KINETIC THEORY OF GASES MCQs for JEE PHYSICS — Medium, Page 8

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Remembering +4 | -1

The rate of diffusion is

View answer and explanation

Correct option: D

Faster in gases than in liquids and solids

Option D (Faster in gases than in liquids and solids) is correct because gas molecules have greater kinetic energy and weaker intermolecular forces, allowing for faster diffusion.

Solution approach

Correct Answer + Core Reason: Faster in gases than in liquids and solids. Gas molecules have higher kinetic energy and weaker intermolecular forces.

Step-by-Step Solution:

Step 1: Define diffusion.

Step 2: Diffusion is the movement of particles from an area of high concentration to an area of low concentration.

Step 3: The rate of diffusion depends on the kinetic energy of the particles and the strength of intermolecular forces.

Step 4: Gases have the highest kinetic energy and the weakest intermolecular forces, allowing for the fastest diffusion.

Why other options are wrong:
- Option A: Faster in solids than in liquids and gases: Solids have very low kinetic energy and strong intermolecular forces, resulting in the slowest diffusion.
- Option B: Faster in liquids than in solids and gases: Liquids have intermediate kinetic energy and intermolecular forces, resulting in a diffusion rate between solids and gases.
- Option C: Equal to solids, liquids and gases: The rate of diffusion varies significantly between solids, liquids, and gases.

Key Concepts:
- Diffusion
- Kinetic energy
- Intermolecular forces

Memory Aids:
- Gases diffuse fastest due to high kinetic energy and weak intermolecular forces.

Quick Tips:
- Remember that diffusion rate is inversely proportional to the strength of intermolecular forces.

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Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules?

View answer and explanation

Correct option: C

No change

Option C is correct because the kinetic energy of an ideal gas depends only on temperature, which is kept constant in this scenario.

Solution approach

Correct Answer + Core Reason: No change. Kinetic energy of an ideal gas depends only on temperature.

Step-by-Step Solution:

Step 1: Define Kinetic Energy of an Ideal Gas: The average kinetic energy of the molecules in an ideal gas is directly proportional to the absolute temperature of the gas. The formula is $KE = \frac{{3}}{{2}}kT$ , where $k$ is the Boltzmann constant and $T$ is the absolute temperature.

Step 2: Analyze Constant Temperature: The problem states that the temperature is kept constant. Therefore, $T$ is constant.

Step 3: Conclusion: Since the kinetic energy depends only on temperature, and the temperature is constant, the kinetic energy of the molecules remains unchanged, regardless of the pressure.

Why other options are wrong:
- Option A: The kinetic energy does not increase because the temperature is constant.
- Option B: The kinetic energy does not decrease because the temperature is constant.
- Option D: The kinetic energy can be determined because it depends only on temperature, which is given as constant.

Key Concepts:
- Kinetic Energy of an Ideal Gas
- Temperature Dependence

Memory Aids:
- KE ∝ T (for ideal gas)

Quick Tips:
- Pressure changes at constant temperature do not affect the kinetic energy of ideal gas molecules.

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Given below are two statements :
Statement I : The average momentum of a molecule in a sample of an ideal gas depends on temperature.
Statement II : The rms speed of oxygen molecules in a gas is v. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become

2v.

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

View answer and explanation

Correct option: D

Statement I is false but Statement II is true

Option D (Statement I is false but Statement II is true) is correct because the average momentum of an ideal gas molecule is zero, while the rms speed calculation is accurate.

Solution approach

Correct Answer + Core Reason: Statement I is false but Statement II is true. The average momentum of an ideal gas molecule is zero, and the rms speed calculation is correct.

Step-by-Step Solution:

Step 1: Analyze Statement I: The average momentum of a molecule in a sample of an ideal gas depends on temperature. The average momentum is zero because molecules move randomly in all directions, so the vector sum of their momenta is zero. Therefore, Statement I is false.

Step 2: Analyze Statement II: The rms speed of oxygen molecules in a gas is v. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become $2v$ .

Step 3: Use 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 4: If temperature is doubled (T' = 2T) and oxygen molecules dissociate into oxygen atoms (M' = M/2), then the new rms speed is: $v'_{rms} = \sqrt{\frac{{3R(2T)}}{{(M/2)}}} = \sqrt{\frac{{12RT}}{{M}}} = 2\sqrt{\frac{{3RT}}{{M}}} = 2v$ .

Step 5: Therefore, Statement II is true.

Why other options are wrong:
- Option A: Both Statement I and Statement II are true: Statement I is false.
- Option B: Both Statement I and Statement II are false: Statement II is true.
- Option C: Statement I is true but Statement II is false: Statement I is false and Statement II is true.

Key Concepts:
- Average Momentum of Ideal Gas
- Root Mean Square (rms) Speed
- Kinetic Theory of Gases

Memory Aids:
- Remember that average momentum is a vector sum and cancels out for random motion.
- The rms speed formula: $v_{rms} = \sqrt{\frac{{3RT}}{{M}}}$

Quick Tips:
- Pay attention to whether you are dealing with average speed or average momentum.

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The

r.m.s.

speed of gas molecules is given by

View answer and explanation

Correct option: B

1.73\sqrt {\frac{{RT}}{M}}

Option B (

1.73\sqrt {\frac{{RT}}{M}}

) is correct because the root-mean-square speed of gas molecules is given by

\sqrt{\frac{{3RT}}{M}}

, and

\sqrt{3} \approx 1.73

.

Solution approach

Correct Answer + Core Reason: $1.73\sqrt {\frac{{RT}}{M}}$ . The r.m.s. speed is calculated using the formula $v_{rms} = \sqrt{\frac{{3RT}}{M}}$ .

Step-by-Step Solution:

Step 1: Recall the formula for the root-mean-square (r.m.s.) speed of gas molecules.

Step 2: The formula is given by: ${v_{rms}} = \sqrt {\frac{{3RT}}{M}}$

Step 3: Simplify the expression by taking the square root of 3.

Step 4: Since $\sqrt{3} \approx 1.73$ , we can write: ${v_{rms}} = 1.73\sqrt {\frac{{RT}}{M}}$

Why other options are wrong:
- Option A: $2.5\sqrt {\frac{{RT}}{M}}$ : This is incorrect because the correct constant is approximately 1.73, not 2.5.
- Option C: $2.5\sqrt {\frac{M}{{RT}}}$ : This is incorrect because the fraction inside the square root is inverted, and the constant is incorrect.
- Option D: $1.73\sqrt {\frac{M}{{RT}}}$ : This is incorrect because the fraction inside the square root is inverted.

