The average thermal energy for a mono-atomic gas is: (${{\rm{k}}_{\rm{B}}}$ is Boltzmann constant and ${\rm{T}}$, absolute temperature)

$\frac{3}{2}\,{{\rm{k}}_{\rm{B}}}{\rm{T}}$

$\frac{1}{2}\,{{\rm{k}}_{\rm{B}}}{\rm{T}}$

$\frac{5}{2}\,{{\rm{k}}_{\rm{B}}}{\rm{T}}$

$\frac{7}{2}\,{{\rm{k}}_{\rm{B}}}{\rm{T}}$

KINETIC THEORY OF GASES MCQs for JEE PHYSICS — Medium, Page 3

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Understanding +4 | -1

According to kinetic theory of gases,
A. The motion of the gas molecules freezes at

0^\circ C

B. The mean free path of gas molecules decreases if the density of molecules is increased.
C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.
D. Average kinetic energy per molecule per degree of freedom is

\frac{3}{2}\,{k_B}T

(for monoatomic gases).
Choose the most appropriate answer from the options given below:

View answer and explanation

Correct option: B

B and C only

Option B (B and C only) is correct because the mean free path decreases with increasing density and increases with increasing temperature at constant pressure.

Solution approach

Correct Answer + Core Reason: B and C only. Mean free path decreases with increasing density and increases with increasing temperature at constant pressure.

Step-by-Step Solution:

Step 1: Recall the kinetic theory of gases.

Step 2: Understand the relationship between temperature and molecular motion.

Step 3: The motion of gas molecules does not freeze at 0°C, but at 0 K.

Step 4: The mean free path ( $\lambda$ ) is inversely proportional to the density (n) of molecules: $\lambda = \frac{{1}}{{\sqrt{{2}} \pi n d^2}}$ , where d is the diameter of the molecules.

Step 5: If temperature increases while pressure is constant, the volume increases (Charles's Law), which increases the mean free path.

Step 6: Average kinetic energy per molecule per degree of freedom is $\frac{{1}}{{2}} k_B T$ .

Why other options are wrong:
- Option A: A and C only: A is incorrect, the motion of gas molecules freezes at 0 K.
- Option C: A and B only: A is incorrect, the motion of gas molecules freezes at 0 K.
- Option D: C and D only: D is incorrect, Average kinetic energy per molecule per degree of freedom is $\frac{{1}}{{2}} k_B T$ .

Key Concepts:
- Kinetic theory of gases
- Mean free path
- Temperature
- Density
- Charles's Law

Memory Aids:
- Remember that mean free path is inversely proportional to density.

Quick Tips:
- Use the kinetic theory of gases to understand the behavior of gases under different conditions.

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Assertion: Each vibrational mode gives two degrees of freedom
Reason: By law of equipartition of energy, the energy for each degree of freedom in thermal equilibrium is

2 k_B T

View answer and explanation

Correct option: C

If Assertion is true and Reason is false

Option C is correct because each vibrational mode contributes two degrees of freedom (kinetic and potential energy), but the energy for each degree of freedom is

\frac{1}{2}k_BT

, not

2k_BT

.

Solution approach

Correct Answer + Core Reason: If Assertion is true and Reason is false. Each vibrational mode contributes two degrees of freedom, but the energy per degree of freedom is not $2k_BT$ .

Step-by-Step Solution:

Step 1: Analyze the assertion.\nEach vibrational mode has both kinetic and potential energy, contributing two degrees of freedom. This is true.

Step 2: Analyze the reason.\nAccording to the law of equipartition of energy, the average energy for each degree of freedom in thermal equilibrium is $\frac{{1}}{{2}}k_BT$ , not $2k_BT$ . This is false.

Step 3: Determine the relationship between the assertion and the reason.\nThe assertion is true, and the reason is false.

Why other options are wrong:
- Option A: The reason is false.
- Option B: The reason is false.
- Option D: The assertion is true.

Key Concepts:
- Degrees of freedom
- Vibrational modes
- Law of equipartition of energy

Memory Aids:
- Each degree of freedom has an average energy of $\frac{{1}}{{2}}k_BT$ .

Quick Tips:
- Remember the law of equipartition of energy.

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At what temperature the rms velocity of helium molecules will be equal to that of hydrogen molecules at NTP?

View answer and explanation

Correct option: C

273

^ \circ C

Option C (273

^ \circ C

) is correct because it is the temperature at which the rms velocity of helium molecules equals that of hydrogen molecules at NTP, calculated using the rms velocity formula.

Solution approach

Correct Answer + Core Reason: 273 $^ \circ C$ . The temperature is calculated by equating the rms velocities of helium and hydrogen molecules.

Step-by-Step Solution:

Step 1: Define the rms velocity formula: ${v_{\mathrm{rms}}} = \sqrt {\frac{{3RT}}{M}}$ , where $R$ is the gas constant, $T$ is the temperature, and $M$ is the molecular weight.

Step 2: Given: ${v_{\mathrm{He}}} = {v_\mathrm{H}}$ , ${T_\mathrm{H}} = 273\;\mathrm{K}$ , ${M_\mathrm{H}} = 2$ , ${M_{\mathrm{He}}} = 4$

Step 3: Equate the rms velocities: $\frac{{{v_\mathrm{H}}}}{{{v_{\mathrm{He}}}}} = \sqrt {\frac{{{T_\mathrm{H}}}}{{{T_{\mathrm{He}}}}} \times \frac{{{M_{\mathrm{He}}}}}{{{M_\mathrm{H}}}}}$

Step 4: Substitute the given values: $1 = \sqrt {\frac{{273}}{{{T_{\mathrm{He}}}}} \times \frac{4}{2}}$

Step 5: Solve for ${T_{\mathrm{He}}}$ : ${T_{\mathrm{He}}} = 546\;\mathrm{K}$

Step 6: Convert to Celsius: ${T_{\mathrm{He}}} = {\left( {546 - 273} \right)^ \circ }C = {273^ \circ }C$

Why other options are wrong:
- Option A (844 K): This is incorrect as it doesn't follow from the correct application of the rms velocity formula and the given conditions.
- Option B (64 K): This is incorrect as it doesn't follow from the correct application of the rms velocity formula and the given conditions.
- Option D (273 K): This is the temperature of Hydrogen, not Helium.

Key Concepts:
- Root Mean Square (RMS) velocity
- Kinetic Theory of Gases
- Temperature conversion (Kelvin to Celsius)

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

Quick Tips:
- Always convert temperature to Kelvin before using it in gas equations.
- Pay attention to the molecular weights of the gases involved.

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KE per unit volume is

E

. The pressure exerted by the gas is given by

View answer and explanation

Correct option: B

\frac{{2E}}{3}

Option B (

\frac{{2E}}{3}

) is correct because the pressure exerted by a gas is related to the kinetic energy per unit volume by the equation

P = \frac{2}{3}E

.

