KINETIC THEORY OF GASES MCQs for JEE PHYSICS — Medium, Page 9

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Analyzing +4 | -1

The internal energy

\left( U \right)

, pressure

\left( p \right)

and volume

\left( V \right)

of an ideal gas are related as

U = 3pV + 4

. The gas is

View answer and explanation

Correct option: B

polyatomic only

Option B is correct because the internal energy equation implies a degree of freedom greater than 6, indicating a polyatomic gas.

Solution approach

Correct Answer + Core Reason: polyatomic only. The given relation between internal energy, pressure, and volume implies a degree of freedom greater than 6, which is characteristic of polyatomic gases.

Step-by-Step Solution:

Step 1: Given: $U = 3pV + 4$ .

Step 2: For an ideal gas, $U = \frac{{f}}{{2}}pV$ , where $f$ is the number of degrees of freedom.

Step 3: Comparing the given equation with the ideal gas equation, we have $\frac{{f}}{{2}}pV = 3pV + 4$ .

Step 4: Rearranging the equation, we get $f = 6 + \frac{{8}}{{pV}}$ .

Step 5: Since $f$ must be an integer, and the term $\frac{{8}}{{pV}}$ is always positive, $f > 6$ .

Step 6: A gas with more than 6 degrees of freedom is polyatomic.

Why other options are wrong:
- Option A: diatomic only: Diatomic gases have 5 degrees of freedom (3 translational, 2 rotational).
- Option C: Either monoatomic or diatomic: Monoatomic gases have 3 degrees of freedom; diatomic gases have 5 degrees of freedom.
- Option D: monoatomic only: Monoatomic gases have 3 degrees of freedom.

Key Concepts:
- Internal energy of an ideal gas
- Degrees of freedom
- Monoatomic, diatomic, and polyatomic gases

Memory Aids:
- Monoatomic: 3 degrees of freedom
- Diatomic: 5 degrees of freedom
- Polyatomic: >6 degrees of freedom

Quick Tips:
- Remember the relationship between internal energy and degrees of freedom for ideal gases.

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Understanding +4 | -1

A polyatomic gas with n degrees of freedom has a mean energy per molecule given by
(

N

is Avogadro’s number)

View answer and explanation

Correct option: C

\frac{{nkT}}{2}

Option C is correct because the mean energy per molecule for a polyatomic gas with n degrees of freedom is given by

\frac{{nkT}}{2}

.

Solution approach

Correct Answer + Core Reason: $\frac{{nkT}}{2}$ . The mean energy per molecule is calculated using the equipartition theorem.

Step-by-Step Solution:

Step 1: Recall the equipartition theorem, which states that 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 mean energy per molecule is $E = n \times \frac{{1}}{{2}}kT$ .

Step 3: Therefore, $E = \frac{{nkT}}{2}$ .

Why other options are wrong:
- Option A: $\frac{{nkT}}{N}$ is incorrect because it divides by Avogadro's number, which is not needed for the energy per molecule.
- Option B: $\frac{{nkT}}{{2N}}$ is incorrect because it divides by Avogadro's number, which is not needed for the energy per molecule.
- Option D: $\frac{{3kT}}{2}$ is incorrect because it only applies to a monoatomic gas (3 degrees of freedom).

Key Concepts:
- Equipartition theorem
- Degrees of freedom
- Mean energy per molecule

Memory Aids:
- Each degree of freedom contributes $\frac{{1}}{{2}}kT$ to the average energy.

Quick Tips:
- Remember the equipartition theorem and how it relates to the degrees of freedom.

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A flask contains Hydrogen and Argon in the ratio

2: 1

by mass. The temperature of the mixture is

30^{\circ} \mathrm{C}

. The ratio of average kinetic energy per molecule of the two gases (

\mathrm{K}

argon/

\mathrm{K}

hydrogen) is:
(Given : Atomic Weight of

\mathrm{Ar}=39.9

)

View answer and explanation

Correct option: B

1

Option B (1) is correct because the average kinetic energy per molecule depends only on temperature, which is the same for both gases.

