root of the ratio of (T/M).

Solution approach

Correct Answer + Core Reason: 2. The ratio of RMS speeds is proportional to the square root of the ratio of (T/M).

Step-by-Step Solution:

Step 1: Given: $\frac{{{T_A}}}{{{M_A}}} = 4\frac{{{T_B}}}{{{M_B}}}$ .

Step 2: The RMS speed is given by $C = \sqrt {\frac{{3RT}}{M}}$ .

Step 3: Therefore, $\frac{{{C_A}}}{{{C_B}}} = \frac{{\sqrt {\frac{{3R{T_A}}}{{{M_A}}}} }}{{\sqrt {\frac{{3R{T_B}}}{{{M_B}}}} }} = \sqrt {\frac{{{T_A}}}{{{M_A}}} \div \frac{{{T_B}}}{{{M_B}}}} = \sqrt {\frac{{{T_A}}}{{{M_A}}} \times \frac{{{M_B}}}{{{T_B}}}}$ .

Step 4: Substituting the given condition, $\frac{{{C_A}}}{{{C_B}}} = \sqrt {4\frac{{{T_B}}}{{{M_B}}} \times \frac{{{M_B}}}{{{T_B}}}} = \sqrt 4 = 2$ .

Why other options are wrong:
- Option B (4): This is incorrect. It is the ratio of $\frac{{{T_A}}}{{{M_A}}}$ to $\frac{{{T_B}}}{{{M_B}}}$ , not the ratio of RMS speeds.
- Option C (1): This is incorrect. It implies that $\frac{{{T_A}}}{{{M_A}}} = \frac{{{T_B}}}{{{M_B}}}$ , which is not the given condition.
- Option D (0.5): This is incorrect. It is the inverse of the correct ratio.

Key Concepts:
- Root mean square speed

Memory Aids:
- $C = \sqrt {\frac{{3RT}}{M}}$

Quick Tips:
- Remember the formula for RMS speed and how it relates to temperature and molecular mass.

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Understanding +4 | -1

The temperature of argon, kept in a vessel, is raised by

{1^ \circ }C

at a constant volume. The total heat supplied to the gas is a combination of translation and rotational energies. Their respective shares are

View answer and explanation

Correct option: D

100% and 0%

Option D (100% and 0%) is correct because argon is a monoatomic gas and only possesses translational kinetic energy.

Solution approach

Correct Answer + Core Reason: 100% and 0%. Argon is a monoatomic gas, hence it possesses only translational energy.

Step-by-Step Solution:

Step 1: Identify the type of gas: Argon is a monoatomic gas.

Step 2: Determine the degrees of freedom: A monoatomic gas has 3 translational degrees of freedom and 0 rotational degrees of freedom.

Step 3: Relate degrees of freedom to energy: The total energy supplied to the gas is distributed among its degrees of freedom.

Step 4: Calculate the energy distribution: Since argon only has translational degrees of freedom, all the supplied heat goes into increasing its translational kinetic energy.

Step 5: Conclusion: Therefore, 100% of the heat supplied goes into translational energy, and 0% goes into rotational energy.

Why other options are wrong:
- Option A: This is incorrect because argon does not have rotational energy.
- Option B: This is incorrect because argon does not have rotational energy.
- Option C: This is incorrect because argon does not have rotational energy.

Key Concepts:
- Degrees of freedom
- Translational kinetic energy
- Rotational kinetic energy
- Monoatomic gases

Memory Aids:
- Remember that monoatomic gases only have translational degrees of freedom.

Quick Tips:
- Know the degrees of freedom for different types of gases (monoatomic, diatomic, polyatomic).

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Analyzing +4 | -1

Two identical containers

\mathrm{A}

and

\mathrm{B}

have frictionless pistons. They contain the same volume of a ideal gas at the same temperature. The mass of the gas in

\mathrm{A}

is

\mathrm{m}_{\mathrm{A}}

and that in

\mathrm{B}

is

\mathrm{mB}

. The gas in each cylinder is now allowed to expand isothermally to double the initial volume. The changes in the pressure in A and B are found to be

\Delta p

and

1.5 \Delta p

respectively.