Solution approach

Correct Answer + Core Reason: $\frac{{2E}}{3}$ . The pressure exerted by the gas is related to kinetic energy per unit volume.

Step-by-Step Solution:

Step 1: Recall the kinetic theory of gases. The pressure exerted by a gas is given by $p = \frac{1}{3} \rho \overline{c^2}$ , where $\rho$ is the density and $\overline{c^2}$ is the mean square speed of the gas molecules.

Step 2: Relate density to mass and volume. Density $\rho = \frac{m}{V}$ , where $m$ is the mass and $V$ is the volume.

Step 3: Express pressure in terms of mass, volume, and mean square speed. $p = \frac{1}{3} \frac{m}{V} \overline{c^2}$ .

Step 4: Relate kinetic energy per unit volume to mass and mean square speed. Kinetic energy per unit volume $E = \frac{1}{2} \frac{m}{V} \overline{c^2}$ .

Step 5: Express pressure in terms of kinetic energy per unit volume. $p = \frac{2}{3} \left( \frac{1}{2} \frac{m}{V} \overline{c^2} \right) = \frac{2}{3} E$ .

Why other options are wrong:
- Option A: This is incorrect because the correct relationship is $p = \frac{2}{3}E$ .
- Option B: This is the correct answer.
- Option C: This is incorrect because the correct relationship is $p = \frac{2}{3}E$ .
- Option D: This is incorrect because the correct relationship is $p = \frac{2}{3}E$ .

Key Concepts:
- Kinetic theory of gases
- Pressure exerted by a gas
- Kinetic energy per unit volume

Memory Aids:
- Pressure = (2/3) * Energy Density

Quick Tips:
- Remember the relationship between pressure and kinetic energy in gases.

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The average thermal energy for a mono-atomic gas is: (

{\mathrm{k}_\mathrm{B}}

is Boltzmann constant and

\mathrm{T}

, absolute temperature)

View answer and explanation

Correct option: B

\frac{3}{2}\,{\mathrm{k}_\mathrm{B}}\mathrm{T}

Option B is correct. A monoatomic gas has 3 degrees of freedom, so the average thermal energy is (3/2) * k_B * T.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}\,{\mathrm{k}_{\mathrm{B}}}\mathrm{T}$ . The average thermal energy for a monoatomic gas is $\frac{3}{2} k_B T$ due to its 3 degrees of freedom.

Step-by-Step Solution:

Step 1: Define average thermal energy.

Step 2: State the equipartition theorem.

Step 3: Apply the theorem to a monoatomic gas with 3 degrees of freedom.

Step 4: Calculate the average thermal energy: $E = \frac{3}{2} k_B T$ .

Why other options are wrong:
- Option A: $\frac{1}{2}\,{\mathrm{k}_{\mathrm{B}}}\mathrm{T}$ is the energy per degree of freedom.
- Option C: $\frac{5}{2}\,{\mathrm{k}_{\mathrm{B}}}\mathrm{T}$ is the average thermal energy for a diatomic gas at certain temperatures.
- Option D: $\frac{7}{2}\,{\mathrm{k}_{\mathrm{B}}}\mathrm{T}$ is not the correct average thermal energy for a monoatomic gas.

Key Concepts:
- Kinetic theory of gases
- Equipartition theorem
- Degrees of freedom

Memory Aids:
- Mono = 3/2

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

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The temperature at which the average translational kinetic energy of a molecule is equal to the energy gained by an electron in accelerating from rest through a potential difference of

1\;volt

is

View answer and explanation

Correct option: D

7.7 \times {10^3}K

Option D (

7.7 \times {10^3}K

) is correct because it is calculated by equating the average translational kinetic energy of a molecule to the energy gained by an electron accelerated through a potential difference of 1 volt.

Solution approach

Correct Answer + Core Reason: $7.7 \times {10^3}K$ . The temperature is found by equating the average translational kinetic energy to the electron's energy gain.

Step-by-Step Solution:

Step 1: The average translational kinetic energy of a molecule is given by $KE = \frac{3}{2} kT$ , where $k$ is the Boltzmann constant and $T$ is the temperature.

Step 2: The energy gained by an electron in accelerating through a potential difference of $1 \,V$ is $E = 1 \,eV = 1.6 \times 10^{-19} \,J$ .

Step 3: Equate the two energies: $\frac{3}{2} kT = 1 \,eV$ .

Step 4: Solve for $T$ : $T = \frac{2}{3} \frac{{eV}}{{k}}$ .

Step 5: Substitute the values: $T = \frac{2}{3} \frac{{1.6 \times 10^{-19} \,J}}{{1.38 \times 10^{-23} \,J/K}}$ .

Step 6: Calculate the temperature: $T = \frac{2}{3} \times \frac{{1.6}}{{1.38}} \times 10^4 \,K \approx 7.7 \times 10^3 \,K$ .

Why other options are wrong:
- **Option A ( $4.6 \times {10^3}K$ ): This is incorrect because it likely involves an error in the calculation or using an incorrect formula.
- Option B ( $11.6 \times {10^3}K$ ): This is incorrect and doesn't follow from the correct application of the equipartition theorem.
- Option C ( $23.2 \times {10^3}K$ ): This is incorrect and doesn't follow from the correct application of the equipartition theorem.

Key Concepts:
- Average translational kinetic energy
- Boltzmann constant
- Electron volt

Memory Aids:**
- Remember the average translational kinetic energy formula: $KE = \frac{3}{2} kT$ .
- Remember the value of the Boltzmann constant: $k = 1.38 \times 10^{-23} \,J/K$ .

Quick Tips:
- Pay attention to the units and make sure they are consistent.
- Remember that $1 \,eV = 1.6 \times 10^{-19} \,J$ .

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If a Vander-Waal’s gas expands freely, then final temperature is

View answer and explanation

Correct option: A

Less than the initial temperature

Option A (Less than the initial temperature) is correct because a van der Waals gas cools during free expansion due to overcoming intermolecular forces.

Solution approach

Correct Answer + Core Reason: Less than the initial temperature. Vander-Waal’s gas cools during free expansion.

Step-by-Step Solution:

Step 1: Understand free expansion: Free expansion is an adiabatic process where a gas expands into a vacuum.

Step 2: Consider the van der Waals gas: In a van der Waals gas, there are intermolecular forces of attraction.

Step 3: Explain the temperature change: During free expansion, the gas has to do work against these intermolecular forces. This work is done at the expense of the internal energy of the gas, causing the temperature to decrease.

Step 4: Final Answer: Less than the initial temperature

Why other options are wrong:
- Option B: Incorrect. The temperature decreases, it doesn't remain the same.
- Option C: Incorrect. The temperature decreases, it doesn't increase.
- Option D: Incorrect. The temperature always decreases in free expansion of Vander waal's gas.