Solution approach

Correct Answer + Core Reason: 1. Average kinetic energy per molecule depends only on temperature.

Step-by-Step Solution:
Step 1: The average kinetic energy per molecule is given by $KE = \frac{3}{2}kT$ , where $k$ is Boltzmann's constant and $T$ is the temperature.
Step 2: Since the temperature is the same for both Hydrogen and Argon, the average kinetic energy per molecule is the same.
Step 3: Therefore, the ratio of average kinetic energy per molecule of Argon to Hydrogen is 1.

Why other options are wrong:
- Option A: 2 - Incorrect because the ratio is not dependent on the mass ratio.
- Option C: 39.9 - Incorrect because the ratio is not dependent on the atomic weight of Argon.
- Option D: $\frac{{39.9}}{2}$ - Incorrect because the ratio is not dependent on the atomic weight of Argon.

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

Memory Aids:
- Average KE per molecule = $\frac{3}{2}kT$ .

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

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Statement-I: Boltzmann's constant is a gas constant for 1 molecule of a gas
Statement-II: Universal gas constant is a gas constant for 1 mole of a gas

View answer and explanation

Correct option: A

Statement-I and statement-II are true

Option A (Statement-I and statement-II are true) is correct because Boltzmann's constant is the gas constant for one molecule, and the universal gas constant is for one mole.

Solution approach

Correct Answer + Core Reason: Statement-I and statement-II are true. Boltzmann's constant applies to single molecules, while the universal gas constant applies to moles of gas.

Step-by-Step Solution:

Step 1: Recall the ideal gas law in terms of number of molecules: $PV = NkT$ , where $k$ is Boltzmann's constant and $N$ is the number of molecules.

Step 2: Recall the ideal gas law in terms of number of moles: $PV = nRT$ , where $R$ is the universal gas constant and $n$ is the number of moles.

Step 3: Compare the two equations. Boltzmann's constant relates pressure and volume to the number of molecules and temperature. The universal gas constant relates pressure and volume to the number of moles and temperature.

Step 4: Conclude that Statement-I and Statement-II are both true.

Why other options are wrong:
- Option B (Statement-I and statement-II are false): Both statements are factually correct.
- Option C (Statement-I is true, but statement-II is false): Statement-II is also true.
- Option D (Statement-I is false but statement-II is true): Statement-I is also true.

Key Concepts:
- Ideal gas law
- Boltzmann's constant
- Universal gas constant
- Moles and molecules

Memory Aids:
- $PV = NkT$ (Boltzmann's constant)
- $PV = nRT$ (Universal gas constant)

Quick Tips:
- Remember the different forms of the ideal gas law.

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If at the same temperature and pressure, the densities of two diatomic gases are

d_1

and

d_2

, respectively. The ratio of mean kinetic energy per molecule of gases will be

View answer and explanation

Correct option: A

1: 1

Option A (1:1) is correct because the mean kinetic energy per molecule depends only on temperature, which is the same for both gases.

Solution approach

Correct Answer + Core Reason: 1:1. The mean kinetic energy per molecule depends only on temperature, which is the same for both gases.

Step-by-Step Solution:

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

Step 2: Since the temperature and pressure are the same for both gases, $T_1 = T_2 = T$ .

Step 3: Calculate the ratio of the mean kinetic energies: $\frac{E_1}{E_2} = \frac{\frac{3}{2}kT_1}{\frac{3}{2}kT_2} = \frac{T_1}{T_2} = \frac{T}{T} = 1$ .

Step 4: Therefore, the ratio of the mean kinetic energy per molecule of the gases is 1:1.

Why other options are wrong:
- Option B (d_1: d_2): This is incorrect because the mean kinetic energy does not depend on the density of the gas.
- **Option C ( $\sqrt{{d_1}}$ : $\sqrt{{d_2}}$ ): This is incorrect because the mean kinetic energy does not depend on the density of the gas.
- Option D ( $\sqrt{{d_2}}$ : $\sqrt{{d_1}}$ ): This is incorrect because the mean kinetic energy does not depend on the density of the gas.

