KINETIC THEORY OF GASES MCQs for JEE PHYSICS — Medium

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Understanding +4 | -1

If

c_s

is the velocity of sound in air and

c

is the rms velocity, then

View answer and explanation

Correct option: C

c_{{s}}=c\left(\frac{\gamma}{3}\right)^{1 / 2}

Option C (

c_{{s}}=c(\frac{\gamma}{3})^{1 / 2}

) is correct because it represents the correct relationship between the speed of sound and the RMS velocity of gas molecules.

Solution approach

Correct Answer + Core Reason: $c_{{s}}=c(\frac{\gamma}{3})^{1 / 2}$ . This formula relates the speed of sound to the RMS velocity.

Step-by-Step Solution:

Step 1: Recall the formula for the speed of sound in a gas: $c_s = \sqrt{\frac{{\gamma p}}{{\rho }}}$ .

Step 2: Recall the formula for the RMS velocity of gas molecules: $c = \sqrt{\frac{{3pV}}{M}} = \sqrt{\frac{{3p}}{{\rho }}}$ .

Step 3: Find the ratio of $c_s$ to $c$ : $\frac{{c_s}}{c} = \sqrt{\frac{\gamma}{3}}$ .

Step 4: Solve for $c_s$ : $c_s = c\sqrt{\frac{\gamma}{3}} = c(\frac{\gamma}{3})^{1/2}$ .

Why other options are wrong:
- Option A: Incorrect because $c_s$ is not simply less than $c$ .
- Option B: Incorrect because $c_s$ is not equal to $c$ .
- Option D: Incorrect because option C is correct.

Key Concepts:
- Speed of sound in a gas
- Root mean square (RMS) velocity
- Adiabatic index ( $\gamma$ )

Memory Aids:
- Remember the formulas for $c_s$ and $c$ .

Quick Tips:
- Pay attention to the relationship between microscopic and macroscopic properties of gases.

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Application +4 | -1

Four particles have velocities

1,0,2,3 \mathrm{~m} / \mathrm{s}

. The root mean square of the particles is (in

\mathrm{m} / \mathrm{s}

)

View answer and explanation

Correct option: B

\sqrt{3.5}

Option B (

\sqrt{3.5}

) is correct because it is calculated using the formula for root mean square velocity.

Solution approach

Correct Answer + Core Reason: $\sqrt{3.5}$ . The root mean square velocity is calculated using the standard formula.

Step-by-Step Solution:

Step 1: Identify the given velocities: $1, 0, 2, 3 \mathrm{~m} / \mathrm{s}$ .

Step 2: Recall the formula for root mean square velocity: ${V_{rms}} = \sqrt{\frac{{\sum v_i^2}}{N}}$

Step 3: Substitute the given velocities into the formula: ${V_{rms}} = \sqrt{\frac{{{1^2} + {0^2} + {2^2} + {3^2}}}{4}}$

Step 4: Calculate the sum of the squares: ${1^2} + {0^2} + {2^2} + {3^2} = 1 + 0 + 4 + 9 = 14$

Step 5: Divide by the number of particles and take the square root: ${V_{rms}} = \sqrt{\frac{{14}}{4}} = \sqrt{3.5}\\ \mathrm{m} / \mathrm{s}$

Final Answer: ${V_{rms}} = \sqrt{3.5}\\ \mathrm{m} / \mathrm{s}$

Why other options are wrong:
- Option A (3.5): This value is incorrect. It might result from not taking the square root at the end.
- Option C (1.5): This value is incorrect. It might result from calculating the arithmetic mean instead of the root mean square.
- **Option D ( $\sqrt{14 / 3}$ ): This value is incorrect. It might result from a calculation error.

Key Concepts:
- Root mean square velocity

Memory Aids:
- Remember the formula for RMS: Square, Mean, then Root.

Quick Tips:**
- Ensure you square each velocity before summing.
- Remember to take the square root at the end.

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Understanding +4 | -1

The translational degrees of freedom

(f_t)

and rotational degrees of freedom

(f_r)

of

{CH}_4

molecule are:

View answer and explanation

Correct option: B

f_t=3

and

f_r=3

Option B (

f_t=3

and

f_r=3

) is correct because methane (

CH_4

) is a non-linear polyatomic molecule, possessing 3 translational and 3 rotational degrees of freedom.