Key Concepts:
- Free expansion
- van der Waals gas
- Intermolecular forces
- Internal energy

Memory Aids:
- Remember that free expansion is an adiabatic process with no external work done, but internal work can be done in a real gas.

Quick Tips:
- For ideal gases, free expansion results in no temperature change. For real gases, temperature typically decreases.

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For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molecular specific heat

\left( {{C_V}} \right)

and gas constant

\left( R \right)

View answer and explanation

Correct option: D

{C_V} = 3R

Option D (

{C_V} = 3R

) is correct because the molecular specific heat is calculated using the equipartition theorem, where each degree of freedom contributes R/2 to the specific heat.

Solution approach

Correct Answer + Core Reason: ${C_V} = 3R$ . The molecular specific heat is calculated using the equipartition theorem.

Step-by-Step Solution:

Step 1: Identify the given information: Number of degrees of freedom $f = 6$ .

Step 2: Use the law of equipartition of energy: Each degree of freedom contributes $\frac{1}{2}kT$ energy per molecule, where $k$ is Boltzmann's constant.

Step 3: Calculate the total energy per molecule: $E = \frac{1}{2}f \times kT = \frac{1}{2} \times 6 \times kT = 3kT$ .

Step 4: Calculate the molar internal energy: $U = 3N_A kT = 3RT$ , where $N_A$ is Avogadro's number and $R = N_A k$ is the gas constant.

Step 5: Calculate the molar specific heat at constant volume: $C_V = \frac{dU}{dT} = \frac{d(3RT)}{dT} = 3R$ .

Step 6: Final Answer: ${C_V} = 3R$

Why other options are wrong:
- **Option A ( ${C_V} = \frac{R}{2}$ ): This is incorrect. It would be the specific heat for a molecule with only one degree of freedom.
- Option B ( ${C_V} = R$ ): This is incorrect. It would be the specific heat for a molecule with two degrees of freedom.
- Option C ( ${C_V} = 2R$ ): This is incorrect. It would be the specific heat for a molecule with four degrees of freedom.

Key Concepts:
- Law of equipartition of energy
- Degrees of freedom
- Molecular specific heat at constant volume
- Gas constant

Memory Aids:**
- $C_V = \frac{1}{2}fR$ (molar specific heat at constant volume)

Quick Tips:
- Remember that each degree of freedom contributes $\frac{1}{2$

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Which of the following is incorrect regarding the assumptions of kinetic theory of gases?

View answer and explanation

Correct option: C

Gas molecules have some intermolecular forces of attraction

Option C is correct because the kinetic theory of gases assumes negligible intermolecular forces, not the presence of some attractive forces.

Solution approach

Correct Answer + Core Reason: Gas molecules have some intermolecular forces of attraction. This contradicts the kinetic theory of gases.

Step-by-Step Solution:

Step 1: Recall the assumptions of the kinetic theory of gases: Elastic collisions, negligible volume, negligible intermolecular forces.

Step 2: Analyze the given options: Option C states that gas molecules have some intermolecular forces of attraction, which contradicts the assumption of negligible intermolecular forces.

Step 3: Conclusion: Option C is incorrect.

Why other options are wrong:
- Option A: Collision between the molecules of gases and with the walls of the container are elastic is a correct assumption.
- Option B: Number of molecules in the sample should be of the order of moles is a reasonable assumption.
- Option D: Gas molecules have negligible volume is a correct assumption.

Key Concepts:
- Kinetic theory of gases
- Assumptions of KTG
- Intermolecular forces

Memory Aids:
- KTG: No intermolecular forces

Quick Tips:
- Remember the key assumptions of the kinetic theory of gases.

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According to the kinetic theory of gases the

r.m.s

velocity of gas molecules is directly proportional to

View answer and explanation

Correct option: B

\sqrt T

Option B (

\sqrt T

) is correct because the root-mean-square (rms) velocity of gas molecules is directly proportional to the square root of the absolute temperature.

Solution approach

Correct Answer + Core Reason: $\sqrt T$ is the correct answer because ${v_{rms}} \propto \sqrt T$ .

Step-by-Step Solution:

Step 1: Recall the formula for the root-mean-square (rms) velocity of gas molecules.

Step 2: The formula is: ${v_{rms}} = \sqrt{\frac{{3RT}}{M}}$ , where $R$ is the ideal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass.

Step 3: Identify the relationship between $v_{rms}$ and $T$ .

Step 4: From the formula, ${v_{rms}} \propto \sqrt T$ .

Why other options are wrong:
- Option A: $T$ is incorrect because $v_{rms}$ is proportional to the square root of $T$ , not $T$ itself.
- Option C: ${T^2}$ is incorrect because $v_{rms}$ is proportional to the square root of $T$ , not $T^2$ .
- Option D: $1/\sqrt T$ is incorrect because $v_{rms}$ is directly proportional to $\sqrt T$ , not inversely proportional.

Key Concepts:
- Kinetic Theory of Gases
- Root-Mean-Square Velocity
- Temperature

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

Quick Tips:
- Remember the relationship between rms velocity and temperature.

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According to the kinetic theory of gases, the pressure exerted by a gas on the walls is measured as

View answer and explanation

Correct option: A

rate of change of momentum imparted to the walls per second per unit area.

Option A is correct. According to the kinetic theory of gases, the pressure exerted by a gas is due to the rate of change of momentum imparted to the walls per second per unit area.

Solution approach

Correct Answer + Core Reason: rate of change of momentum imparted to the walls per second per unit area. Pressure is defined as force per unit area, and force is the rate of change of momentum.

Step-by-Step Solution:

Step 1: Recall the kinetic theory of gases: Gas particles are in constant random motion.

Step 2: Understand the origin of pressure: Pressure arises from the collisions of gas particles with the walls of the container.

Step 3: Relate force and momentum: Force is the rate of change of momentum, $F = \frac{{dp}}{{dt}}$

Step 4: Define pressure: Pressure is force per unit area, $P = \frac{F}{A}$

Step 5: Combine the concepts: $P = \frac{1}{A} \frac{{dp}}{{dt}}$ , which means pressure is the rate of change of momentum per unit area.

Step 6: Conclusion: Therefore, the pressure exerted by a gas is the rate of change of momentum imparted to the walls per second per unit area.

Why other options are wrong:
- Option B: Momentum imparted to the walls per unit area does not account for the rate of change, which is crucial for defining force and pressure.
- Option C: Change of momentum imparted to the walls per unit area is not a rate, and thus does not fully describe the pressure.
- Option D: Change in momentum per unit volume is related to density and velocity, but not directly to pressure.