Key Concepts:
- Mean kinetic energy of a gas molecule
- Boltzmann constant
- Temperature

Memory Aids:
- The mean kinetic energy of a gas molecule is directly proportional to the absolute temperature.

Quick Tips:**
- Remember that the mean kinetic energy of a gas molecule depends only on temperature, not on pressure or density.

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Application +4 | -1

The absolute temperature of a gas is increases 3 times. The root mean square velocity of the moelcules will be

View answer and explanation

Correct option: D

\sqrt{3}

times

Option D (

\sqrt{3}

times) is correct because the root mean square velocity is proportional to the square root of the absolute temperature.

Solution approach

Correct Answer + Core Reason: $\sqrt{3}$ times. The root mean square velocity is proportional to the square root of the absolute temperature.

Step-by-Step Solution:

Step 1: Recall the formula for root mean square velocity: $v_{rms} = \sqrt{\frac{{3kT}}{{m}}}$ , where $k$ is Boltzmann's constant, $T$ is absolute temperature, and $m$ is the mass of the molecule.

Step 2: Let the initial temperature be $T$ and the final temperature be $3T$ .

Step 3: The initial root mean square velocity is $v_{rms} = \sqrt{\frac{{3kT}}{{m}}}$ .

Step 4: The final root mean square velocity is $v'_{rms} = \sqrt{\frac{{3k(3T)}}{{m}}} = \sqrt{\frac{{9kT}}{{m}}}$ .

Step 5: Find the ratio of the final to initial root mean square velocities: $\frac{{v'_{rms}}}{{v_{rms}}} = \frac{{\sqrt{\frac{{9kT}}{{m}}}}}{{\sqrt{\frac{{3kT}}{{m}}}}} = \sqrt{\frac{{9kT}}{{3kT}}} = \sqrt{3}$ .

Step 6: Therefore, $v'_{rms} = \sqrt{3} v_{rms}$ . The root mean square velocity increases by a factor of $\sqrt{3}$ .

Why other options are wrong:
- Option A: 3 times is incorrect. The rms velocity is proportional to the square root of temperature, not directly proportional.
- Option B: 9 times is incorrect. This would be the case if the rms velocity was proportional to the square of the temperature.
- Option C: $1/3$ times is incorrect. This would be the case if the rms velocity was inversely proportional to the temperature.

Key Concepts:
- Root mean square velocity
- Kinetic theory of gases
- Temperature and molecular speed

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

Quick Tips:
- The rms velocity is proportional to the square root of the absolute temperature.
- Ensure the temperature is in Kelvin.

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Assertion: The root mean square and most probable speeds of the molecules in a gas are the same
Reason: The Maxwell's distribution for the speed of molecules in a gas is symmetrical.

View answer and explanation

Correct option: D

If both Assertion and Reason are false

Option D is correct because the root mean square and most probable speeds are different, and Maxwell's distribution is asymmetrical.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are false. RMS and most probable speeds are different, and the distribution is asymmetrical.

Step-by-Step Solution:

Step 1: The root mean square speed ( $V_{rms}$ ) and most probable speed ( $V_{mp}$ ) of gas molecules are not the same.

Step 2: Maxwell's distribution for the speed of molecules in a gas is asymmetrical.

Why other options are wrong:
- Option A: Both Assertion and Reason are false.
- Option B: Both Assertion and Reason are false.
- Option C: Assertion is false.

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

Memory Aids:
- $V_{rms} \neq V_{mp}$
- Maxwell's Distribution: Asymmetrical

Quick Tips:
- Remember the difference between RMS and most probable speeds.

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Application +4 | -1

In a vertical open vessel an ideal gas is filled and it is enclosed by a piston of mass

4 \mathrm{~kg}

and cross-section area is

2 \mathrm{~cm}^2

then find out pressure applied by gas on piston in equilibrium.