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

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

Quick Tips:
- Pay attention to the units and ensure they are consistent.

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At O K which of the following properties of a gas will be zero

View answer and explanation

Correct option: A

Kinetic energy

Option A (Kinetic energy) is correct because at 0 K, all molecular motion ceases, resulting in zero kinetic energy.

Solution approach

Correct Answer + Core Reason: Kinetic energy. At absolute zero (0 K), all molecular motion stops, resulting in zero kinetic energy.

Step-by-Step Solution:

Step 1: Define absolute zero. Absolute zero (0 K or -273.15 °C) is the lowest possible temperature, where all thermal motion ceases.

Step 2: Relate temperature to kinetic energy. The kinetic energy of gas molecules is directly proportional to the absolute temperature: $KE = \frac{{3}}{{2}}kT$ , where k is the Boltzmann constant and T is the absolute temperature.

Step 3: Apply the condition at 0 K. At T = 0 K, the kinetic energy is: $KE = \frac{{3}}{{2}}k(0) = 0$ .

Why other options are wrong:
- Option B (Potential energy): Potential energy can exist even at 0 K, depending on the intermolecular forces and the state of the substance.
- Option C (Vibrational energy): Vibrational energy is a form of kinetic energy, and it approaches zero at 0 K, but the question asks which property will be zero. While vibrational energy approaches zero, kinetic energy is fundamentally zero.
- Option D (Density): Density is mass per unit volume and is not necessarily zero at 0 K. It depends on the substance and its state.

Key Concepts:
- Kinetic energy
- Absolute zero
- Temperature and molecular motion

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

Quick Tips:
- At 0 K, all translational, rotational, and vibrational motion ideally ceases.

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The equation of state for

3 \mathrm{~g}

of oxygen at a pressure

P

and temperature

T

, when occupying a volume

V

, will be:

View answer and explanation

Correct option: B

P V=\frac{3}{32} {RT}

Option B (

P V=\frac{3}{32} {RT}

) is correct because it uses the ideal gas law with the correct number of moles calculated from the given mass and molecular mass of oxygen.

Solution approach

Correct Answer + Core Reason: $P V=\frac{3}{32} {RT}$ . The equation of state is found using the ideal gas law, with the number of moles calculated correctly.

Step-by-Step Solution:

Step 1: State the ideal gas law: $PV = nRT$

Step 2: Calculate the number of moles (n) of oxygen: $n = \frac{m}{\text{molecular mass}}$ , where $m = 3 \text{ g}$ and the molecular mass of oxygen ( $O_2$ ) is $32 \text{ g/mol}$ .

Step 3: Substitute the values: $n = \frac{3}{32}$ moles

Step 4: Substitute the number of moles into the ideal gas law: $PV = \frac{3}{32}RT$

Why other options are wrong:
- Option A: $P V=\frac{5}{32} {RT}$ : This is incorrect because it uses an incorrect number of moles.
- Option C: $P V=\frac{32}{3} {RT}$ : This is incorrect because it uses the inverse of the correct number of moles.
- Option D: $P V=\frac{31}{32} {RT}$ : This is incorrect because it uses an incorrect number of moles.

Key Concepts:
- Ideal gas law: $PV = nRT$
- Calculation of number of moles: $n = \frac{m}{M}$

Memory Aids:
- Remember the ideal gas law and how to calculate the number of moles.

Quick Tips:
- Always use consistent units when applying the ideal gas law.

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An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true?
A. The mean free path of the molecules decreases.
B. The mean collision time between the molecules decreases.
C. The mean free path remains unchanged.
D. The mean collision time remains unchanged.

View answer and explanation

Correct option: A

B and C

Option A (B and C) is correct because in a closed container, the mean free path remains constant, and as temperature increases, the mean collision time decreases.

Solution approach

Correct Answer + Core Reason: B and C. In a closed container, the mean free path remains constant, and as temperature increases, the mean collision time decreases.

Step-by-Step Solution:

Step 1: Recall the formula for mean free path: $\lambda = \frac{V}{{\sqrt{2} \pi d^2 N}} = \frac{{kT}}{{\sqrt{2} \pi d^2 p}}$ , where V is volume, N is the number of molecules, d is the diameter of the molecule, k is Boltzmann's constant, T is temperature, and p is pressure.

Step 2: Since the container is closed, the volume V and the number of molecules N are constant. Also, from the ideal gas equation, $\frac{P}{T} = \frac{{nR}}{V}$ . For constant n and V, $\frac{p}{T}$ = constant.

Step 3: Therefore, the mean free path $\lambda$ does not depend on p or T and remains unchanged. So, statement C is correct, and statement A is incorrect.

Step 4: Recall the formula for mean collision time: $\tau = \frac{{\text{{mean free path}}}}{{V_{avg}}} = \frac{\lambda}{{V_{avg}}}$ .

Step 5: The average velocity $V_{avg} \propto \sqrt{T}$ . Therefore, $\tau \propto \frac{1}{{\sqrt{T}}}$ .

Step 6: As the temperature increases, the mean collision time $\tau$ decreases. So, statement B is correct, and statement D is incorrect.

Why other options are wrong:
- Option B: Statement A is incorrect.
- Option C: Statement D is incorrect.
- Option D: Statement A is incorrect and statement D is incorrect.

Key Concepts:
- Mean free path
- Mean collision time
- Ideal gas equation

Memory Aids:
- Remember the relationship between mean free path, temperature, and pressure.
- Remember the relationship between mean collision time and temperature.

Quick Tips:
- Always consider the conditions of the container (closed or open) when analyzing mean free path and collision time.

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The ideal gas equation connecting Pressure (P) Volume (V) and absolute temperature (T) is (Where

{K_B}

is the Blotzmann Constant and N is the number of molecules)

View answer and explanation

Correct option: C

\;PV = {K_B}NT

Option C (

\;PV = {K_B}NT

) is correct because this is the ideal gas equation using Boltzmann's constant and the number of molecules.

Solution approach

Correct Answer + Core Reason: $\;PV = {K_B}NT$ . This is the ideal gas equation expressed in terms of Boltzmann constant and number of molecules.

Step-by-Step Solution:

Step 1: Recall the ideal gas equation.

Step 2: The ideal gas equation relates pressure (P), volume (V), number of molecules (N), Boltzmann constant ( $K_B$ ), and absolute temperature (T).