Key Concepts:
- Kinetic theory of gases
- Pressure as force per unit area
- Force as rate of change of momentum

Memory Aids:
- Remember that pressure is related to the continuous collisions of gas particles with the walls.

Quick Tips:
- Understand the fundamental relationship between pressure, force, and momentum in the context of kinetic theory.

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The ratio of mean kinetic energy of hydrogen and oxygen at a given temperature is

View answer and explanation

Correct option: D

1\;:1

Option D (1 : 1) is correct because the average kinetic energy of a molecule depends only on the temperature and is independent of the type of gas.

Solution approach

Correct Answer + Core Reason: $1\:1$ . The average kinetic energy of a molecule depends only on temperature.

Step-by-Step Solution:

Step 1: Recall the equipartition theorem: The average kinetic energy of a molecule is given by $\frac{3}{2}kT$ , where $k$ is Boltzmann's constant and $T$ is the temperature.

Step 2: The kinetic energy depends only on the temperature and is independent of the type of gas.

Step 3: Since the temperature is the same for both hydrogen and oxygen, their mean kinetic energies are equal.

Step 4: Therefore, the ratio of their mean kinetic energies is 1:1.

Step 5: Final Answer: $1\:1$

Why other options are wrong:
- Option A: $1\:16$ : This is incorrect because it considers the ratio of molar masses, which is not relevant to the mean kinetic energy at a given temperature.
- Option B: $1\:8$ : This is incorrect because it considers the ratio of molar masses, which is not relevant to the mean kinetic energy at a given temperature.
- Option C: $1\:4$ : This is incorrect because it considers the ratio of molar masses, which is not relevant to the mean kinetic energy at a given temperature.

Key Concepts:
- Kinetic Theory of Gases
- Equipartition Theorem
- Average Kinetic Energy

Memory Aids:
- "KE depends only on T"

Quick Tips:
- Remember that the average kinetic energy of a gas molecule is directly proportional to the absolute temperature.

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Assertion (A) : In equation

p=\frac{1}{3} \alpha v_{\text {rms }}^2

, the term

\alpha

represents density of gas.

Reason (R) :

v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}

.

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 (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because both the assertion and the reason are true statements, but the reason does not explain the assertion.

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: Analyze the Assertion.

Step 2: The assertion states that in the equation $p = \frac{1}{3} \alpha v_{\text{rms}}^2$ , the term $\alpha$ represents the density of the gas. This is true because $\alpha = \frac{mN}{V}$ , where m is the mass of a molecule, N is the number of molecules, and V is the volume, which is the density.

Step 3: Analyze the Reason.

Step 4: The reason states that $v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}$ . This is also true, as it is the formula for the root-mean-square speed of gas molecules.

Step 5: Determine if the Reason explains the Assertion.

Step 6: The reason (the formula for $v_{\text{rms}}$ ) does not explain why $\alpha$ represents the density of the gas. The assertion is based on the definition of density, while the reason is about the root-mean-square speed.

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

Key Concepts:
- Pressure of a gas
- Root-mean-square speed
- Density of a gas

Memory Aids:
- Remember the formula for the pressure of a gas and the root-mean-square speed.

Quick Tips:
- Understand the definitions of pressure, density, and root-mean-square speed in the context of gases.

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Assertion (A) Assertion The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume.

Reason (R) The molecules of gas collide with each other and the velocities of the molecules change due to the collision.

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 because both the assertion and reason are true, but the reason does not explain the assertion.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. The assertion is based on the kinetic theory of gases, and the reason is a general property of gas molecules.

Step-by-Step Solution:

Step 1: Analyze the Assertion: The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume. This is true because $KE = \frac{3}{2}PV$ .

Step 2: Analyze the Reason: The molecules of gas collide with each other and the velocities of the molecules change due to the collision. This is also true, as gas molecules are in constant random motion and collide with each other.

Step 3: Determine the relationship: The reason does not explain the assertion. The assertion is a direct consequence of the kinetic theory of gases, while the reason is a general property of gas molecules.

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

Key Concepts:
- Kinetic Theory of Gases
- Translational Kinetic Energy
- Ideal Gas Law

Memory Aids:
- Remember the formula: $KE = \frac{3}{2}PV$ .

Quick Tips:
- Understand the relationship between kinetic energy, pressure, and volume in ideal gases.

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The root mean square speed of the molecules of a gas is

View answer and explanation

Correct option: C

Independent of its pressure but directly proportional to the square root of its Kelvin temperature

Option C (Independent of its pressure but directly proportional to the square root of its Kelvin temperature) is correct because the RMS speed is proportional to the square root of temperature and independent of pressure.

Solution approach

Correct Answer + Core Reason: Independent of its pressure but directly proportional to the square root of its Kelvin temperature. The root mean square speed depends on temperature and molar mass, not pressure.

Step-by-Step Solution:
Step 1: Recall the formula for root mean square speed: $v_{rms} = \sqrt{\frac{{3RT}}{{M}}}$ , where R is the gas constant, T is the Kelvin temperature, and M is the molar mass.
Step 2: Analyze the formula: The formula shows that $v_{rms}$ is directly proportional to the square root of the Kelvin temperature (T) and inversely proportional to the square root of the molar mass (M). It does not depend on pressure.
Step 3: Conclusion: The root mean square speed is independent of pressure but directly proportional to the square root of its Kelvin temperature.

Why other options are wrong:
- Option A: The RMS speed is directly proportional to the square root of the Kelvin temperature, not directly proportional to the Kelvin temperature.
- Option B: The RMS speed is not directly proportional to the square root of its pressure.
- Option D: The RMS speed is not directly proportional to its pressure.

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

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

Quick Tips:
- When dealing with RMS speed problems, always ensure the temperature is in Kelvin and consider the molar mass of the gas.

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Assertion: The ratio

C_p / C_v

is more for helium gas than for hydrogen gas.
Reason: All metals acts as black bodies.

View answer and explanation

Correct option: D

If both assertion and reason are false

Option D is correct because both the assertion and the reason are false statements regarding the ratio of specific heats and the properties of metals as black bodies.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are false. The ratio of specific heats is higher for hydrogen than helium, and metals are not perfect black bodies.

Step-by-Step Solution:
Step 1: Analyze the assertion: The ratio $C_p / C_v$ ( $\gamma$ ) is more for helium gas than for hydrogen gas.
Step 2: Recall the formula for $\gamma$ : $\gamma = 1 + \frac{{2}}{{f}}$ , where $f$ is the degrees of freedom.
Step 3: Determine the degrees of freedom for Helium (monatomic): $f_{He} = 3$ . Therefore, $\gamma_{He} = 1 + \frac{{2}}{{3}} = 1.67$ .
Step 4: Determine the degrees of freedom for Hydrogen (diatomic): $f_{H_2} = 5$ . Therefore, $\gamma_{H_2} = 1 + \frac{{2}}{{5}} = 1.4$ .
Step 5: Compare the values: $\gamma_{He} > \gamma_{H_2}$ . Thus, the assertion is false.
Step 6: Analyze the reason: All metals act as black bodies.
Step 7: Recall the definition of a black body: A black body absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. Metals reflect a significant portion of incident radiation, especially in the visible spectrum. Thus, the reason is false.