View answer and explanation

Correct option: B

3 P_0

Option B (

3 P_0

) is correct because the gas pressure equals atmospheric pressure plus the pressure due to the piston's weight.

Solution approach

Correct Answer + Core Reason: $3 P_0$ . The pressure exerted by the gas is the sum of atmospheric pressure and the pressure due to the piston's weight.

Step-by-Step Solution:

Step 1: Calculate the force due to the piston's weight: $F = mg = 4 \text{{ kg}} \times 10 \text{{ m/s}}^2 = 40 \text{{ N}}$ .

Step 2: Calculate the area of the piston: $A = 2 \text{{ cm}}^2 = 2 \times 10^{-4} \text{{ m}}^2$ .

Step 3: Calculate the pressure due to the piston's weight: $P_{piston} = \frac{F}{A} = \frac{{40}}{{2 \times 10^{-4}}} = 2 \times 10^5 \text{{ N/m}}^2 = 2 \text{{ atm}} = 2P_0$ .

Step 4: Calculate the total pressure exerted by the gas: $P_{gas} = P_0 + P_{piston} = P_0 + 2P_0 = 3P_0$ .

Final Answer: $3 P_0$

Why other options are wrong:
- Option A: Incorrect calculation of the gas pressure.
- Option C: Incorrect calculation of the gas pressure.
- Option D: Incorrect calculation of the gas pressure.

Key Concepts:
- Pressure: $P = \frac{F}{A}$
- Atmospheric pressure

Memory Aids:
- Remember that the total pressure is the sum of all pressures acting on the piston.

Quick Tips:
- Convert all units to SI units before calculations.

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Application +4 | -1

An ideal gas is allowed to expands from volume

V

to

2 V

according to the law

V P^2=

Constant. If initial temperature of the gas is

T

then its final temperature will be

View answer and explanation

Correct option: B

T \sqrt{2}

Option B (

T \sqrt{2}

) is correct because the final temperature is derived from the given relation

VP^2 = constant

and the ideal gas law.

Solution approach

Correct Answer + Core Reason: $T \sqrt{2}$ . The final temperature is calculated using the given relation and the ideal gas law.

Step-by-Step Solution:

Step 1: Write the given relation: $VP^2 = K$ (constant).

Step 2: Write the ideal gas law: $PV = nRT$ , where $n$ is the number of moles and $R$ is the gas constant.

Step 3: Express $P$ in terms of $V$ and $T$ using the ideal gas law: $P = \frac{nRT}{V}$ .

Step 4: Substitute this expression for $P$ into the given relation: $V(\frac{nRT}{V})^2 = K \Rightarrow \frac{n^2R^2T^2}{V} = K \Rightarrow T^2 \propto V$ .

Step 5: Therefore, $T \propto \sqrt{V}$ .

Step 6: Write the ratio of final temperature ( $T'$ ) to initial temperature ( $T$ ) in terms of the ratio of final volume ( $2V$ ) to initial volume ( $V$ ): $\frac{T'}{T} = \sqrt{\frac{2V}{V}} = \sqrt{2}$ .

Step 7: Solve for $T'$ : $T' = T\sqrt{2}$ .

Why other options are wrong:
- Option A (T): This implies the temperature remains constant, which is incorrect given the expansion and the relation $VP^2 = constant$ .
- **Option C (T/ $\sqrt{2}$ ): This would be the case if the volume decreased instead of increased.
- Option D (2T): This is an incorrect relationship between the initial and final temperatures based on the given conditions.

Key Concepts:
- Ideal Gas Law
- Thermodynamics
- Proportionality

Memory Aids:
- PV = nRT

Quick Tips:**
- Combine the given relation with the ideal gas law to find the relationship between temperature and volume.

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Application +4 | -1

The volume of vessel

A

is twice the volume of another vessel

B

, and both of them are filled with the same gas. If the gas in

A

is at twice the temperature and twice the pressure in comparison to the gas in

B

, then the ratio of the gas molecules in

A

to that of

B

View answer and explanation

Correct option: A

\frac{2}{1}

Option C (Equal to or greater than

{\sin ^{ - 1}}\left( {\frac{3}{4}} \right)

) is correct because for total internal reflection to occur, the angle of incidence must be greater than or equal to the critical angle.