Solution approach

Correct Answer + Core Reason: $f_t=3$ and $f_r=3$ . Methane is a non-linear polyatomic molecule.

Step-by-Step Solution:

Step 1: Identify the type of molecule: $CH_4$ is a non-linear polyatomic molecule.

Step 2: Determine the translational degrees of freedom ( $f_t$ ). All molecules have 3 translational degrees of freedom.
$f_t = 3$

Step 3: Determine the rotational degrees of freedom ( $f_r$ ). For a non-linear molecule, $f_r = 3$ .
$f_r = 3$

Why other options are wrong:
- Option A: Incorrect values for both translational and rotational degrees of freedom.
- Option C: Incorrect value for rotational degrees of freedom.
- Option D: Incorrect values for both translational and rotational degrees of freedom.

Key Concepts:
- Degrees of freedom
- Translational degrees of freedom
- Rotational degrees of freedom
- Non-linear polyatomic molecules

Memory Aids:
- Non-linear molecules have 3 rotational degrees of freedom.

Quick Tips:
- Remember that all molecules have 3 translational degrees of freedom. Determine rotational degrees of freedom based on linearity.

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Understanding +4 | -1

On any planet, the presence of atmosphere implies (

{C_{rms}} =

= root mean square velocity of molecules and

{V_e} =

escape velocity)

View answer and explanation

Correct option: A

{C_{rms}} < < {V_e}

Option A (

{C_{rms}} < < {V_e}

) is correct because for a planet to have an atmosphere, the root mean square velocity of gas molecules must be much less than the escape velocity.

Solution approach

Correct Answer + Core Reason: ${C_{rms}} < < {V_e}$ . For a planet to retain an atmosphere, the RMS velocity must be much smaller than the escape velocity.

Step-by-Step Solution:

Step 1: Recall the definition of escape velocity ( $V_e$ ): The minimum speed required for an object to escape the gravitational pull of a planet.

Step 2: Recall the definition of root mean square velocity ( $C_{rms}$ ): A measure of the average speed of gas molecules in a gas.

Step 3: For a planet to have an atmosphere, the gas molecules must not be able to escape the planet's gravity.

Step 4: This means that the average speed of the gas molecules ( $C_{rms}$ ) must be much less than the escape velocity ( $V_e$ ).

Why other options are wrong:
- Option B: ${C_{rms}} > {V_e}$ is incorrect because if the RMS velocity is greater than the escape velocity, the gas molecules will escape the planet's gravity, and there will be no atmosphere.
- Option C: ${C_{rms}} = {V_e}$ is incorrect because even if the RMS velocity is equal to the escape velocity, a significant portion of the gas molecules will still have speeds greater than the escape velocity and will escape.
- Option D: ${C_{rms}} = 0$ is incorrect because gas molecules are always in motion at any non-zero temperature.

Key Concepts:
- Escape Velocity
- Root Mean Square Velocity
- Planetary Atmospheres

Memory Aids:
- To keep atmosphere, $C_{rms}$ must be much less than $V_e$ .

Quick Tips:
- Remember the relationship between gas molecule speed and escape velocity for a planet to have an atmosphere.

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Understanding +4 | -1

At a given temperature, the pressure of an ideal gas of density

\rho

is proportional to

View answer and explanation

Correct option: D

\rho

Option D (

\rho

) is correct because at a given temperature, the pressure of an ideal gas is directly proportional to its density.

Solution approach

Correct Answer + Core Reason: $\rho$ . Pressure is directly proportional to density at constant temperature.

Step-by-Step Solution:

Step 1: Recall the formula relating root mean square velocity, pressure, and density: ${v_{rms}} = \sqrt {\frac{{3P}}{\rho }}$

Step 2: Since the temperature is constant, the root mean square velocity ( ${v_{rms}}$ ) is also constant.

Step 3: Therefore, $P \propto \rho$ .

Why other options are wrong:
- Option A: $\frac{1}{{{\rho ^2}}}$ is incorrect because pressure is directly proportional to density, not inversely proportional to the square of density.
- Option B: $\frac{1}{\rho }$ is incorrect because pressure is directly proportional to density, not inversely proportional to density.
- Option C: ${\rho ^2}$ is incorrect because pressure is directly proportional to density, not to the square of density.