View answer and explanation

Correct option: C

3 \mathrm{~mA}_{\mathrm{A}}=2 \mathrm{mB}

Option C is correct because the ratio of masses is inversely proportional to the ratio of pressure changes, leading to the relationship 3mA = 2mB.

Solution approach

Correct Answer + Core Reason: $3 \mathrm{~mA}_{\mathrm{A}}=2 \mathrm{mB}$ . The ratio of masses is inversely proportional to the ratio of pressure changes.

Step-by-Step Solution:

Step 1: Let $p_A$ and $p_B$ be the initial pressures in containers A and B, respectively.

Step 2: After isothermal expansion to double the volume, the pressures become $\frac{p_A}{2}$ and $\frac{p_B}{2}$ .

Step 3: The change in pressure in A is $\Delta p = p_A - \frac{p_A}{2} = \frac{p_A}{2}$ , so $p_A = 2\Delta p$ .

Step 4: The change in pressure in B is $1.5\Delta p = p_B - \frac{p_B}{2} = \frac{p_B}{2}$ , so $p_B = 3\Delta p$ .

Step 5: Therefore, $\frac{p_A}{p_B} = \frac{2\Delta p}{3\Delta p} = \frac{2}{3}$ .

Step 6: Using the ideal gas law, $pV = \frac{m}{M}RT$ , where $m$ is the mass and $M$ is the molar mass. Since the volume and temperature are the same, $p \propto m$ .

Step 7: Thus, $\frac{p_A}{p_B} = \frac{m_A}{m_B} = \frac{2}{3}$ , which means $3m_A = 2m_B$ .

Why other options are wrong:
- Option A: Incorrect relationship between the masses.
- Option B: Incorrect relationship between the masses.
- Option D: Incorrect relationship between the masses.

Key Concepts:
- Ideal gas law
- Isothermal process
- Pressure and volume relationship

Memory Aids:
- Ideal gas law: $pV = nRT$

Quick Tips:
- Remember the relationship between pressure, volume, and temperature in an isothermal process.
- Use the ideal gas law to relate pressure and mass.

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Analyzing +4 | -1

The equation of state of a gas is given by

\left( {P + \frac{{a{T^2}}}{V}} \right){V^c} = \left( {RT + b} \right),

where

a,b,c

and

R

are constants. The isotherms can be represented by

P = A{V^m} - B{V^n},

where

A

and

B

depend only on temperature and

View answer and explanation

Correct option: A

m = - c

and

n = - 1

Option A is correct because by rearranging the given equation of state and comparing it with the target form, we find that m = -c and n = -1.

Solution approach

Correct Answer + Core Reason: $m = - c$ and $n = - 1$ . By rearranging the equation of state and comparing coefficients.

Step-by-Step Solution:
Step 1: Start with the given equation of state: $\left( {P + \frac{{a{T^2}}}{V}} \right){V^c} = \left( {RT + b} \right)$ .
Step 2: Expand the equation: $P{V^c} + a{T^2}{V^{c - 1}} = RT + b$ .
Step 3: Isolate P: $P{V^c} = RT + b - a{T^2}{V^{c - 1}}$ .
Step 4: Divide by ${V^c}$ : $P = \frac{{RT + b}}{{{V^c}}} - \frac{{a{T^2}{V^{c - 1}}}}{{{V^c}}}$ .
Step 5: Simplify: $P = (RT + b){V^{ - c}} - a{T^2}{V^{ - 1}}$ .
Step 6: Compare with $P = A{V^m} - B{V^n}$ . We have $A = RT + b$ , $B = a{T^2}$ , $m = - c$ , and $n = - 1$ .

Why other options are wrong:
- Option B: $m = c$ and $n = 1$ is incorrect because the powers of V are negative.
- Option C: $m = - c$ and $n = 1$ is incorrect because the power of the second term should be -1.
- Option D: $m = c$ and $n = - 1$ is incorrect because the power of the first term should be -c.

Key Concepts:
- Equation of state
- Isotherms

Memory Aids:
- Rearrange and compare coefficients.

Quick Tips:
- Always isolate the variable you are solving for.