Solution approach

Correct Answer + Core Reason: Equal to or greater than ${\sin ^{ - 1}}\left( {\frac{3}{4}} \right)$ . Total internal reflection occurs when the angle of incidence is greater than the critical angle.

Step-by-Step Solution:

Step 1: Find the refractive indices of the two media: $\mu_1 = \frac{c}{v_1} = \frac{{3 \times 10^8}}{{1.5 \times 10^8}} = 2$ and $\mu_2 = \frac{c}{v_2} = \frac{{3 \times 10^8}}{{2.0 \times 10^8}} = 1.5 = \frac{3}{2}$

Step 2: Apply Snell's law at the critical angle: $\mu_1 \sin i_c = \mu_2 \sin 90^\circ$

Step 3: Substitute the values: $2 \sin i_c = \frac{3}{2} \times 1$

Step 4: Solve for the critical angle: $\sin i_c = \frac{3}{4}$ , so $i_c = \sin^{-1}(\frac{3}{4})$

Step 5: For total internal reflection, $i \ge i_c$ , so $i \ge \sin^{-1}(\frac{3}{4})$

Why other options are wrong:
- **Option A (Equal to ${\sin ^{ - 1}}\left( {\frac{2}{3}} \right)$ ): This is incorrect because it calculates the critical angle incorrectly.
- Option B (Equal to or less than ${\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$ ): This is incorrect because the angle of incidence must be greater than or equal to the critical angle for total internal reflection.
- Option D (Less than ${\sin ^{ - 1}}\left( {\frac{2}{3}} \right)$ ): This is incorrect because the angle of incidence must be greater than or equal to the critical angle for total internal reflection.

Key Concepts:
- Total internal reflection
- Snell's law
- Critical angle
- Refractive index

Memory Aids:**
- Remember Snell's law: $\mu_1 \sin \theta_1 = \mu_2 \sin \theta_2$
- Remember that total internal reflection occurs when light travels from a denser to a rarer medium.

Quick Tips:
- Ensure you correctly calculate the refractive indices.
- Remember the condition for total internal reflection: $i \ge i_c$ .

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Understanding +4 | -1

Which statements are correct about degrees of freedom?
(A) A molecule with

n

degrees of freedom has

n^2

different ways of storing energy.
(B) Each degree of freedom is associated with

\frac{1}{2} R T

average energy per mole.
(C) A monatomic gas molecule has 1 rotational degree of freedom whereas diatomic molecule has 2 rotational degrees of freedom.
(D)

\mathrm{CH}_4

has a total of 6 degrees of freedom.
Choose the correct answer from the option given below:

View answer and explanation

Correct option: B

(B) and (D) only

Option B is correct because, according to Einstein's photoelectric equation, the kinetic energy of emitted electrons depends on the frequency of the incident light.

Solution approach

Correct Answer + Core Reason: Frequency of incident light. The kinetic energy of emitted electrons depends on the frequency of the incident light, according to the photoelectric equation.

Step-by-Step Solution:

Step 1: Recall Einstein's photoelectric equation: $E_K = h\nu - W$ , where $E_K$ is the kinetic energy of the emitted electrons, $h$ is Planck's constant, $\nu$ is the frequency of the incident light, and $W$ is the work function of the metal.

Step 2: Analyze the equation: The kinetic energy $E_K$ is directly proportional to the frequency $\nu$ of the incident light.

Step 3: Understand that the intensity of light affects the number of photoelectrons emitted, but not their kinetic energy.

Why other options are wrong:
- Option A: Intensity of light affects the number of photoelectrons, not their kinetic energy.
- Option C: Velocity of incident light is related to frequency, but the primary dependence is on frequency.
- Option D: Intensity affects the number of electrons, and velocity is indirectly related through frequency, but the direct dependence is on frequency.