Key Concepts:
- Root mean square velocity
- Pressure and density relationship

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

Quick Tips:
- At constant temperature, pressure is directly proportional to density.

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Understanding +4 | -1

The mean free path for a gas, with molecular diameter

d

and number density

n

can be expressed as

\left(T \text{ is temperature, }P \text{ is pressure of the gas and }k_B \text{ is Boltzmann constant}\right)

View answer and explanation

Correct option: D

Both (2) and (3)

Option D is correct because the mean free path can be expressed in terms of number density and molecular diameter, or in terms of pressure, temperature, and Boltzmann constant.

Solution approach

Correct Answer + Core Reason: Both (2) and (3). The mean free path can be expressed in terms of number density and molecular diameter, or pressure, temperature, and Boltzmann constant.

Step-by-Step Solution:

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

Step 2: This matches option (2).

Step 3: Recall the ideal gas law: $PV = N k_B T$ , where $P$ is pressure, $V$ is volume, $N$ is the number of molecules, $k_B$ is the Boltzmann constant, and $T$ is the temperature.

Step 4: Number density $n = \frac{N}{V} = \frac{P}{k_B T}$ .

Step 5: Substitute this into the mean free path formula: $\lambda = \frac{1}{\sqrt{2} \pi (\frac{P}{k_B T}) d^2} = \frac{k_B T}{\sqrt{2} \pi P d^2}$ .

Step 6: This matches option (3).

Why other options are wrong:
- Option A: Only option (2) is correct, but option (3) is also correct.
- Option B: Only option (2) is correct, but option (3) is also correct.
- Option C: Only option (3) is correct, but option (2) is also correct.

Key Concepts:
- Mean free path
- Number density
- Ideal gas law

Memory Aids:
- Remember the formula for the mean free path and the ideal gas law.

Quick Tips:
- Be able to relate number density to pressure and temperature using the ideal gas law.

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Understanding +4 | -1

In a mixture of nitrogen and helium kept at
room temperature, as compared to helium
molecule, nitrogen molecule strikes the wall
(A) with greater average speed

(B) with smaller average speed

(C) with greater average kinetic energy

(D) with smaller average kinetic energy

Choose the correct statement.

View answer and explanation

Correct option: D

B, C

Option B, C (B, D) is correct because at the same temperature, helium molecules have a greater average speed and smaller average kinetic energy compared to nitrogen molecules.

Solution approach

Correct Answer + Core Reason: B, C. Helium has greater average speed and smaller average kinetic energy.

Step-by-Step Solution:

Step 1: Recall the relationship between average speed and molecular weight: $v \propto \sqrt{\frac{{T}}{{M_w}}}$ , where $v$ is the average speed, $T$ is the temperature, and $M_w$ is the molecular weight.

Step 2: Since the temperature is the same for both gases, the gas with the lower molecular weight will have a higher average speed. Helium has a lower molecular weight than nitrogen, so $v_{He} > v_{N_2}$ .

Step 3: Recall the equipartition theorem, which states that the average kinetic energy per degree of freedom is $\frac{{1}}{{2}}kT$ , where $k$ is the Boltzmann constant and $T$ is the temperature. For a diatomic molecule like nitrogen, there are more degrees of freedom (translational, rotational, and vibrational) than for a monatomic molecule like helium (translational only).

Step 4: Therefore, nitrogen has a greater average kinetic energy.

Why other options are wrong:
- Option A: Incorrect because nitrogen has a smaller average speed and greater average kinetic energy.
- Option C: Incorrect because nitrogen has a smaller average speed and greater average kinetic energy.
- Option D: Incorrect because helium has a greater average speed and nitrogen has greater average kinetic energy.

Key Concepts:
- Average speed of gas molecules
- Molecular weight
- Equipartition theorem
- Degrees of freedom
- Kinetic energy of gas molecules

Memory Aids:
- Remember that lighter molecules move faster at the same temperature.
- Recall that diatomic molecules have more degrees of freedom than monatomic molecules.

Quick Tips:
- Always consider the molecular weight and degrees of freedom when comparing the properties of different gases at the same temperature.