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Assertion:

{V}_{\mathrm{rms}}

and

{V}_{\mathrm{mean}}

of gaseous molecules is nearly of the order of velocity of sound.
Reason: The sound travels in air because of vibrational molecular motion.

View answer and explanation

Correct option: B

If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion

Option B is correct. Both the assertion and reason are true, but the reason doesn't explain why the speeds are of the same order.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion. Both statements are correct, but the reason doesn't explain the assertion.

Step-by-Step Solution:

Step 1: Understand the assertion: $V_{rms}$ and $V_{mean}$ of gaseous molecules are nearly of the order of the velocity of sound.

Step 2: Understand the reason: Sound travels in air because of vibrational molecular motion.

Step 3: The assertion is true because the speed of sound in a gas is related to the average speed of the gas molecules.

Step 4: The reason is also true because sound propagation involves the transfer of energy through molecular vibrations.

Step 5: However, the reason does not explain why the speeds are of the same order. The relationship between the speed of sound and the molecular speeds is more complex and involves factors like the adiabatic index.

Why other options are wrong:
- Option A: The reason is not a correct explanation.
- Option C: The assertion is true.
- Option D: Both assertion and reason are true.

Key Concepts:
- Root-mean-square velocity
- Mean velocity
- Speed of sound in a gas
- Molecular motion

Memory Aids:
- Remember that the speed of sound is related to the average speed of gas molecules.

Quick Tips:
- Understand the relationship between molecular speeds and the speed of sound.

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Understanding +4 | -1

The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies

{E_1}\;\mathrm{and}\;{E_2}

respectively. Then

View answer and explanation

Correct option: A

{E_1} = {E_2}

Option A (

{E_1} = {E_2}

) is correct because at the same temperature, the average kinetic energy depends only on the temperature and degrees of freedom, which are the same for both diatomic gases.

Solution approach

Correct Answer + Core Reason: ${E_1} = {E_2}$ . The gases carbon monoxide (CO) and nitrogen ( ${\mathrm{N}_2})$ are diatomic, so both have equal kinetic energy $\frac{5}{2}kT,\;ie.{E_1} = {E_2}$ .

Step-by-Step Solution:

Step 1: Recall the equipartition theorem: Each degree of freedom has an average energy of $\frac{1}{2}kT$ , where k is Boltzmann's constant and T is the temperature.

Step 2: Both CO and N2 are diatomic molecules. At normal temperatures, they have 5 degrees of freedom (3 translational and 2 rotational).

Step 3: Therefore, the average kinetic energy of both gases is $\frac{5}{2}kT$ .

Step 4: Since the temperature is the same, the kinetic energies are equal.

Why other options are wrong:
- Option B: ${E_1} > {E_2}$ - The kinetic energies are equal at the same temperature.
- Option C: ${E_1} < {E_2}$ - The kinetic energies are equal at the same temperature.
- Option D: ${E_1}\;$ and ${E_2}$ cannot be compared - The kinetic energies can be compared and are equal.

Key Concepts:
- Equipartition Theorem
- Kinetic Energy of Gases
- Degrees of Freedom

Memory Aids:
- Average kinetic energy depends only on temperature and degrees of freedom.

Quick Tips:
- Remember the degrees of freedom for different types of molecules.

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Application +4 | -1

Ten moles of a diatomic perfect gas are allowed to expand at constant pressure. The initial volume and temperature are

V_0

and

T_0

, respectively. If

\frac{7}{2} R T_0

heat is transferred to the gas, then the final volume and temperature are

View answer and explanation

Correct option: A

1.1 V_0, 1.1 T_0

Option A (

1.1 V_0, 1.1 T_0

) is correct because the final volume and temperature are calculated using the heat transferred at constant pressure.

Solution approach

Correct Answer + Core Reason: $1.1 V_0, 1.1 T_0$ . The final volume and temperature are calculated using the heat transferred at constant pressure.

Step-by-Step Solution:

Step 1: Use the formula $Q=n C_p \Delta T$ .

Step 2: Substitute the given values: $\frac{7}{2} R T_0 =(10)\left(\frac{7}{2} R\right)\T_f-T_0)$ .

Step 3: Solve for $T_f$ : $T_f-T_0 = \frac{T_0}{10} \Rightarrow T_f = 1.1 T_0$ .