Key Concepts:
- Photoelectric effect
- Einstein's photoelectric equation
- Frequency of light
- Intensity of light
- Work function

Memory Aids:
- $E_K = h\nu - W$

Quick Tips:
- Remember that the kinetic energy of photoelectrons depends on the frequency of the incident light, not its intensity.

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Analyzing +4 | -1

Assertion(A): If initial and final volumes are equal, then work done by gas is zero.

Reason(R): In isochoric process, initial and final volumes are equal and work done by gas is zero.

View answer and explanation

Correct option: D

If Assertion is incorrect but Reason is correct.

Option D is correct. The assertion is incorrect because work done is zero only if the volume remains constant *throughout* the process, not just at the initial and final states. The reason is correct because it accurately describes an isochoric process.

Solution approach

Correct Answer + Core Reason: If Assertion is incorrect but Reason is correct.

Step-by-Step Solution:

Step 1: Analyze the Assertion: The assertion states that if initial and final volumes are equal, the work done by the gas is zero. This is not always true. The volume must remain constant *throughout* the process for the work to be zero.

Step 2: Analyze the Reason: The reason states that in an isochoric process, initial and final volumes are equal, and work done by the gas is zero. This is a correct definition of an isochoric process.

Step 3: Determine the relationship: The reason correctly describes a specific case (isochoric process) where the assertion's condition is met, but the assertion is not universally true.

Step 4: Conclusion: The assertion is incorrect, but the reason is correct.

Why other options are wrong:
- Option A: Incorrect because the assertion is false.
- Option B: Incorrect because the assertion is false.
- Option C: Incorrect because the reason is correct.
- Option D: Correct because the assertion is false, and the reason is true.

Key Concepts:
- Work done by a gas: $W = \int P dV$
- Isochoric process: Constant volume process.

Memory Aids:
- Isochoric: "Iso-" means same, "-choric" relates to volume. Same volume.

Quick Tips:
- Remember that work done depends on the entire process, not just the initial and final states.

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Understanding +4 | -1

In the absence of intermolecular forces of attraction, the observed pressure

p

will be

View answer and explanation

Correct option: C

> p

Option C (

> p

) is correct because in the absence of intermolecular forces, the molecules exert more pressure due to the lack of attractive forces reducing the impact.

Solution approach

Correct Answer + Core Reason: $> p$ . Without intermolecular forces, the molecules will not stick together, leading to a higher observed pressure.

Step-by-Step Solution:

Step 1: Consider the effect of intermolecular forces on pressure.

Step 2: Intermolecular forces cause molecules to attract each other, reducing the force with which they hit the walls of the container.

Step 3: If intermolecular forces are absent, the molecules will exert a greater force on the walls.

Step 4: Therefore, the observed pressure will be greater than the pressure when intermolecular forces are present.

Why other options are wrong:
- Option A: $p$ - This is incorrect because the pressure would be different in the absence of intermolecular forces.
- Option B: $< p$ - This is incorrect because the pressure would be higher, not lower, in the absence of intermolecular forces.
- Option D: Zero - This is incorrect because the molecules still have kinetic energy and will exert pressure.

Key Concepts:
- Ideal Gas Law
- Intermolecular Forces
- Pressure

Memory Aids:
- Absence of attraction leads to higher impact.

Quick Tips:
- Remember that intermolecular forces reduce the observed pressure.

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Application +4 | -1

The rms speed of an ideal diatomic gas at temperature

T

is

V

. When gas dissociates into atoms then its new rms speed becomes double. The temperature at which the gas dissociated into atoms are:

View answer and explanation

Correct option: D

2 T

Option D (2 T) is correct because when the diatomic gas dissociates, the molar mass halves, and the rms speed doubles, leading to a doubling of the temperature.

Solution approach

Correct Answer + Core Reason: $2 T$ . The temperature doubles when the gas dissociates because the rms speed doubles and the molar mass halves.

Step-by-Step Solution:

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

Step 2: For the diatomic gas, $v = \sqrt{\frac{3RT}{M}}$ .