Why other options are wrong:
- Option A: Incorrect because both assertion and reason are false.
- Option B: Incorrect because both assertion and reason are false.
- Option C: Incorrect because both assertion and reason are false.

Key Concepts:
- Ratio of specific heats ( $\gamma$ )
- Degrees of freedom
- Black body radiation

Memory Aids:
- Monatomic gases have 3 degrees of freedom.
- Diatomic gases have 5 degrees of freedom.

Quick Tips:
- Remember the degrees of freedom for monatomic and diatomic gases to quickly determine the ratio of specific heats.

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If Maxwell distribution is valid and

v_p

denotes the most probable speed,

v_{av}

is the average speed and

v_{rms}

is the root-mean-square speed, then

View answer and explanation

Correct option: C

v_p < v_{av} < v_{rms}

Option C (

v_p < v_{av} < v_{rms}

) is correct because the most probable speed is the lowest, followed by the average speed, and the root-mean-square speed is the highest.

Solution approach

Correct Answer + Core Reason: $v_p < v_{av} < v_{rms}$ . This is the correct order of the speeds in Maxwell distribution.

Step-by-Step Solution:

Step 1: Recall the formulas for the most probable speed ( $v_p$ ), average speed ( $v_{av}$ ), and root-mean-square speed ( $v_{rms}$ ):
- $v_p = \sqrt{{\frac{{2RT}}{{M}}}}$
- $v_{av} = \sqrt{{\frac{{8RT}}{{\pi M}}}}$
- $v_{rms} = \sqrt{{\frac{{3RT}}{{M}}}}$

Step 2: Compare the speeds:
- $v_p : v_{av} : v_{rms} = \sqrt{{2}} : \sqrt{{\frac{{8}}{{\pi}}}} : \sqrt{{3}}$
- Since $\sqrt{{2}} \approx 1.414$ , $\sqrt{{\frac{{8}}{{\pi}}}} \approx 1.596$ , and $\sqrt{{3}} \approx 1.732$ , we have $v_p < v_{av} < v_{rms}$ .

Step 3: Conclusion: The correct order is $v_p < v_{av} < v_{rms}$ .

Why other options are wrong:
- Option A: This is incorrect because the average speed is greater than the most probable speed.
- Option B: This is incorrect because the root-mean-square speed is greater than both the average and most probable speeds.
- Option D: This is incorrect because the average speed is less than the root-mean-square speed.

Key Concepts:
- Maxwell distribution
- Most probable speed
- Average speed
- Root-mean-square speed

Memory Aids:
- Remember the order: most probable < average < root-mean-square.

Quick Tips:
- The speeds are related to the square root of the temperature and inversely proportional to the square root of the molar mass.

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Understanding +4 | -1

Relation between the ratio of specific heats

\left( \gamma \right)

of gas and degree of freedom 'f' will be

View answer and explanation

Correct option: C

f = 2/\left( {\gamma - 1} \right)

Option C (

f = 2/(\gamma - 1)

) is correct because it represents the relationship between the ratio of specific heats and the degree of freedom of a gas.

Solution approach

Correct Answer + Core Reason: $f = 2/(\gamma - 1)$ . This formula directly relates the degrees of freedom to the ratio of specific heats.

Step-by-Step Solution:

Step 1: Start with the definition of the ratio of specific heats: $\gamma = \frac{{C_P}}{{C_V}}$

Step 2: Express $C_P$ in terms of $C_V$ : $C_P = C_V + R$

Step 3: Substitute into the equation for $\gamma$ : $\gamma = \frac{{C_V + R}}{{C_V}}$

Step 4: Express $C_V$ in terms of the degrees of freedom, f: $C_V = \frac{1}{2}fR$

Step 5: Substitute $C_V$ into the equation for $\gamma$ : $\gamma = \frac{{\frac{1}{2}fR + R}}{{\frac{1}{2}fR}} = \frac{{f + 2}}{f}$

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

Why other options are wrong:
- Option A: This is incorrect because it adds f and 2 directly, which is not the correct relationship.
- Option B: This is incorrect because it involves reciprocals and does not lead to the correct relationship.
- Option D: This is incorrect because it multiplies $(\gamma - 1)$ by 2, which is not the correct relationship.

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

Memory Aids:
- Remember that $\gamma$ is always greater than 1.

Quick Tips:
- Start with the definition of $\gamma$ and work your way through the substitutions.

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The specific heat of an ideal gas is

View answer and explanation

Correct option: D

Independent of

T

Option D is correct. For an ideal gas, the specific heat is independent of temperature, assuming constant degrees of freedom.

Solution approach

Correct Answer + Core Reason: Independent of $T$ . The specific heat of an ideal gas is independent of temperature under ideal conditions.

Step-by-Step Solution:

Step 1: Define specific heat.

Step 2: State the properties of an ideal gas.

Step 3: Explain that for an ideal gas, specific heat is constant with respect to temperature.

Why other options are wrong:
- Option A: Specific heat is not proportional to $T$ for an ideal gas.
- Option B: Specific heat is not proportional to ${T^2}$ for an ideal gas.
- Option C: Specific heat is not proportional to ${T^3}$ for an ideal gas.

Key Concepts:
- Specific heat
- Ideal gas law

Memory Aids:
- Ideal = Independent

Quick Tips:
- Remember the assumptions made for ideal gases.

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The average translational kinetic energy of

{O}_2

(molar mass 32) at a particular temperature is

0.048 \mathrm{~eV}

. The average translational kinetic energy of

{N}_2

(molar mass 28) molecules (in

\mathrm{eV}

) at the same temperature is

View answer and explanation

Correct option: C

0.048

Option C (0.048) is correct because the average translational kinetic energy depends only on temperature, which is the same for both gases.

Solution approach

Correct Answer + Core Reason: 0.048. The average translational kinetic energy depends only on temperature.

Step-by-Step Solution:

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

Step 2: Note the dependence on temperature. The kinetic energy depends only on temperature and not on the molar mass of the gas.

Step 3: Compare the temperatures. Since the temperature is the same for both oxygen and nitrogen, their average translational kinetic energies are equal.

Step 4: Conclude the kinetic energy of nitrogen. Therefore, the average translational kinetic energy of N₂ is also 0.048 eV.