Step 4: Since the pressure is constant, use Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow \frac{V_0}{T_0} = \frac{V_f}{1.1 T_0}$ .

Step 5: Solve for $V_f$ : $V_f = 1.1 V_0$ .

Why other options are wrong:
- Option B: This option ( $0.9 V_0, 0.9 T_0$ ) is incorrect because it implies a decrease in both volume and temperature, which is not consistent with heat being added to the gas.
- Option C: This option ( $1.1 V_0, \frac{10}{11} T_0$ ) is incorrect because it has an incorrect final temperature.
- Option D: This option ( $0.9 V_0, \frac{10}{9} T_0$ ) is incorrect because it has an incorrect final volume.

Key Concepts:
- Thermodynamics
- Heat transfer at constant pressure
- Charles's Law

Memory Aids:
- Remember the formula for heat transfer at constant pressure: $Q = nC_p \Delta T$ .

Quick Tips:
- Use Charles's Law when pressure is constant.
- Pay attention to the sign of the heat transfer (positive for heat added, negative for heat removed).

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Application +4 | -1

The root mean square speed of molecules of nitrogen gas at

27^{\circ} \mathrm{C}

is approximately: (Given mass of a nitrogen molecule

=4.6 \times 10^{-26} \mathrm{~kg}

and take Boltzmann constant

\mathrm{k}_{\mathrm{B}}=1.4 \times 10^{-23} \mathrm{JK}^{-1}

)

View answer and explanation

Correct option: D

523 \mathrm{~m} / \mathrm{s}

Option D (523 m/s) is correct because the RMS speed is calculated using the formula Vrms =

\sqrt{{3kT/m}}

.

Solution approach

Correct Answer + Core Reason: 523 m/s. The RMS speed is calculated using the formula Vrms = $\sqrt{{3kT/m}}$ .

Step-by-Step Solution:

Step 1: Given: Temperature, $T = 27^{\circ} \text{C} = 300 \text{K}$ ; Mass of nitrogen molecule, $m = 4.6 \times 10^{-26} \text{kg}$ ; Boltzmann constant, $k_B = 1.4 \times 10^{-23} \text{JK}^{-1}$ .

Step 2: Recall the formula for RMS speed: $V_{rms} = \sqrt{\frac{3kT}{m}}$ .

Step 3: Substitute the given values: $V_{rms} = \sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}$ .

Step 4: Calculate the RMS speed: $V_{rms} = \sqrt{\frac{1260 \times 10^{-23}}{4.6 \times 10^{-26}}} = \sqrt{27391.3} \approx 165 \sqrt{1.0} \approx 523 \text{m/s}$ .

Step 5: Final Answer: $523 \text{m/s}$ .

Why other options are wrong:
- Option A (27.4 m/s): This is incorrect. It is a result of an incorrect calculation or misunderstanding of the formula.
- Option B (91 m/s): This is incorrect. It is a result of an incorrect calculation or misunderstanding of the formula.
- Option C (1260 m/s): This is incorrect. It is a result of an incorrect calculation or misunderstanding of the formula.

Key Concepts:
- Root mean square speed of gas molecules
- Kinetic theory of gases

Memory Aids:
- $V_{rms} = \sqrt{\frac{3kT}{m}}$

Quick Tips:
- Remember to convert temperature to Kelvin before calculation.

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Understanding +4 | -1

Diatomic molecules like hydrogen have energies due to
bothtranslationalas well as rotational motion. From the
equation in kinetic theory

pV= \frac{2}{3}E, \, E

is

View answer and explanation

Correct option: C

Only the translational part of the energy, because during
collisions with the wall, pressure related to change in
linear momentum

Option C (Only the translational part of the energy, because during collisions with the wall, pressure related to change in linear momentum) is correct because the pressure exerted on the walls of the container is due to the change in linear momentum of the molecules.

Solution approach

Correct Answer + Core Reason: Only the translational part of the energy, because during collisions with the wall, pressure related to change in linear momentum. The pressure is related to the change in linear momentum.

Step-by-Step Solution:

Step 1: Recall the kinetic theory of gases equation: $pV = \frac{2}{3}E$ .