Step 3: When the gas dissociates into atoms, the molar mass becomes $M' = \frac{M}{2}$ , and the rms speed becomes $v' = 2v$ . Let the new temperature be $T'$ .\nSo, $2v = \sqrt{\frac{3RT'}{M/2}} = \sqrt{\frac{6RT'}{M}}$ .

Step 4: Divide the second equation by the first equation: $\frac{2v}{v} = \frac{\sqrt{\frac{6RT'}{M}}}{\sqrt{\frac{3RT}{M}}} \Rightarrow 2 = \sqrt{\frac{6RT'}{3RT}} = \sqrt{\frac{2T'}{T}}$ .

Step 5: Square both sides: $4 = \frac{2T'}{T} \Rightarrow T' = 2T$ .

Why other options are wrong:
- Option A: $T$ - Incorrectly assumes the temperature remains the same.
- Option B: $\sqrt{2} T$ - Incorrectly relates the temperature change to the rms speed change.
- Option C: $\frac{T}{2}$ - Incorrectly assumes the temperature halves.

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

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

Quick Tips:
- When a diatomic gas dissociates into atoms, its molar mass is halved.

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Understanding +4 | -1

According to the kinetic theory of gases, the temperature of a gas is a measure of average

View answer and explanation

Correct option: C

Kinetic energies of its molecules

Option C (Kinetic energies of its molecules) is correct because temperature is directly proportional to the average kinetic energy of the gas molecules.

Solution approach

Correct Answer + Core Reason: Kinetic energies of its molecules. Temperature is a measure of the average kinetic energy of the gas molecules.

Step-by-Step Solution:

Step 1: Recall the Kinetic Theory of Gases: The kinetic theory of gases relates macroscopic properties of gases (like temperature and pressure) to the microscopic behavior of the gas molecules.

Step 2: State the Relationship: According to the kinetic theory, the average kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas: $KE_{avg} = \frac{3}{2}kT$ , where $k$ is the Boltzmann constant and $T$ is the absolute temperature.

Step 3: Conclusion: Therefore, the temperature of a gas is a measure of the average kinetic energies of its molecules.

Why other options are wrong:
- Option A: Incorrect. Temperature is related to the average kinetic energy, not just the velocities.
- Option B: Incorrect. Temperature is related to kinetic energy, which depends on the square of the velocity (and thus momentum squared), not linear momentum directly.
- Option D: Incorrect. The kinetic theory primarily deals with translational kinetic energy, not angular momentum.

Key Concepts:
- Kinetic Theory of Gases
- Average Kinetic Energy
- Temperature
- Boltzmann Constant

Memory Aids:
- Remember that higher temperature means faster-moving molecules, which means higher kinetic energy.

Quick Tips:
- The kinetic theory of gases provides a microscopic explanation for macroscopic gas properties.

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Application +4 | -1

Two moles of an ideal gas with

\frac{{{C_p}}}{{{C_V}}} = \frac{5}{3}

are mixed with 3 mol of another ideal gas with

\frac{{{C_p}}}{{{C_V}}} = \frac{4}{3}.

The value of

\frac{{{C_p}}}{{{C_V}}}

for the mixture is

View answer and explanation

Correct option: A

1.42

Option A (1.42) is correct because it accurately calculates the specific heat ratio for the mixture of ideal gases.

Solution approach

Correct Answer + Core Reason: 1.42. The specific heat ratio for the mixture is calculated using the number of moles and specific heat ratios of individual gases.