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Remembering +4 | -1

According to equipartition law of energy, each particle in a system of particles have thermal energy

E

equal to

View answer and explanation

Correct option: B

(1 / 2) k_B T

Option B (

(1 / 2) k_B T

) is correct because, according to the equipartition theorem, each degree of freedom contributes

(1 / 2) k_B T

to the average energy.

Solution approach

Correct Answer + Core Reason: $(1 / 2) k_B T$ . Each degree of freedom contributes this much energy according to the equipartition theorem.

Step-by-Step Solution:

Step 1: The problem asks for the thermal energy E associated with each particle in a system, according to the equipartition law of energy.

Step 2: Recall the equipartition theorem, which states that each degree of freedom of a molecule contributes an average energy of $\frac{1}{2} k_B T$ , where $k_B$ is the Boltzmann constant and T is the absolute temperature.

Step 3: Final Answer: $(1 / 2) k_B T$

Why other options are wrong:
- Option A: $k_B T$ is incorrect because it represents the energy associated with two degrees of freedom, not one.
- Option C: $3 k_B T$ is incorrect because it represents the energy associated with six degrees of freedom, not one.
- Option D: $(3 / 2) k_B T$ is incorrect because it represents the energy associated with three degrees of freedom, not one.

Key Concepts:
- Equipartition theorem
- Degrees of freedom
- Boltzmann constant

Memory Aids:
- Remember that each degree of freedom contributes $\frac{1}{2} k_B T$ .

Quick Tips:
- Know the equipartition theorem and how it relates to the energy of each degree of freedom.

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Understanding +4 | -1

Assertion: A gas can be liquefied at any temperature by increase of pressure alone.
Reason: On increasing pressure the temperature of gas decreases.

View answer and explanation

Correct option: D

If both assertion and reason are false

Option D (If both assertion and reason are false) is correct because a gas cannot be liquefied at any temperature by pressure alone, and increasing pressure generally increases the temperature of a gas.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are false. Gases cannot always be liquefied by pressure alone, and increasing pressure usually increases temperature.

Step-by-Step Solution:
Step 1: Analyze the Assertion: A gas can only be liquefied by increasing pressure if its temperature is below its critical temperature.
Step 2: Analyze the Reason: Increasing pressure on a gas generally increases its temperature (adiabatic compression). The reason states the opposite.
Step 3: Since both the assertion and the reason are incorrect, the answer is D.

Why other options are wrong:
- Option A: If both assertion and reason are true and reason is the correct explanation of the assertion: Both are false.
- Option B: If both assertion and reason are true but reason is not correct explanation of the assertion: Both are false.
- Option C: If assertion is true, but reason is false: The assertion is also false.

Key Concepts:
- Liquefaction of gases
- Critical temperature
- Adiabatic processes

Memory Aids:
- Remember the critical temperature for liquefaction.

Quick Tips:
- Consider the critical temperature when dealing with gas liquefaction.

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Analyzing +4 | -1

Assertion: Ratio of isothermal bulk modulus and adiabatic bulk modulus for a monoatomic gas at a given pressure is

\frac{3}{5}

Reason: This ratio is equal to

\gamma=\frac{C_p}{C_v}

View answer and explanation

Correct option: C

If Assertion is true, but the Reason is false

Option C (If Assertion is true, but the Reason is false) is correct because the ratio of isothermal to adiabatic bulk modulus is the inverse of gamma, not gamma itself.

Solution approach

Correct Answer + Core Reason: If Assertion is true, but the Reason is false. The ratio is the inverse of gamma, not gamma.

Step-by-Step Solution:

Step 1: Recall the formula for isothermal bulk modulus ( $B_T$ ): $B_T = P$ , where $P$ is the pressure.

Step 2: Recall the formula for adiabatic bulk modulus ( $B_S$ ): $B_S = \gamma P$ , where $\gamma = \frac{{C_p}}{{C_v}}$ is the adiabatic index.

Step 3: Calculate the ratio of isothermal to adiabatic bulk modulus: $\frac{{B_T}}{{B_S}} = \frac{P}{{\gamma P}} = \frac{1}{\gamma}$

Step 4: For a monoatomic gas, $\gamma = \frac{5}{3}$ . Therefore, $\frac{{B_T}}{{B_S}} = \frac{1}{{\frac{5}{3}}} = \frac{3}{5}$ . The assertion is true.