Step 3: The equation is given by: $PV = N{K_B}T$

Why other options are wrong:
- Option A: $PV = \frac{{{K_B}N}}{T}$ : This is incorrect because the temperature should be multiplied, not divided.
- Option B: $PV = \frac{{{K_B}N}}{{VT}}$ : This is incorrect because both volume and temperature should be multiplied.
- Option D: $P = \frac{{{K_B}NV}}{T}$ : This is incorrect because the volume should be multiplied with the pressure.

Key Concepts:
- Ideal gas equation
- Boltzmann constant
- Number of molecules

Memory Aids:
- Remember the equation: $PV = N{K_B}T$

Quick Tips:
- Know the different forms of the ideal gas equation.

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In the expression for

V_{r m s}

of a mixture of 2 gases at same temperature,

V_{r m s}=\sqrt{\frac{3 R T}{M}}

, the molecular weight

\mathrm{M}

is

View answer and explanation

Correct option: B

Harmonic mean of molecular weights

Option B is correct because the molecular weight M in the Vrms expression for a gas mixture is the harmonic mean of the individual molecular weights.

Solution approach

Correct Answer + Core Reason: Harmonic mean of molecular weights. The RMS velocity uses the harmonic mean for gas mixtures.

Step-by-Step Solution:

Step 1: Recall the formula for $V_{rms}$ of a gas: $V_{rms} = \sqrt{\frac{3RT}{M}}$ .

Step 2: For a mixture of gases, the effective molecular weight $M$ is not a simple arithmetic mean.

Step 3: The correct way to calculate the effective molecular weight for $V_{rms}$ is to use the harmonic mean.

Step 4: The harmonic mean is given by: $M = \frac{\sum N_i}{\sum \frac{N_i}{M_i}}$ , where $N_i$ is the number of molecules of gas $i$ and $M_i$ is its molecular weight.

Why other options are wrong:
- Option A: The arithmetic mean is used for calculating the average kinetic energy, not directly in the $V_{rms}$ formula.
- Option C: The geometric mean is not used in this context.
- Option D: The larger molecular weight alone does not represent the effective molecular weight of the mixture.

Key Concepts:
- Root-mean-square velocity ( $V_{rms}$ )
- Harmonic mean
- Kinetic theory of gases

Memory Aids:
- Remember that $V_{rms}$ involves the harmonic mean of molecular weights.

Quick Tips:
- Be careful when dealing with gas mixtures; the effective molecular weight is not always a simple average.

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With increase in pressure at constant temperature, rms speed of gases

View answer and explanation

Correct option: C

Remains same

Option C (Remains same) is correct because the RMS speed of gas molecules depends only on temperature, and the temperature is held constant.

Solution approach

Correct Answer + Core Reason: Remains same. The root-mean-square (rms) speed of gas molecules depends only on temperature.

Step-by-Step Solution:

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

Step 2: Analyze the given conditions. The problem states that the temperature is constant.

Step 3: Determine the effect of pressure on rms speed. From the formula, we can see that the rms speed depends only on temperature and molar mass. Since the temperature is constant and the gas is the same, the rms speed remains constant, regardless of the pressure.

Why other options are wrong:
- Option A: Increases: This is incorrect because the rms speed depends only on temperature, not pressure.
- Option B: Decreases: This is incorrect because the rms speed depends only on temperature, not pressure.
- Option D: First decreases then increases: This is incorrect because the rms speed remains constant as long as the temperature is constant.

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

Memory Aids:
- $v_{rms}$ depends only on $T$ .

Quick Tips:
- Remember the formula for rms speed and its dependence on temperature.

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Pressure of a gas at constant volume is proportional to

View answer and explanation

Correct option: D

Total internal energy of the gas

The correct answer is D because at constant volume, the pressure of a gas is directly proportional to its total internal energy.

Solution approach

Correct Answer + Core Reason: Total internal energy of the gas. Pressure is directly proportional to the internal energy at constant volume.

Step-by-Step Solution:
Step 1: Recall the relationship between pressure and internal energy: $P = \frac{{2}}{{3}} \frac{{E}}{{V}}$ , where P is pressure, E is internal energy, and V is volume.
Step 2: Analyze the relationship at constant volume: If volume (V) is constant, then $P \propto E$ .
Step 3: Conclude that pressure is proportional to the total internal energy of the gas.

Why other options are wrong:
- Option A: Temperature of the gas: While pressure is related to temperature, the direct proportionality is with internal energy.
- Option B: Average kinetic energy of the molecules: Average kinetic energy is related to temperature and internal energy, but internal energy is the direct proportional factor.
- Option C: Average potential energy of the molecules: Potential energy is not directly related to pressure at constant volume in an ideal gas.

Key Concepts:
- Internal Energy of a Gas
- Pressure of a Gas
- Kinetic Theory of Gases
- Relationship: $P = \frac{{2}}{{3}} \frac{{E}}{{V}}$

Memory Aids:
- P = (2/3)(E/V) helps remember the direct proportionality between P and E at constant V.

Quick Tips:
- Remember the relationship between pressure, volume, and internal energy for ideal gases.

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Application +4 | -1

Calculate the ratio of rms speeds of oxygen gas molecules to that of hydrogen gas molecules kept at the same temperature.

View answer and explanation

Correct option: A

1:4

Option A (

1:4

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

Solution approach

Correct Answer + Core Reason: $1:4$ . The ratio is determined by the inverse square root of the ratio of molar masses.

Step-by-Step Solution:

Step 1: Recall the formula for the root mean square (rms) speed of a gas molecule: $v_{rms} = \sqrt{\frac{{3RT}}{M}}$ , where $R$ is the ideal gas constant, $T$ is the temperature, and $M$ is the molar mass.

Step 2: Since the temperature is the same for both gases, the ratio of their rms speeds is inversely proportional to the square root of their molar masses: $\frac{{(v_{rms})_{O_2}}}{{(v_{rms})_{H_2}}} = \sqrt{\frac{{M_{H_2}}}{{M_{O_2}}}}$ .

Step 3: The molar mass of oxygen ( $O_2$ ) is 32 g/mol, and the molar mass of hydrogen ( $H_2$ ) is 2 g/mol. Substitute these values into the ratio: $\frac{{(v_{rms})_{O_2}}}{{(v_{rms})_{H_2}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$ .

Step 4: Therefore, the ratio of rms speeds of oxygen to hydrogen is $1:4$ .