Step-by-Step Solution:

Step 1: Recall the formula for the specific heat ratio of a mixture:
$\frac{C_{p(mix)}}{C_{V(mix)}} = \frac{n_1 C_{p_1} + n_2 C_{p_2}}{n_1 C_{V_1} + n_2 C_{V_2}}$

Step 2: Use the relationship $C_p = \frac{\gamma}{\gamma - 1}R$ and $C_V = \frac{R}{\gamma - 1}$ :
$\frac{C_{p(mix)}}{C_{V(mix)}} = \frac{n_1 \frac{\gamma_1}{\gamma_1 - 1}R + n_2 \frac{\gamma_2}{\gamma_2 - 1}R}{n_1 \frac{R}{\gamma_1 - 1} + n_2 \frac{R}{\gamma_2 - 1}} = \frac{n_1 \gamma_1 (\gamma_2 - 1) + n_2 \gamma_2 (\gamma_1 - 1)}{n_1 (\gamma_2 - 1) + n_2 (\gamma_1 - 1)}$

Step 3: Substitute the given values: $n_1 = 2, \gamma_1 = \frac{5}{3}, n_2 = 3, \gamma_2 = \frac{4}{3}$
$\frac{C_{p(mix)}}{C_{V(mix)}} = \frac{2 \times \frac{5}{3} (\frac{4}{3} - 1) + 3 \times \frac{4}{3} (\frac{5}{3} - 1)}{2 (\frac{4}{3} - 1) + 3 (\frac{5}{3} - 1)} = \frac{2 \times \frac{5}{3} \times \frac{1}{3} + 3 \times \frac{4}{3} \times \frac{2}{3}}{2 \times \frac{1}{3} + 3 \times \frac{2}{3}} = \frac{\frac{10}{9} + \frac{24}{9}}{\frac{2}{3} + \frac{6}{3}} = \frac{\frac{34}{9}}{\frac{8}{3}} = \frac{34}{9} \times \frac{3}{8} = \frac{17}{12} \approx 1.42$

Step 4: Final Answer: The specific heat ratio for the mixture is approximately 1.42.

Why other options are wrong:
- Option B: This option is incorrect as it overestimates the specific heat ratio. This could be due to a miscalculation in the formula.
- Option C: This option is incorrect as it overestimates the specific heat ratio. This might be due to a miscalculation in the formula.
- Option D: This option is incorrect as it overestimates the specific heat ratio. This could be due to a miscalculation in the formula.

Key Concepts:
- Specific heat ratio
- Ideal gas mixture

Memory Aids:
- Remember the formula for the specific heat ratio of a mixture.

Quick Tips:
- Ensure correct substitution of values in the formula.
- Pay attention to the relationships between $C_p$ , $C_V$ , and $\gamma$ .

257 Medium

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Understanding +4 | -1

Statement-I: Moon has no atmosphere
Statement-II: The root mean square velocity of air molecules on moon is greater than the escape velocity of air molecules on moon.

View answer and explanation

Correct option: A

Statement-I and statement-II are true

Option A (Statement-I and statement-II are true) is correct because the Moon lacks an atmosphere due to the high RMS velocity of gas molecules exceeding the escape velocity.

Solution approach

Correct Answer + Core Reason: Statement-I and statement-II are true. The Moon has no atmosphere because the RMS velocity of air molecules is greater than the escape velocity, causing them to escape into space.

Step-by-Step Solution:

Step 1: Understand why planets have atmospheres.\nPlanets retain atmospheres if the average speed of gas molecules is significantly less than the escape velocity.

Step 2: Consider the Moon's gravity and temperature.\nThe Moon has a low gravitational pull and relatively high surface temperature.

Step 3: Calculate or recall typical RMS velocities of gas molecules at lunar temperatures.\nThe RMS velocity of common gases is higher than the Moon's escape velocity.

Step 4: Compare RMS velocity and escape velocity.\nSince $V_{rms} > V_e$ , gas molecules escape the Moon's gravity.

Why other options are wrong:
- Option B: Statement-I and statement-II are false. Both statements are true.
- Option C: Statement-I is true, but statement-II is false. Statement II is also true.
- Option D: Statement-I is false but statement-II is true. Statement I is also true.

Key Concepts:
- Escape velocity
- Root mean square (RMS) velocity
- Kinetic theory of gases
- Planetary atmospheres

Memory Aids:
- Higher temperature → higher RMS velocity
- Lower gravity → lower escape velocity

Quick Tips:
- If RMS velocity > escape velocity, the gas will escape.

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