Step 2: Understand that $E$ in this equation represents the total translational kinetic energy of the gas molecules.

Step 3: The pressure exerted by the gas on the walls of the container is due to the change in linear momentum of the gas molecules during collisions with the walls.

Step 4: Rotational and vibrational energies do not directly contribute to the pressure exerted on the walls.

Why other options are wrong:
- Option A: The total energy per unit volume. This is incorrect because the equation relates to translational energy, not total energy.
- Option B: Only the translational part of energy, because rotational energy is very small compared to thetranslational energy. While rotational energy might be smaller, the primary reason is the mechanism of pressure generation.
- Option D: The translational part of the energy, because rotational energies of molecules can be of either sign and its average over all the molecules is zero. The sign of rotational energies is not the primary reason.

Key Concepts:
- Kinetic theory of gases
- Translational kinetic energy
- Pressure exerted by a gas

Memory Aids:
- Remember that pressure is related to the change in linear momentum.

Quick Tips:
- Understand the assumptions and limitations of the kinetic theory of gases.

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Remembering +4 | -1

If

{n}

is the number density and

{d}

is the diameter of the molecule, then the average distance covered by a molecule between two successive collisions (i.e. mean free path) is represented by:

View answer and explanation

Correct option: C

\frac{1}{\sqrt{2} n \pi d^2}

Option C (

\frac{1}{\sqrt{2} n \pi d^2}

) is correct because this is the standard formula for the mean free path of a molecule in a gas.

Solution approach

Correct Answer + Core Reason: $\frac{1}{\sqrt{2} n \pi d^2}$ . This is the formula for the mean free path.

Step-by-Step Solution:

Step 1: Define mean free path: The average distance a molecule travels between successive collisions.

Step 2: Recall the formula for mean free path: $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$ , where $n$ is the number density and $d$ is the diameter of the molecule.

Why other options are wrong:
- Option A: This option is missing the square root in the denominator.
- Option B: This option has the inverse relationship and incorrect placement of variables.
- Option D: This option has the number density and diameter squared incorrectly.

Key Concepts:
- Mean free path
- Number density
- Molecular diameter

Memory Aids:
- Remember the formula for mean free path.

Quick Tips:
- Know the factors that affect the mean free path.

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Understanding +4 | -1

Statement-I: Carbon dioxide molecule has six degrees of freedom.
Statement-II: Water molecule in water vapour has six degrees of freedom

View answer and explanation

Correct option: C

Statement-I is true, but statement-II is false

Option C is correct. Carbon dioxide (CO2) molecule has 6 degrees of freedom, while a water molecule in water vapor has more than 6 degrees of freedom due to vibrational modes at high temperatures.

Solution approach

Correct Answer + Core Reason: Statement-I is true, but statement-II is false. Statement-I is true and statement-II is false. \nCarbondioxide molecules (triatomic gas molecule) has 6 degrees of freedom at room temperature. But at high temperature, water molecule (triatomic gas molecule) has $(6+2 {V})$ degrees of freedom where ${V}$ is vibrational modes.

Step-by-Step Solution:

Step 1: Carbon dioxide (CO2) is a linear triatomic molecule. At room temperature, it has 3 translational and 2 rotational degrees of freedom, totaling 5. However, at higher temperatures, it also has 1 vibrational mode, contributing 1 vibrational kinetic energy and 1 vibrational potential energy, making a total of 6 degrees of freedom. Statement I is thus considered true.

Step 2: Water (H2O) is a non-linear triatomic molecule. It has 3 translational and 3 rotational degrees of freedom, totaling 6. At higher temperatures, it also has vibrational modes. Thus, the total degrees of freedom are more than 6. Statement II is false.

Why other options are wrong:
- Option A: Statement II is false.
- Option B: Statement II is false.
- Option D: Statement I is true.

Key Concepts:
- Degrees of freedom
- Translational degrees of freedom
- Rotational degrees of freedom
- Vibrational degrees of freedom

Memory Aids:
- Linear molecule: 3 translational + 2 rotational.
- Non-linear molecule: 3 translational + 3 rotational.

Quick Tips:
- Remember to consider vibrational modes at high temperatures.