Why other options are wrong:
- Option B: This option is wrong because the ratio is not $1:8$ . It seems like the square root was not correctly calculated or the molar mass ratio was inverted.
- Option C: This option is wrong because the ratio is not $1:2$ . This might be obtained if the square root was taken of 1/4 incorrectly as 1/ $\sqrt{2}$ .
- Option D: This option is wrong because the ratio is not $1:6$ . There is no clear calculation error that would lead to this value.

Key Concepts:
- Root mean square speed
- Kinetic theory of gases
- Molar mass

Memory Aids:
- $v_{rms} \propto \frac{1}{{\sqrt{M}}}$ : rms speed is inversely proportional to the square root of molar mass.

Quick Tips:
- Remember the formula for rms speed and its dependence on temperature and molar mass.
- Always use consistent units for molar mass (e.g., g/mol or kg/mol).

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The average kinetic energy per mole of hydrogen at given temperature is

View answer and explanation

Correct option: A

equal of that of oxygen

Option A (equal of that of oxygen) is correct because the average kinetic energy per mole of an ideal gas depends only on the temperature and is independent of the type of gas.

Solution approach

Correct Answer + Core Reason: equal of that of oxygen. For an ideal gas, the mean kinetic energy is proportional to the absolute temperature. It does not depend on the kind of gas.

Step-by-Step Solution:

Step 1: The average kinetic energy per mole of an ideal gas is given by $KE = \frac{{3}}{{2}}RT$ , where $R$ is the ideal gas constant and $T$ is the absolute temperature.

Step 2: This formula shows that the kinetic energy depends only on the temperature and is independent of the type of gas.

Step 3: Therefore, at a given temperature, the average kinetic energy per mole of hydrogen is equal to that of oxygen.

Why other options are wrong:
- Option B: Incorrect because the kinetic energy depends only on temperature.
- Option C: Incorrect because the kinetic energy depends only on temperature.
- Option D: Incorrect because the kinetic energy depends only on temperature.

Key Concepts:
- Kinetic theory of gases
- Average kinetic energy

Memory Aids:
- KE = (3/2)RT

Quick Tips:
- Remember that the average kinetic energy of an ideal gas depends only on temperature.

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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, thus having 3 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. Translational degrees of freedom refer to the number of independent directions in which a molecule can move freely.

Step 2: Consider the motion of a diatomic gas molecule. A diatomic gas molecule can move along the x-axis, the y-axis, and the z-axis.

Step 3: Count the number of translational degrees of freedom. Since the molecule can move independently in three directions, it has 3 translational degrees of freedom.

Step 4: Final Answer: A diatomic gas has 3 translational degrees of freedom.

Why other options are wrong:
- Option A (2): This is an incorrect number of translational degrees of freedom.
- Option C (5): This is the total number of degrees of freedom for a diatomic gas (3 translational + 2 rotational), but the question asks only for translational.
- Option D (6): This is an incorrect number of translational degrees of freedom.

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

Memory Aids:
- Remember that all gases have 3 translational degrees of freedom.

Quick Tips:
- Translational degrees of freedom are the same for all types of gases.

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Analyzing +4 | -1

Assertion: A gas is kept in an insulated cylinder with a movable piston, in compressed state. As the piston is released, temperature of the gas decreases.
Reason: According to the kinetic theory of gas, a molecule colliding with the piston must rebound with less speed than it had before the collision. Hence average speed of the molecules is reduced.

View answer and explanation

Correct option: A

If both assertion and reason are true and reason is the correct explanation of the assertion

The assertion and reason are both true, and the reason is the correct explanation of the assertion because the gas does work on the piston, reducing the average speed of the molecules and thus the temperature.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true and reason is the correct explanation of the assertion because the gas does work on the piston, reducing the average speed of the molecules.

Step-by-Step Solution:

Step 1: Consider an insulated cylinder with a movable piston containing a compressed gas.

Step 2: When the piston is released, the gas expands, doing work on the piston. Since the cylinder is insulated, there is no heat exchange with the surroundings (adiabatic process).

Step 3: According to 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. In this case, $Q = 0$ , so $\Delta U = -W$ .

Step 4: Since the gas is doing work (W > 0), the change in internal energy is negative ( $\Delta U < 0$ ).

Step 5: The internal energy of an ideal gas is directly proportional to its temperature. Therefore, if the internal energy decreases, the temperature also decreases.

Step 6: According to the kinetic theory of gases, the temperature is related to the average kinetic energy of the molecules, which is related to the average speed of the molecules.

Step 7: When a molecule collides with the receding piston, it loses some of its kinetic energy, and its speed decreases. This leads to a decrease in the average speed of the molecules, and hence a decrease in temperature.

Why other options are wrong:
- Option B: This option is incorrect because the reason *is* the correct explanation of the assertion.
- Option C: This option is incorrect because both the assertion and the reason are true.
- Option D: This option is incorrect because both the assertion and the reason are true.

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

Memory Aids:
- Remember the first law of thermodynamics: $\Delta U = Q - W$ .

Quick Tips:
- When dealing with adiabatic processes, remember that $Q = 0$ .

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The temperature of a gas is held constant, while its volume is decreased. The pressure exerted by its volume is decreased. The pressure exerted by the gas on the walls of the container increases, because its molecules

View answer and explanation

Correct option: A

Strike the walls more frequently

Option A (Strike the walls more frequently) is correct because decreasing the volume at constant temperature increases the frequency of molecular collisions with the container walls.

Solution approach

Correct Answer + Core Reason: Strike the walls more frequently. Decreasing the volume at constant temperature increases the frequency of molecular collisions with the container walls.

Step-by-Step Solution:

Step 1: Recall the ideal gas law: $PV = nRT$ , where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is the temperature.

Step 2: Since the temperature is held constant, and the number of moles is constant, $PV = constant$ .

Step 3: If the volume is decreased, the pressure must increase to maintain the constant value of $PV$ .

Step 4: The pressure exerted by a gas on the walls of the container is due to the collisions of the gas molecules with the walls.

Step 5: When the volume is decreased, the molecules have less space to move around, so they collide with the walls more frequently. This increased collision frequency results in a higher pressure.

Why other options are wrong:
- Option B (Strike the walls with higher velocities): Since the temperature is constant, the average kinetic energy of the molecules remains constant, so their average velocity does not change.
- Option C (Are in contact with the walls for a shorter time): The time of contact during a collision is not significantly affected by the volume.
- Option D (Strike the walls with larger force): Since the temperature is constant, the average kinetic energy of the molecules remains constant, so the average force exerted during each collision does not change.