Why other options are wrong:
- Option A (0.0015): This is incorrect because it assumes a dependence on molar mass, which is not present in the formula for average translational kinetic energy.
- Option B (0.003): This is incorrect because it assumes a dependence on molar mass, which is not present in the formula for average translational kinetic energy.
- Option D (0.768): This is incorrect because it assumes a dependence on molar mass, which is not present in the formula for average translational kinetic energy.

Key Concepts:
- Average translational kinetic energy
- Boltzmann constant
- Temperature

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

Quick Tips:
- Remember that average translational kinetic energy depends only on temperature.

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Application +4 | -1

Find the ratio of specific heat at constant pressure to the specific heat constant volume for

N{H_3}

View answer and explanation

Correct option: C

1.28

Option C (1.28) is correct because the ratio of specific heats for a polyatomic gas like NH3 is less than 1.33, and 1.28 is the closest value.

Solution approach

Correct Answer + Core Reason: 1.28. The ratio of specific heats for $NH_3$ is less than 1.33 due to its polyatomic nature.

Step-by-Step Solution:

Step 1: Identify the gas type: $NH_3$ (Ammonia) is a polyatomic gas.

Step 2: Recall the formula for $\gamma$ for polyatomic gases: $\gamma_{poly} = \frac{{4 + f_{vib}}}{{3 + f_{vib}}}$ , where $f_{vib}$ is the number of vibrational degrees of freedom.

Step 3: Consider the limit as $f_{vib}$ approaches infinity: As the number of vibrational degrees of freedom increases, $\gamma_{poly}$ approaches 1.

Step 4: Consider the case when $f_{vib} = 0$ : When there are no vibrational degrees of freedom, $\gamma_{poly} = \frac{4}{3} \approx 1.33$ .

Step 5: Determine the range of $\gamma_{poly}$ : Since $NH_3$ is a polyatomic gas with some vibrational degrees of freedom, its $\gamma$ value will be between 1 and 1.33.

Step 6: Choose the closest value: Among the given options, 1.28 is the only value within the expected range.

Why other options are wrong:
- Option A (1.33): This is the value of $\gamma$ for a polyatomic gas with no vibrational degrees of freedom. $NH_3$ has vibrational degrees of freedom, so its $\gamma$ will be less than 1.33.
- Option B (1.44): This value is too high for a polyatomic gas like $NH_3$ .
- Option D (1.67): This is the value of $\gamma$ for a monatomic gas, which is not applicable to $NH_3$ .

Key Concepts:
- Specific heat ratio ( $\gamma$ )
- Degrees of freedom (vibrational)
- Polyatomic gases

Memory Aids:
- Remember that as the atomicity of a gas increases, the value of $\gamma$ decreases.

Quick Tips:
- For polyatomic gases, the value of $\gamma$ is always between 1 and 1.33.
- The more complex the molecule, the lower the value of $\gamma$ .

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Application +4 | -1

The R.M.S. speed of oxygen molecules at temperature

{T}

(in kelvin) is

{v} \,{ms}^{-1}

. As the temperature becomes

4 {~T}

and the oxygen gas dissociated into atomic oxygen, then the speed of atomic oxygen

View answer and explanation

Correct option: D

Becomes

2 \sqrt{2} \,v

Option D (Becomes

2 \sqrt{2} \\,v

) is correct because the RMS speed is proportional to the square root of temperature and inversely proportional to the square root of the molar mass.

Solution approach

Correct Answer + Core Reason: Becomes $2 \sqrt{2} \\,v$ . The RMS speed changes due to temperature and dissociation.

Step-by-Step Solution:

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

Step 2: Initial RMS speed of oxygen molecules: $v_1 = \sqrt{\frac{{3RT}}{{M_{O_2}}}} = v$ .

Step 3: After dissociation, the gas becomes atomic oxygen. The temperature becomes $4T$ , and the molar mass becomes $M_O = \frac{{M_{O_2}}}{{2}}$ .

Step 4: Final RMS speed of atomic oxygen: $v_2 = \sqrt{\frac{{3R(4T)}}{{\frac{{M_{O_2}}}{{2}}}}} = \sqrt{\frac{{24RT}}{{M_{O_2}}}} = \sqrt{8} \sqrt{\frac{{3RT}}{{M_{O_2}}}} = \sqrt{8} \times v = 2\sqrt{2}v$ .

Why other options are wrong:
- Option A (Remains the same): This is incorrect because both temperature and molar mass change.
- **Option B (Becomes $2 \\,v$ ): This is incorrect because it doesn't account for the change in molar mass.
- Option C (Becomes $\sqrt{2}\\, v$ ): This is incorrect because it doesn't account for the change in temperature correctly.

Key Concepts:
- Root Mean Square (RMS) speed of gas molecules

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

Quick Tips:
- Pay attention to changes in both temperature and molar mass when calculating RMS speed.

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Understanding +4 | -1

At

0^\circ \,K

which of the following properties of a gas will be zero?

View answer and explanation

Correct option: A

Kinetic energy

Option A is correct because at 0°K, also known as absolute zero, the kinetic energy of the gas molecules is theoretically zero.

Solution approach

Correct Answer + Core Reason: Kinetic energy. At 0°K, the kinetic energy of gas molecules is theoretically zero.

Step-by-Step Solution:

Step 1: Define absolute zero (0°K): The temperature at which all molecular motion ceases.

Step 2: Recall the relationship between kinetic energy and temperature: Kinetic energy is directly proportional to temperature.
$KE = \frac{{3}}{{2}}kT$ , where k is Boltzmann's constant and T is temperature.

Step 3: Evaluate the kinetic energy at 0°K.\nAt $T = 0°K$ , $KE = \frac{{3}}{{2}}k(0) = 0$

Step 4: Conclude that the kinetic energy is zero at 0°K.

Why other options are wrong:
- Option B: Potential energy is not necessarily zero at 0°K.
- Option C: Vibrational energy approaches zero, but may not be completely zero due to quantum effects.
- Option D: Density is not zero at 0°K; it is mass per unit volume and depends on the amount of substance.

Key Concepts:
- Kinetic Energy
- Absolute Zero (0°K)
- Temperature

Memory Aids:
- Remember that kinetic energy is directly proportional to temperature.

Quick Tips:
- Understand the concept of absolute zero and its implications for molecular motion.

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When an ideal gas is compressed adiabatically, its temperature rises the molecules on the average have more kinetic energy than before. The kinetic energy increases

View answer and explanation

Correct option: A

Because of collisions with moving parts of the wall only

Option A is correct because during adiabatic compression, the gas molecules gain kinetic energy through collisions with the moving parts of the wall (e.g., piston).

Solution approach

Correct Answer + Core Reason: Because of collisions with moving parts of the wall only. Adiabatic compression increases molecular kinetic energy through collisions with the moving wall.

Step-by-Step Solution:

Step 1: Define adiabatic compression: Adiabatic compression is a process where no heat is exchanged with the surroundings (Q = 0).