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Remembering +4 | -1

The number of translational degrees of freedom for a diatomic gas is

View answer and explanation

Correct option: B

3

Option B (3) is correct because a diatomic gas, like any gas, can move freely in three-dimensional space, corresponding to three translational degrees of freedom.

Solution approach

Correct Answer + Core Reason: 3. A diatomic gas has 3 translational degrees of freedom.

Step-by-Step Solution:

Step 1: Define translational degrees of freedom: The number of independent ways a molecule can move in space.

Step 2: A molecule can move along the x, y, and z axes.

Step 3: Therefore, a diatomic gas has 3 translational degrees of freedom.

Why other options are wrong:
- Option A: 2 is incorrect; there are 3 translational degrees of freedom.
- Option C: 5 is the total number of degrees of freedom (3 translational + 2 rotational) for a diatomic gas at high temperatures, but the question specifically asks for translational degrees of freedom.
- Option D: 6 is incorrect; it does not represent the translational degrees of freedom.

Key Concepts:
- Degrees of freedom
- Translational motion
- Diatomic gas

Memory Aids:
- Remember that any gas has 3 translational degrees of freedom.

Quick Tips:
- Know the different types of degrees of freedom and how to calculate them.

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Analyzing +4 | -1

A gas is enclosed in a closed pot. On keeping this pot in a train moving with high speed, the temperature of the gas:

View answer and explanation

Correct option: A

Will increase

Option A is correct because the temperature of the gas will increase due to the increase in kinetic energy of the gas molecules as the train moves with high speed.

Solution approach

Correct Answer + Core Reason: Will increase. The temperature increases due to the increased kinetic energy of the gas molecules.

Step-by-Step Solution:
Step 1: When the pot is kept in a train moving with high speed, the gas molecules inside the pot gain kinetic energy due to the motion of the train.
Step 2: This increase in kinetic energy translates to an increase in the average kinetic energy of the gas molecules.
Step 3: Since temperature is directly proportional to the average kinetic energy of the gas molecules, the temperature of the gas increases.

Why other options are wrong:
- Option B: The temperature increases, not decreases.
- Option C: The temperature changes, it does not remain the same.
- Option D: The temperature change depends on the speed, not the nature of the gas (although the specific heat capacity would influence how much the temperature changes for a given energy input).

Key Concepts:
- Kinetic theory of gases
- Temperature and kinetic energy

Memory Aids:
- Remember that temperature is a measure of the average kinetic energy of the molecules.

Quick Tips:
- Consider the frame of reference when analyzing the motion of gas molecules.

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Understanding +4 | -1

Consider a mixture of gas molecule of types A, B and C having masses

{m_A} < {m_B} < {m_C}

. The ratio of their root mean square speeds at normal temperature and pressure is

View answer and explanation

Correct option: D

\frac{1}{{{\mathrm{V}_A}}} < \frac{1}{{{\mathrm{V}_B}}} < \frac{1}{{{\mathrm{V}_C}}}

Option D (

\frac{1}{{{\mathrm{V}_A}}} < \frac{1}{{{\mathrm{V}_B}}} < \frac{1}{{{\mathrm{V}_C}}}

) is correct because the root mean square speed is inversely proportional to the square root of the mass.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{{\mathrm{V}_A}}} < \frac{1}{{{\mathrm{V}_B}}} < \frac{1}{{{\mathrm{V}_C}}}$ . RMS speed is inversely proportional to mass.

Step-by-Step Solution:

Step 1: Recall the formula for root mean square (rms) speed: ${\mathrm{V}_{rms}} = \sqrt {\frac{{3RT}}{m}}$ , where R is the gas constant, T is the temperature, and m is the molar mass.