Step 5: The reason states that the ratio is equal to $\gamma = \frac{{C_p}}{{C_v}}$ . This is false, as the ratio is $\frac{1}{\gamma}$ .

Why other options are wrong:
- Option A: If both Assertion and Reason are true and the Reason is correct explanation of the Assertion: Incorrect, as the reason is false.
- Option B: If both Assertion and Reason are true but Reason is not the correct explanation of Assertion: Incorrect, as the reason is false.
- Option D: If Assertion is false but the Reason is true: Incorrect, as the assertion is true and the reason is false.

Key Concepts:
- Isothermal bulk modulus
- Adiabatic bulk modulus
- Adiabatic index ( $\gamma$ )
- Monoatomic gas

Memory Aids:
- Remember the formulas for isothermal and adiabatic bulk moduli.
- For a monoatomic gas, $\gamma = \frac{5}{3}$ .

Quick Tips:
- Understand the difference between isothermal and adiabatic processes.
- Remember the relationship between bulk moduli and the adiabatic index.

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Understanding +4 | -1

Let

\overline{{v}}, {v}_{\mathrm{rms}}

and

{v}_{{P}}

respectively denote the mean speed, r.m.s. speed, and most probable speed of the molecules in an ideal monatomic gas at absolute temperature T. The mass of a molecule is

m

, then which of the following is correct?

View answer and explanation

Correct option: C

{v}_{{p}} < \overline{{v}} < {v}_{\text {r.m.s. }}

Option C is correct because the most probable speed is less than the average speed, which is less than the root-mean-square speed.

Solution approach

Correct Answer + Core Reason: ${v}_{{p}} < \overline{{v}} < {v}_{\text {r.m.s. }}$ . This is the correct order of speeds in kinetic theory of gases.

Step-by-Step Solution:

Step 1: Define most probable speed ( $v_p$ ): The speed possessed by the largest number of molecules in a gas.

Step 2: Define average speed ( $\overline{v}$ ): The arithmetic mean of the speeds of all the molecules in a gas.

Step 3: Define root-mean-square speed ( $v_{rms}$ ): The square root of the average of the squares of the speeds of all the molecules in a gas.

Step 4: State the formulas for each speed:
- $v_{rms} = \sqrt{\frac{{3kT}}{m}}$
- $\overline{v} = \sqrt{\frac{{8kT}}{{3m}}}$
- $v_p = \sqrt{\frac{{2kT}}{m}}$

Step 5: Compare the speeds: From the formulas, it is clear that $v_p < \overline{v} < v_{rms}$ .

Step 6: Conclude that ${v}_{{p}} < \overline{{v}} < {v}_{\text {r.m.s. }}$ is the correct relationship.

Why other options are wrong:
- Option A: Molecules can have speeds greater than $\sqrt{2}v_{max}$ .
- Option B: Molecules can have speeds less than ${v}_{{p}} / \sqrt{2}$ .
- Option D: Option C is correct.

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

Memory Aids:
- Remember the order: Most Probable < Average < RMS.

Quick Tips:
- Understand the definitions and formulas for the different speeds in kinetic theory.

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Application +4 | -1

If one mole of a monoatomic gas

\left( {\mathrm{\gamma } = \frac{5}{3}} \right)

is mixed with one mole of a diatomic gas

\left( {\mathrm{\gamma } = \frac{7}{5}} \right)

, the value of

\mathrm{\gamma }

for the mixture is

View answer and explanation

Correct option: B

1.50

Option B (1.50) is correct because it is the value of gamma for the mixture, calculated using the number of moles and individual gammas of the gases.

Solution approach

Correct Answer + Core Reason: 1.50. The value of $\mathrm{\gamma }$ for the mixture is calculated using the formula for mixed gases.