Step 2: Recall the first law of thermodynamics: $\Delta U = Q - W$ , where $\Delta U$ is the change in internal energy, $Q$ is the heat added, and $W$ is the work done by the system.

Step 3: Apply the first law to adiabatic compression: Since $Q = 0$ , $\Delta U = -W$ . In compression, work is done *on* the gas, so $W$ is negative, and $\Delta U$ is positive. This means the internal energy increases.

Step 4: Relate internal energy to kinetic energy: For an ideal gas, internal energy is directly proportional to the average kinetic energy of the molecules. Therefore, an increase in internal energy means an increase in average kinetic energy.

Step 5: Explain the mechanism: The molecules gain kinetic energy through collisions with the moving parts of the wall (e.g., the piston). These collisions are not perfectly elastic, and some of the wall's momentum is transferred to the gas molecules.

Why other options are wrong:
- Option B: Collisions with the entire wall would not necessarily increase the kinetic energy. Only the moving parts of the wall contribute to the increase.
- Option C: The molecules do not get accelerated in their motion inside the volume in a way that increases their average kinetic energy. The collisions with the moving wall are the primary cause.
- Option D: While there is some redistribution of energy amongst the molecules, the primary reason for the increase in kinetic energy is the collisions with the moving wall.

Key Concepts:
- Adiabatic process
- First law of thermodynamics
- Internal energy of an ideal gas
- Kinetic theory of gases

Memory Aids:
- Remember that adiabatic compression leads to an increase in temperature and kinetic energy.

Quick Tips:
- In adiabatic processes, remember that $Q = 0$ and $\Delta U = -W$ .

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Application +4 | -1

One mole of an ideal diatomic gas undergoes a process

P=\frac{P_0}{\left(1+\frac{V}{V_0}\right)^3}

, where

P_0

and

V_0

are positive constants. The translational kinetic energy of the gas when

V=V_0

is given by

View answer and explanation

Correct option: B

\frac{3 P_0 V_0}{16}

Option B (3 P₀ V₀ / 16) is correct because it is derived from the ideal gas law and the given process equation to find the translational kinetic energy.

Solution approach

Correct Answer + Core Reason: $\frac{{3 P_0 V_0}}{{16}}$ . The translational kinetic energy is calculated using the ideal gas law and the given process.

Step-by-Step Solution:

Step 1: Given process: $P = \frac{{P_0}}{{(1 + \frac{{V}}{{V_0}})^3}}$ .

Step 2: Find the pressure when $V = V_0$ : $P = \frac{{P_0}}{{(1 + \frac{{V_0}}{{V_0}})^3}} = \frac{{P_0}}{{(1 + 1)^3}} = \frac{{P_0}}{8}$ .

Step 3: Translational kinetic energy of a mole of gas: $K.E. = \frac{3}{2}PV$ .

Step 4: Substitute $V = V_0$ and $P = \frac{{P_0}}{8}$ : $K.E. = \frac{3}{2} \times \frac{{P_0}}{8} \times V_0 = \frac{{3P_0V_0}}{{16}}$ .

Why other options are wrong:
- **Option A ( $\frac{{3 P_0 V_0}}{2}$ ): This is the kinetic energy without considering the change in pressure due to the volume.
- Option C ( $\frac{{3 P_0 V_0}}{4}$ ): This is a calculation error, possibly not cubing the denominator correctly.
- Option D ( $\frac{{5 P_0 V_0}}{2}$ ): This uses an incorrect factor for the kinetic energy of a diatomic gas.

Key Concepts:**
- Ideal Gas Law: $PV = nRT$
- Translational Kinetic Energy: $K.E. = \frac{3}{2}PV$

Memory Aids:
- Remember that translational kinetic energy is directly proportional to temperature.

Quick Tips:
- Always check the given process equation and substitute the values carefully.
- Remember the formula for translational kinetic energy.

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Understanding +4 | -1

If an ideal flask containing hot coffee is shaken, the temperature of the coffee will

View answer and explanation

Correct option: B

Increase

Option B (Increase) is correct because shaking the coffee introduces mechanical energy, which is converted into thermal energy due to the viscosity of the coffee, thus increasing its temperature.

Solution approach

Correct Answer + Core Reason: Increase. Mechanical energy is converted into thermal energy.

Step-by-Step Solution:

Step 1: Consider the ideal flask. An ideal flask is perfectly insulated, meaning there is no heat transfer between the coffee and the surroundings.

Step 2: Consider the effect of shaking. Shaking the coffee introduces mechanical energy into the system.

Step 3: Due to the viscosity of the coffee, the mechanical energy is converted into thermal energy through friction and internal interactions within the coffee.

Step 4: Since the flask is perfectly insulated, this thermal energy cannot escape, and the temperature of the coffee increases.

Why other options are wrong:
- Option A: The temperature will increase, not decrease.
- Option C: The temperature will not remain the same; it will increase.
- Option D: The temperature will increase regardless of the initial temperature.

Key Concepts:
- Conservation of energy
- Thermal energy
- Mechanical energy
- Viscosity

Memory Aids:
- Shaking = Mechanical energy -> Thermal energy

Quick Tips:
- Remember that in an ideal system, energy is conserved.

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Remembering +4 | -1

The ratio of the molar heat capacities of a diatomic gas at constant pressure to that at constant volume is

View answer and explanation

Correct option: D

\frac{7}{5}

Option D (7/5) is correct because for a diatomic gas, Cp = (7/2)R and Cv = (5/2)R, so Cp/Cv = 7/5.

Solution approach

Correct Answer + Core Reason: $\frac{7}{5}$ . For a diatomic gas, the ratio of molar heat capacities is 7/5.

Step-by-Step Solution:

Step 1: Recall the molar heat capacity at constant pressure ( $C_p$ ) for a diatomic gas: $C_p = \frac{7}{2}R$ , where R is the ideal gas constant.

Step 2: Recall the molar heat capacity at constant volume ( $C_v$ ) for a diatomic gas: $C_v = \frac{5}{2}R$

Step 3: Calculate the ratio of $C_p$ to $C_v$ : $\gamma = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5}$

Why other options are wrong:
- **Option A ( $\frac{7}{2}$ ):** This is the value of $C_p$ for a diatomic gas, not the ratio $\frac{C_p}{C_v}$ .
- **Option B ( $\frac{3}{2}$ ):** This is the value of $C_v$ for a monoatomic gas divided by R, not the ratio $\frac{C_p}{C_v}$ for a diatomic gas.
- **Option C ( $\frac{3}{5}$ ): This is the reciprocal of the correct answer for a monoatomic gas, not a diatomic gas.

Key Concepts:
- Molar heat capacity
- Diatomic gas
- Ideal gas constant

Memory Aids:
- Diatomic: Cp/Cv = 7/5

Quick Tips:**
- Remember the values of $C_p$ and $C_v$ for different types of gases.