Key Concepts:
- Ideal gas law
- Pressure
- Volume
- Temperature
- Kinetic theory of gases

Memory Aids:
- $PV = nRT$
- Pressure is due to molecular collisions.

Quick Tips:
- Remember that at constant temperature, decreasing the volume increases the pressure.
- Consider the relationship between pressure, volume, and molecular collisions.

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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 closer to 1.28 considering vibrational degrees of freedom.

Solution approach

Correct Answer + Core Reason: 1.28. The ratio of specific heats depends on the degrees of freedom.

Step-by-Step Solution:

Step 1: Identify that $NH_3$ is a polyatomic gas.

Step 2: Recall the formula for $\gamma$ for a polyatomic gas: $\gamma = \frac{{f + 2}}{{f}}$ , where f is the number of degrees of freedom.

Step 3: For $NH_3$ , the number of translational degrees of freedom is 3, and the number of rotational degrees of freedom is 3. It also has vibrational degrees of freedom. The formula considering vibrational degrees of freedom is ${\gamma _{poly}} = \frac{{\left( {4 + {f_{vib}}} \right)}}{{\left( {3 + {f_{vib}}} \right)}}$

Step 4: Since $f_{vib} > 0$ , ${\gamma _{poly}} < \frac{4}{3} = 1.33$ .

Step 5: Among the given options, the value closest to and less than 1.33 is 1.28.

Why other options are wrong:
- Option A (1.33): This is the value of $\gamma$ for a gas with only translational and rotational degrees of freedom, neglecting vibrational modes. It's an overestimate.
- 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 monoatomic gas, which is not applicable here.

Key Concepts:
- Degrees of freedom
- Ratio of specific heats
- Polyatomic gases

Memory Aids:
- Remember that as atomicity increases, $\gamma$ decreases.

Quick Tips:
- For polyatomic gases, consider vibrational degrees of freedom.

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Application +4 | -1

For a gas at

300 \mathrm{~K}

temperature energy is

300 \mathrm{~J}

and speed is

300 \mathrm{~m} /

sec. If temperature is made 2 times of initial then find out the final energy and speed.

View answer and explanation

Correct option: A

300 \sqrt{2} \mathrm{~m} / \mathrm{sec}

Option A (

300 \sqrt{2} \mathrm{~m} / \mathrm{sec}

) is correct because the final speed is

300 \sqrt{2} \mathrm{~m} / \mathrm{sec}

, calculated by using the relation between speed and temperature.

Solution approach

Correct Answer + Core Reason: $300 \sqrt{2} \mathrm{~m} / \mathrm{sec}$ . The final speed is calculated using the temperature relationship.

Step-by-Step Solution:

Step 1: The energy U is proportional to the temperature T: $U \propto T$ .

Step 2: If the temperature is doubled, the final energy is $U_f = 2U_i = 2(300 \text{{ J}}) = 600 \text{{ J}}$ .

Step 3: The speed v is proportional to the square root of the temperature: $v \propto \sqrt{T}$ .

Step 4: Therefore, $\frac{{v_i}}{{v_f}} = \sqrt{\frac{{T_i}}{{T_f}}} \Rightarrow \frac{{300}}{{v_f}} = \sqrt{\frac{1}{2}}$ .

Step 5: Solve for $v_f$ : $v_f = 300 \sqrt{2} \text{{ m/s}}$ .

Why other options are wrong:
- Option B: Incorrect, this is not the correct final speed.
- Option C: Incorrect, this is not the correct final speed.
- Option D: Incorrect, this is not the correct final speed.

Key Concepts:
- Relationship between energy and temperature
- Relationship between speed and temperature

Memory Aids:
- Remember that $v \propto \sqrt{T}$ .

Quick Tips:
- Understand the relationship between temperature, energy, and speed for gases.

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Understanding +4 | -1

The value of molar specific heat at constant volume for 1 mole of polyatomic gas having

n

number of degrees of freedom at temperature

T \, K

is (here,

R=

universal gas constant)

View answer and explanation

Correct option: B

\frac{n R}{2}

Option B (

\frac{n R}{2}

) is correct because the molar specific heat at constant volume for a polyatomic gas with n degrees of freedom is given by

\frac{nR}{2}

.

Solution approach

Correct Answer + Core Reason: $\frac{n R}{2}$ . The molar specific heat at constant volume for a polyatomic gas with n degrees of freedom is $\frac{nR}{2}$ .

Step-by-Step Solution:

Step 1: Recall the equipartition theorem: Each degree of freedom contributes $\frac{1}{2}kT$ to the average energy per molecule.

Step 2: For a polyatomic gas with $n$ degrees of freedom, the average energy per molecule is $\frac{n}{2}kT$ .

Step 3: The total internal energy for 1 mole of the gas is $U = \frac{n}{2}N_AkT = \frac{n}{2}RT$ , where $R$ is the universal gas constant.

Step 4: The molar specific heat at constant volume is $C_V = \frac{dU}{dT} = \frac{d}{dT}(\frac{n}{2}RT) = \frac{nR}{2}$ .

Why other options are wrong:
- Option A: $\frac{n R}{2 T}$ : This is incorrect because the temperature is not in the denominator.
- Option C: $\frac{n R T}{2}$ : This is incorrect because the temperature is not in the expression for molar specific heat.
- Option D: $2 n R T$ : This is incorrect because the correct expression is $\frac{nR}{2}$ .

Key Concepts:
- Equipartition Theorem
- Degrees of Freedom
- Molar Specific Heat at Constant Volume

Memory Aids:
- $C_V = \frac{nR}{2}$

Quick Tips:
- Remember the formula for molar specific heat at constant volume for different types of gases.

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Understanding +4 | -1

Molecules of a gas behave like

View answer and explanation

Correct option: C

Perfectly elastic rigid sphere

Option C (Perfectly elastic rigid sphere) is correct because the kinetic theory of gases assumes that gas molecules undergo perfectly elastic collisions and are rigid spheres.

Solution approach

Correct Answer + Core Reason: Perfectly elastic rigid sphere. This is an assumption of the kinetic theory of gases.

Step-by-Step Solution:

Step 1: Recall the assumptions of the kinetic theory of gases.

Step 2: One of the key assumptions is that gas molecules are perfectly elastic rigid spheres.

Step 3: Perfectly elastic means that when molecules collide, kinetic energy is conserved.

Step 4: Rigid sphere means that the molecules do not deform during collisions.