Step 2: Analyze the relationship between rms speed and mass: ${\mathrm{V}_{rms}} \propto \frac{1}{{\sqrt m }}$

Step 3: Given ${m_A} < {m_B} < {m_C}$ , then ${\mathrm{V}_A} > {\mathrm{V}_B} > {\mathrm{V}_C}$

Step 4: Therefore, $\frac{1}{{{\mathrm{V}_A}}} < \frac{1}{{{\mathrm{V}_B}}} < \frac{1}{{{\mathrm{V}_C}}}$

Why other options are wrong:
- Option A: ${\mathrm{V}_A}\mathrm{ = }{\mathrm{V}_B}\mathrm{ = }{\mathrm{V}_C} = 0$ is incorrect because rms speed is not zero at normal temperature and pressure.
- Option B: $\frac{1}{{{\mathrm{V}_A}}}\mathrm{ > }\frac{1}{{{\mathrm{V}_B}}} > \frac{1}{{{\mathrm{V}_C}}}$ is incorrect because it implies ${\mathrm{V}_A} < {\mathrm{V}_B} < {\mathrm{V}_C}$ , which is the opposite of the correct relationship.
- Option C: ${\mathrm{V}_A}\mathrm{ = }{\mathrm{V}_B} \ne \mathrm{ }{\mathrm{V}_C}$ is incorrect because the rms speeds are all different due to different masses.

Key Concepts:
- Root Mean Square Speed
- Kinetic Theory of Gases

Memory Aids:
- RMS speed is inversely proportional to the square root of mass.

Quick Tips:
- Remember the formula for rms speed and its relationship to mass and temperature.

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Understanding +4 | -1

A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per

{O_2}

molecule to that per

{N_2}

molecule is

View answer and explanation

Correct option: A

0.042361111

Option A (01:01) is correct because the average rotational kinetic energy depends only on temperature and degrees of freedom, which are the same for both gases.

Solution approach

Correct Answer + Core Reason: 01:01. Average rotational kinetic energy depends only on temperature and degrees of freedom.

Step-by-Step Solution:

Step 1: Recall the equipartition theorem: Each degree of freedom has an average energy of $\frac{{1}}{{2}}kT$ , where k is Boltzmann's constant and T is the temperature.

Step 2: For a diatomic molecule, there are 2 rotational degrees of freedom.

Step 3: The average rotational kinetic energy for each molecule is $2 \times \frac{{1}}{{2}}kT = kT$ .

Step 4: Since both oxygen and nitrogen are diatomic and at the same temperature, their average rotational kinetic energies are the same.

Step 5: Therefore, the ratio of their average rotational kinetic energies is 1:1.

Why other options are wrong:
- Option B: This ratio is incorrect because the temperature and degrees of freedom are the same for both gases.
- Option C: This ratio is incorrect because the temperature and degrees of freedom are the same for both gases.
- Option D: The moments of inertia do not directly affect the *average* rotational kinetic energy at a given temperature; they affect the rotational speed for a given energy.

Key Concepts:
- Equipartition Theorem
- Rotational Kinetic Energy
- Degrees of Freedom

Memory Aids:
- Remember that average kinetic energy depends on temperature.

Quick Tips:
- For average kinetic energy problems, focus on temperature and degrees of freedom.

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Understanding +4 | -1

According to Maxwell’s law of distribution of velocities of molecules, the most probable velocity is

View answer and explanation

Correct option: D

Less than the root mean square velocity

Option D is correct because, according to Maxwell's distribution, the most probable velocity is less than the root mean square velocity.

Solution approach

Correct Answer + Core Reason: Less than the root mean square velocity. The most probable velocity is the lowest of the three velocities.

Step-by-Step Solution:

Step 1: Define Maxwell's distribution: Maxwell's distribution describes the distribution of molecular speeds in a gas.

Step 2: Define the three velocities: The three characteristic velocities are the most probable velocity ( $v_{mp}$ ), the average velocity ( $v$ ), and the root mean square velocity ( $v_{rms}$ ).

Step 3: State the relationship: The relationship between these velocities is $v_{rms} > v > v_{mp}$ .

Step 4: Conclusion: Therefore, the most probable velocity is less than the root mean square velocity.

Why other options are wrong:
- Option A: The most probable velocity is not greater than the mean velocity.
- Option B: The most probable velocity is not equal to the mean velocity.
- Option C: The most probable velocity is not equal to the root mean square velocity.

Key Concepts:
- Maxwell's distribution
- Molecular speeds

Memory Aids:
- Remember the order: $v_{rms} > v > v_{mp}$ .

Quick Tips:
- Understand the definitions of the three characteristic velocities.

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