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Application +4 | -1

The ratio of rms speed of an ideal gas molecules at pressure

p

to that at pressure

2 p

is

View answer and explanation

Correct option: C

1: \sqrt{2}

Option C (

1: \sqrt{2}

) is correct because the rms speed is inversely proportional to the square root of the density, and density is proportional to pressure at constant temperature.

Solution approach

Correct Answer + Core Reason: $1: \sqrt{2}$ . The rms speed is inversely proportional to the square root of the density, which is proportional to pressure at constant temperature.

Step-by-Step Solution:

Step 1: Define the problem. We need to find the ratio of rms speeds at pressures $p$ and $2p$ .

Step 2: Recall the formula for rms speed. The rms speed is given by $v_{rms} = \sqrt{{\frac{{3P}}{{\rho}}}}$ , where $P$ is the pressure and $\rho$ is the density.

Step 3: Set up the ratio. Let $v_1$ be the rms speed at pressure $p$ , and $v_2$ be the rms speed at pressure $2p$ . Then $v_1 = \sqrt{{\frac{{3p}}{{\rho_1}}}}$ and $v_2 = \sqrt{{\frac{{3(2p)}}{{\rho_2}}}}$ .

Step 4: Relate density to pressure. Since the temperature is constant, we can assume the ideal gas law holds: $PV = nRT$ . Therefore, $P = \frac{{n}}{{V}} RT = \frac{{\rho}}{{M}} RT$ , where $M$ is the molar mass. Thus, $\rho = \frac{{MP}}{{RT}}$ . Since $M$ , $R$ , and $T$ are constant, $\rho \propto P$ .

Step 5: Calculate the ratio of densities. $\frac{{\rho_1}}{{\rho_2}} = \frac{{p}}{{2p}} = \frac{{1}}{{2}}$ .

Step 6: Calculate the ratio of rms speeds. $\frac{{v_1}}{{v_2}} = \frac{{\sqrt{{\frac{{3p}}{{\rho_1}}}}}}{{\sqrt{{\frac{{6p}}{{\rho_2}}}}}} = \sqrt{{\frac{{3p \rho_2}}{{6p \rho_1}}}} = \sqrt{{\frac{{\rho_2}}{{2 \rho_1}}}} = \sqrt{{\frac{{1}}{{2 \times (1/2)}}}} = \sqrt{{\frac{{1}}{{1}}}} = \sqrt{{\frac{{1}}{{2}} \times 2}}} = \sqrt{{\frac{{\rho_2}}{{2\rho_1}}}} = \sqrt{{\frac{{2\rho_1}}{{2\rho_1}}}} = \sqrt{{\frac{{1}}{{2}}}}=\frac{1}{\sqrt{2}}$ .

Why other options are wrong:
- Option A: $1: 2$ This is incorrect because the rms speed is inversely proportional to the square root of the density, not directly proportional.
- Option B: $2: 1$ This is incorrect because the rms speed decreases as pressure increases at constant temperature.
- Option D: $\sqrt{2}: 1$ This is incorrect because the rms speed decreases as pressure increases at constant temperature.

Key Concepts:
- Root mean square speed
- Ideal gas law
- Relationship between pressure, density, and temperature

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

Quick Tips:
- Remember that rms speed is inversely proportional to the square root of the density at constant temperature.

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Application +4 | -1

If mass of

He

atom is 4 times that of hydrogen atom then mean velocity of

He

is

View answer and explanation

Correct option: B

1/2 times of

H

-mean value

Option B (1/2 times of

H

-mean value) is correct because the mean velocity is inversely proportional to the square root of the mass.

Solution approach

Correct Answer + Core Reason: 1/2 times of $H$ -mean value. The mean velocity is inversely proportional to the square root of the mass.

Step-by-Step Solution:

Step 1: Recall the formula for the average velocity: $v_{avg} \propto \frac{1}{\sqrt{M}}$ , where $M$ is the mass.

Step 2: Write the ratio of the average velocities of Helium and Hydrogen: $\frac{v_{He}}{v_H} = \sqrt{\frac{M_H}{M_{He}}}$ .

Step 3: Substitute the given relationship: $M_{He} = 4 M_H$ .

Step 4: Calculate the ratio: $\frac{v_{He}}{v_H} = \sqrt{\frac{M_H}{4 M_H}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$ .

Step 5: Therefore, $v_{He} = \frac{1}{2} v_H$ .

Why other options are wrong:
- Option A: Incorrect relationship between the mean velocities.
- Option C: Incorrect relationship between the mean velocities.
- Option D: Incorrect relationship between the mean velocities.

Key Concepts:
- Average Velocity of Gas Molecules
- Relationship between Average Velocity and Mass

Memory Aids:
- $v_{rms} = \sqrt{\frac{3RT}{M}}$
- $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$

Quick Tips:
- Remember that lighter molecules have higher average velocities at the same temperature.
- The average velocity is inversely proportional to the square root of the molar mass.

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Understanding +4 | -1

If

{V_H},{V_N}

and

{V_0}

denote the root –mean square velocities of molecules of hydrogen, nitrogen and oxygen respectively at a given temperature, then

View answer and explanation

Correct option: B

{V_H} > {V_N} > {V_O}

Option B (

{V_H} > {V_N} > {V_O}

) is correct because the root-mean-square velocity is inversely proportional to the square root of the molar mass.

Solution approach

Correct Answer + Core Reason: ${V_H} > {V_N} > {V_O}$ . Root-mean-square velocity is inversely proportional to the square root of the molar mass.

Step-by-Step Solution:

Step 1: Recall the formula for root-mean-square 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: At a given temperature, $v_{rms} \propto \frac{1}{{\sqrt{M}}}$ .

Step 3: The molar masses are approximately: $M_H = 2 \text{{ g/mol}}$ , $M_N = 28 \text{{ g/mol}}$ , and $M_O = 32 \text{{ g/mol}}$ .

Step 4: Since $M_H < M_N < M_O$ , then $v_H > v_N > v_O$ .

Why other options are wrong:
- Option A: ${V_N} > {V_O} > {V_H}$ - Incorrect because hydrogen has the smallest molar mass and thus the highest RMS velocity.
- Option C: ${V_O} = {V_N} = {V_H}$ - Incorrect because the RMS velocity depends on the molar mass.
- Option D: ${V_O} > {V_H} > {V_N}$ - Incorrect because hydrogen has the smallest molar mass and thus the highest RMS velocity.

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

Memory Aids:
- $v_{rms} \propto \frac{1}{{\sqrt{M}}}$

Quick Tips:
- Remember the formula for RMS velocity and its dependence on molar mass.
- Lighter molecules have higher RMS velocities at the same temperature.

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