Why other options are wrong:
- Option A: Inelastic rigid sphere is incorrect because the collisions are assumed to be perfectly elastic, not inelastic.
- Option B: Perfectly elastic non-rigid sphere is incorrect because the molecules are assumed to be rigid, not non-rigid.
- Option D: Inelastic non-rigid sphere is incorrect because the collisions are assumed to be perfectly elastic, and the molecules are assumed to be rigid.

Key Concepts:
- Kinetic theory of gases
- Elastic collisions
- Rigid spheres

Memory Aids:
- Remember the assumptions of the kinetic theory of gases.

Quick Tips:
- The kinetic theory of gases provides a simplified model for understanding the behavior of gases.

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Application +4 | -1

Two gases occupy two containers

\mathrm{A}

and

\mathrm{B}

the gas in

\mathrm{A}

, of volume

0.10 \mathrm{~m}^3

, exerts a pressure of

1.40 \mathrm{MPa}

and that in

\mathrm{B}

of volume

0.15 \mathrm{~m}^3

exerts a pressure

0.7 \mathrm{MPa}

. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in

\mathrm{MPa}

)

View answer and explanation

Correct option: B

0.98

Option B (0.98 MPa) is correct because the final pressure is calculated using Boyle's law, considering the combined volume and constant temperature.

Solution approach

Correct Answer + Core Reason: 0.98 MPa. The final pressure is calculated using Boyle's law for a constant temperature process.

Step-by-Step Solution:

Step 1: Identify the given information: $P_A = 1.40 \text{{ MPa}}$ , $V_A = 0.10 \text{{ m}}^3$ , $P_B = 0.7 \text{{ MPa}}$ , $V_B = 0.15 \text{{ m}}^3$ .

Step 2: Apply Boyle's Law: Since the temperature is constant, $P_1V_1 + P_2V_2 = P_f(V_1 + V_2)$ .

Step 3: Substitute the given values: $(1.40 \text{{ MPa}})(0.10 \text{{ m}}^3) + (0.7 \text{{ MPa}})(0.15 \text{{ m}}^3) = P_f(0.10 \text{{ m}}^3 + 0.15 \text{{ m}}^3)$ .

Step 4: Simplify: $0.14 + 0.105 = P_f(0.25)$ .

Step 5: Calculate the final pressure: $P_f = \frac{{0.245}}{{0.25}} = 0.98 \text{{ MPa}}$ .

Why other options are wrong:
- Option A (0.7): This is the pressure of gas B only and doesn't account for the contribution of gas A.
- Option C (1.4): This is the pressure of gas A only and doesn't account for the contribution of gas B.
- Option D (210): This is likely a calculation error, possibly multiplying the pressures and volumes incorrectly.

Key Concepts:
- Boyle's Law: $P_1V_1 = P_2V_2$ (at constant temperature)

Memory Aids:
- Boyle's Law: Think of it as Pressure and Volume are inversely related at constant temperature.

Quick Tips:
- Ensure consistent units before applying Boyle's Law.
- Remember that when gases mix, the total volume changes.

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Analyzing +4 | -1

Assertion : Vibrational energy of particle at temperature T is kT.

Reason : For every particle, vibrational degree of freedom is 2.

View answer and explanation

Correct option: D

If both Assertion & Reason are False

Option D is correct because the vibrational energy of a particle at temperature T is not simply kT, and a particle can have more or less than 2 vibrational degrees of freedom depending on the molecular structure.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are False. The vibrational energy is not kT, and the number of vibrational degrees of freedom is not always 2.

Step-by-Step Solution:
Step 1: The average vibrational energy of a particle at temperature T is given by kT only for each mode when the temperature is high enough. At lower temperatures, quantum effects become important.
Step 2: The number of vibrational degrees of freedom depends on the molecule. For a diatomic molecule, there is 1 vibrational mode, corresponding to 2 degrees of freedom (kinetic and potential energy). For more complex molecules, the number of vibrational modes is different.
Step 3: Therefore, both the assertion and the reason are false.

Why other options are wrong:
- Option A: Both assertion and reason are false, so this option is incorrect.
- Option B: Both assertion and reason are false, so this option is incorrect.
- Option C: Both assertion and reason are false, so this option is incorrect.

Key Concepts:
- Kinetic theory of gases
- Degrees of freedom
- Vibrational energy

Memory Aids:
- Remember that the number of degrees of freedom depends on the type of molecule.

Quick Tips:
- Always consider the specific molecule when determining the number of vibrational degrees of freedom.

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Understanding +4 | -1

Increase in temperature of a gas filled in a container would lead to:

View answer and explanation

Correct option: B

Increase in its kinetic energy

Option B is correct because increasing the temperature of a gas increases the average kinetic energy of its molecules.

Solution approach

Correct Answer + Core Reason: Increase in its kinetic energy. Temperature is directly proportional to the average kinetic energy of gas molecules.

Step-by-Step Solution:

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

Step 2: Consider the effect of increasing temperature: Increasing the temperature increases the average kinetic energy of the gas molecules.

Step 3: $U = \frac{{f}}{{2}}nRT$ , where U is internal energy, f is degrees of freedom, n is number of moles, R is gas constant, and T is temperature.

Why other options are wrong:
- Option A: The mass of the gas remains constant.
- Option C: Increasing temperature generally increases pressure (if volume is constant).
- Option D: Increasing temperature generally increases intermolecular distance (if pressure is constant).

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

Memory Aids:
- Higher temperature means faster molecules.

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

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Application +4 | -1

If three molecules of oxygen at

27^{\circ} \mathrm{C}

have velocities

0.5 \mathrm{~km} / \mathrm{s}, 1 \mathrm{~km} / \mathrm{s}

and

2 \mathrm{~km} / \mathrm{s}

respectively. Then their root mean square speed will be

View answer and explanation

Correct option: B

1.32 \mathrm{~km} / \mathrm{s}

Option B (1.32 km/s) is correct because the root mean square speed is calculated as the square root of the average of the squares of the velocities.

Solution approach

Correct Answer + Core Reason: $1.32 \mathrm{~km} / \mathrm{s}$ . The root mean square speed is calculated as the square root of the average of the squares of the velocities.

Step-by-Step Solution:

Step 1: Given velocities: $v_1 = 0.5 \mathrm{~km/s}$ , $v_2 = 1 \mathrm{~km/s}$ , $v_3 = 2 \mathrm{~km/s}$ .

Step 2: Calculate the squares of the velocities: $v_1^2 = (0.5)^2 = 0.25 \mathrm{~km^2/s^2}$ , $v_2^2 = (1)^2 = 1 \mathrm{~km^2/s^2}$ , $v_3^2 = (2)^2 = 4 \mathrm{~km^2/s^2}$ .

Step 3: Calculate the average of the squares: $\frac{v_1^2 + v_2^2 + v_3^2}{3} = \frac{0.25 + 1 + 4}{3} = \frac{5.25}{3} = 1.75 \mathrm{~km^2/s^2}$ .

Step 4: Calculate the root mean square speed: $v_{rms} = \sqrt{1.75} \approx 1.32 \mathrm{~km/s}$ .

Step 5: Final Answer: The root mean square speed is $1.32 \mathrm{~km/s}$ .

Why other options are wrong:
- Option A: $1.7 \mathrm{~km} / \mathrm{s}$ : This is incorrect because it is the square root of the average of the squared velocities before taking the square root.
- Option C: $1.13 \mathrm{~km} / \mathrm{s}$ : This is incorrect due to a calculation error.
- Option D: $1.17 \mathrm{~km} / \mathrm{s}$ : This is incorrect due to a calculation error.

Key Concepts:
- Root Mean Square Speed

Memory Aids:
- Remember the formula for RMS speed: $v_{rms} = \sqrt{\frac{v_1^2 + v_2^2 + ... + v_n^2}{n}}$ .

Quick Tips:
- Square the velocities first, then average, and finally take the square root.
- Double-check your calculations to avoid errors.

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Understanding +4 | -1

Speed of sound in a gas is

v

and

r.m.s.

velocity of the gas molecules is

c.

The ratio of

v

to

c

is

View answer and explanation

Correct option: D

\sqrt {\frac{\gamma }{3}}

Option D (

\sqrt{\frac{\gamma}{3}}

) is correct because it represents the ratio of the speed of sound in a gas to the RMS velocity of the gas molecules.

Solution approach

Correct Answer + Core Reason: $\sqrt {\frac{\gamma }{3}}$ . The ratio of the speed of sound to the RMS velocity is derived from their respective formulas.

Step-by-Step Solution:

Step 1: Write the formula for the speed of sound in a gas: $v = \sqrt{\frac{{\gamma P}}{\rho}}$ , where $\gamma$ is the adiabatic index, $P$ is the pressure, and $\rho$ is the density.

Step 2: Write the formula for the RMS velocity of gas molecules: $c = \sqrt{\frac{{3P}}{\rho}}$

Step 3: Calculate the ratio $\frac{v}{c}$ : $\frac{v}{c} = \frac{{\sqrt{\frac{{\gamma P}}{\rho}}}}{{\sqrt{\frac{{3P}}{\rho}}}} = \sqrt{\frac{{\gamma P}}{{\rho}} \times \frac{{\rho}}{{3P}}} = \sqrt{\frac{\gamma}{3}}$

Why other options are wrong:
- **Option A ( $\frac{3}{\gamma }$ ): This is the inverse of the square of the correct answer.
- Option B ( $\frac{\gamma }{3}$ ): This is the square of the correct answer.
- Option C ( $\sqrt {\frac{3}{\gamma }}$ ): This is the inverse of the correct answer.

Key Concepts:
- Speed of sound in a gas
- RMS velocity of gas molecules
- Adiabatic index

Memory Aids:
- Remember the formulas for the speed of sound and RMS velocity.

Quick Tips:**
- Pay attention to the definitions of the variables in the formulas.

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Understanding +4 | -1

For Boyle’s law to hold good, the gas should be

View answer and explanation

Correct option: B

Perfect and of constant mass and temperature

Boyle's law states that for a perfect gas with constant mass and temperature, pressure and volume are inversely proportional.

Solution approach

Correct Answer + Core Reason: Perfect and of constant mass and temperature. Boyle's law applies to ideal gases at constant temperature and mass.

Step-by-Step Solution:

Step 1: State Boyle's Law: Boyle's law states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional: $P \propto \frac{1}{V}$ or $PV = constant$ .

Step 2: Identify the conditions for Boyle's Law: Boyle's law holds true for:
- Perfect (Ideal) Gas: The gas should behave ideally, meaning it follows the ideal gas law ( $PV = nRT$ ).
- Constant Mass: The amount of gas (number of moles) should remain constant.
- Constant Temperature: The temperature of the gas should remain constant.

Step 3: Conclusion: For Boyle's law to hold good, the gas should be perfect and of constant mass and temperature.

Why other options are wrong:
- Option A: Perfect and at constant temperature but variable mass. Boyle's law requires constant mass.
- Option B: Perfect and of constant mass and temperature. Correct.
- Option C: Real and at constant temperature but variable mass. Boyle's law is best approximated by perfect gases, not real gases, and requires constant mass.
- Option D: Real and of constant mass and temperature. Boyle's law is best approximated by perfect gases, not real gases.

Key Concepts:
- Boyle's Law
- Ideal Gas Law
- Thermodynamics

Memory Aids:
- Boyle's Law: PV = constant (at constant temperature and mass)

Quick Tips:
- Remember that Boyle's law is an approximation that works best for ideal gases at relatively low pressures and high temperatures.

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Understanding +4 | -1

A cylinder rolls without slipping down an inclined plane, the number of degrees of freedom it has, is

View answer and explanation

Correct option: A

2

Option A is correct because a cylinder rolling without slipping on an inclined plane has one translational and one rotational degree of freedom.

Solution approach

Correct Answer + Core Reason: 2. A rolling cylinder has two degrees of freedom: one translational and one rotational.

Step-by-Step Solution:

Step 1: Define Degrees of Freedom: The number of independent variables required to describe the motion of a body completely.

Step 2: Analyze the motion of a rolling cylinder: A cylinder rolling down an inclined plane has both translational and rotational motion.

Step 3: Determine the independent variables: We need one variable to describe the translational position along the inclined plane (e.g., the distance traveled) and one variable to describe the rotational position (e.g., the angle of rotation).

Step 4: Conclude the number of degrees of freedom: Therefore, the cylinder has 2 degrees of freedom.

Why other options are wrong:
- Option B: 3 is incorrect because it would imply an additional independent motion, which is not present in pure rolling.
- Option C: 5 is incorrect as it overcounts the necessary independent variables.
- Option D: 1 is incorrect because it only considers either translation or rotation, not both.

Key Concepts:
- Degrees of Freedom
- Rolling Motion
- Translational Motion
- Rotational Motion

Memory Aids:
- Rolling = Translation + Rotation

Quick Tips:
- For rolling without slipping, the translational and rotational motions are related, but they are still independent degrees of freedom.

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Understanding +4 | -1

What is the average velocity of the molecules of an ideal gas?

View answer and explanation

Correct option: D

Zero

Option D (Zero) is correct because the average velocity considers the vector sum of all molecular velocities, which cancels out to zero in an ideal gas.

Solution approach

Correct Answer + Core Reason: Zero. The average velocity of gas molecules is zero due to random motion in all directions.

Step-by-Step Solution:

Step 1: Define average velocity: Average velocity is the displacement divided by the time interval.

Step 2: Consider the motion of gas molecules: In an ideal gas, molecules move randomly in all possible directions. This means that for every molecule moving in one direction, there is another molecule moving in the opposite direction.

Step 3: Calculate the vector sum of velocities: When we calculate the average velocity, we need to consider the vector sum of all the molecular velocities. Because of the random motion, the vector components in each direction cancel each other out.

Step 4: Conclusion: The vector sum of all molecular velocities is zero, and therefore the average velocity is zero.

Why other options are wrong:
- Option A: Infinity is incorrect because the molecules have finite speeds, and their average cannot be infinite.
- Option B: Constant is incorrect because the velocities of individual molecules are constantly changing due to collisions.
- Option C: Unstable is incorrect because while individual molecular velocities are changing, the average velocity is well-defined and zero.

Key Concepts:
- Average velocity
- Ideal gas
- Random motion

Memory Aids:
- Remember that the average *speed* of gas molecules is not zero, but the average *velocity* is zero.

Quick Tips:
- Distinguish between average speed and average velocity. Average speed is the total distance traveled divided by the time interval, while average velocity is the displacement divided by the time interval.

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Understanding +4 | -1

According to the kinetic theory of gases, at absolute temperature

View answer and explanation

Correct option: C

Molecular motion stops

The correct answer is C (Molecular motion stops) because at absolute zero temperature, the kinetic energy of the molecules becomes zero, resulting in no motion.

Solution approach

Correct Answer + Core Reason: Molecular motion stops. At absolute zero temperature, the kinetic energy and thus the motion of molecules ceases.

Step-by-Step Solution:

Step 1: Recall the relationship between temperature and molecular motion: Higher temperature implies greater molecular motion.

Step 2: Define absolute temperature: Absolute temperature is 0 Kelvin, the lowest possible temperature.

Step 3: Relate absolute temperature to molecular motion: At absolute zero, molecular motion stops.

Step 4: State the formula for rms velocity: ${v_{rms}} = \sqrt {\frac{{3RT}}{M}}$

Step 5: Substitute T=0: ${v_{rms}} = \sqrt {\frac{{3R(0)}}{M}} = 0$

Why other options are wrong:
- Option A: Water freezes at 273.15 K, not absolute zero.
- Option B: Liquid helium freezes at a temperature above absolute zero.
- Option D: Liquid hydrogen freezes at a temperature above absolute zero.

Key Concepts:
- Absolute Zero Temperature
- Kinetic Theory of Gases
- Molecular Motion

Memory Aids:
- Absolute zero = no molecular motion

Quick Tips:
- Remember that absolute zero is the theoretical lowest possible temperature where all molecular motion ceases.

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Application +4 | -1

At identical temperatures, the rms speed of hydrogen molecules is 4 times that for oxygen molecules. In a mixture of these in mass ratio

\mathrm{H}_2: \mathrm{O}_2=1: 8

, the rms speed of all molecules is

\mathrm{n}

times the rms speed for

\mathrm{O}_2

molecules, where

\mathrm{n}

is

View answer and explanation

Correct option: D

(11)^{1 / 2}

Option D (

\sqrt{{11}}

) is correct because it correctly calculates the root mean square speed of the mixture based on the number ratio and individual rms speeds.

Solution approach

Correct Answer + Core Reason: $(11)^{1 / 2}$ . The rms speed of the mixture is calculated using the weighted average of the squared rms speeds of the individual gases.

Step-by-Step Solution:

Step 1: Define the problem. We need to find the rms speed of the mixture in terms of the rms speed of oxygen.

Step 2: Find the number ratio of hydrogen to oxygen. Given the mass ratio $H_2 : O_2 = 1:8$ , we have $m_{H_2} = 1$ and $m_{O_2} = 8$ . The number of moles of $H_2$ is $n_{H_2} = \frac{{1}}{{2}}$ , and the number of moles of $O_2$ is $n_{O_2} = \frac{{8}}{{32}} = \frac{{1}}{{4}}$ . The ratio of the number of molecules is $\frac{{N_{H_2}}}{{N_{O_2}}} = \frac{{n_{H_2}}}{{n_{O_2}}} = \frac{{1/2}}{{1/4}} = 2$ . Thus, $N_{H_2} = 2 N_{O_2}$ .

Step 3: Calculate the rms speed of the mixture. Let $v_{O_2}$ be the rms speed of oxygen. Then the rms speed of hydrogen is $v_{H_2} = 4 v_{O_2}$ . The mean squared speed of the mixture is given by:
$v_{rms}^2 = \frac{{N_{H_2} v_{H_2}^2 + N_{O_2} v_{O_2}^2}}{{N_{H_2} + N_{O_2}}} = \frac{{2 N_{O_2} (4 v_{O_2})^2 + N_{O_2} v_{O_2}^2}}{{2 N_{O_2} + N_{O_2}}} = \frac{{32 N_{O_2} v_{O_2}^2 + N_{O_2} v_{O_2}^2}}{{3 N_{O_2}}} = \frac{{33 v_{O_2}^2}}{{3}} = 11 v_{O_2}^2$ .\nTherefore, $v_{rms} = \sqrt{{11}} v_{O_2}$ . So, $n = \sqrt{{11}}$ .

Why other options are wrong:
- Option A: $3$ This is incorrect because it doesn't account for the correct weighting of the rms speeds based on the number of molecules of each gas.
- Option B: $4 / 3$ This is incorrect because it doesn't account for the correct weighting of the rms speeds based on the number of molecules of each gas.
- Option C: $(8 / 3)^{1 / 2}$ This is incorrect because it doesn't account for the correct weighting of the rms speeds based on the number of molecules of each gas.

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

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

Quick Tips:
- Remember to use the number of molecules when calculating the average squared speed, not the mass.

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