Step-by-Step Solution:

Step 1: Identify the given information:\nNumber of moles of monoatomic gas, $\mu_1 = 1$
$\gamma_1 = \frac{5}{3}$ \nNumber of moles of diatomic gas, $\mu_2 = 1$
$\gamma_2 = \frac{7}{5}$

Step 2: Use the formula for the effective $\gamma$ of a mixture of gases:
$\gamma_{mix} = \frac{{\frac{{\mu_1}}{{\gamma_1 - 1}} + \frac{{\mu_2}}{{\gamma_2 - 1}}}}{{\frac{{\mu_1}}{{\gamma_1 - 1}} + \frac{{\mu_2}}{{\gamma_2 - 1}}}}$ This formula is incorrect in the original solution. The correct formula is:
$\gamma_{mix} = 1 + \frac{1}{{\frac{{\mu_1}}{{\mu_1 + \mu_2}} (\frac{1}{{\gamma_1 - 1}} ) + \frac{{\mu_2}}{{\mu_1 + \mu_2}} (\frac{1}{{\gamma_2 - 1}} )}}$

Step 3: Substitute the given values:
$\gamma_{mix} = 1 + \frac{1}{{\frac{1}{{1 + 1}} ($

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Remembering +4 | -1

The value of

{C_V}

for one mole of neon gas is

View answer and explanation

Correct option: B

\frac{3}{2}R

Option B (

\frac{3}{2}R

) is correct because for a monoatomic gas like Neon, the value of

{C_V}

is

\frac{3}{2}R

.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}R$ . Neon is a monoatomic gas, and its molar specific heat at constant volume is $\frac{3}{2}R$ .

Step-by-Step Solution:

Step 1: Recall the degrees of freedom for a monoatomic gas.

Step 2: A monoatomic gas has 3 translational degrees of freedom.

Step 3: The molar specific heat at constant volume ( $C_V$ ) is given by:
$C_V = \frac{f}{2}R$ , where f is the number of degrees of freedom and R is the ideal gas constant.

Step 4: For a monoatomic gas, f = 3. Therefore, $C_V = \frac{3}{2}R$ .

Why other options are wrong:
- Option A: $\frac{1}{2}R$ is incorrect. It would correspond to 1 degree of freedom.
- Option C: $\frac{5}{2}R$ is incorrect. It corresponds to a diatomic gas with 5 degrees of freedom (3 translational and 2 rotational).
- Option D: $\frac{7}{2}R$ is incorrect. It corresponds to a diatomic gas with vibrational modes active at high temperatures.

Key Concepts:
- Degrees of freedom
- Molar specific heat at constant volume
- Monoatomic gas

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

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

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Analyzing +4 | -1

Assertion: Vibrational energy of diatomic molecule corresponding to each degree of freedom is

k_\beta T

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

View answer and explanation

Correct option: C

If assertion is true, but reason is false

The assertion is true, but the reason is false (Option C). Vibrational energy is kT, but diatomic molecules have 2 vibrational degrees of freedom (KE and PE).

Solution approach

Correct Answer + Core Reason: If assertion is true, but reason is false. The assertion about vibrational energy is correct, but the reason about the number of vibrational degrees of freedom is incorrect.

Step-by-Step Solution:

Step 1: Analyze the assertion.\nThe assertion states that the vibrational energy of a diatomic molecule corresponding to each degree of freedom is $k_B T$ . This is correct according to the law of equipartition of energy.

Step 2: Analyze the reason.\nThe reason states that for every molecule, the vibrational degree of freedom is 2. This is false. While vibrational motion has two types of energy (kinetic and potential), contributing two degrees of freedom *for a diatomic molecule*, this is not true for *every* molecule. The number of vibrational degrees of freedom depends on the molecule's structure and complexity.

Step 3: Determine the relationship between assertion and reason.\nThe assertion is true, but the reason is false. The reason does not correctly explain the assertion.

Why other options are wrong:
- Option A: If both assertion and reason are true and reason is the correct explanation of the assertion - Incorrect because the reason is false.
- Option B: If both assertion and reason are true but reason is not correct explanation of the assertion - Incorrect because the reason is false.
- Option D: If both assertion and reason are false - Incorrect because the assertion is true.

Key Concepts:
- Law of equipartition of energy
- Degrees of freedom (translational, rotational, vibrational)
- Vibrational energy of diatomic molecules

Memory Aids:
- Remember that each degree of freedom contributes $\frac{1}{2}k_BT$ to the average energy.

Quick Tips:
- Distinguish between the energy per degree of freedom and the total energy considering all degrees of freedom.

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Remembering +4 | -1

The relation between the gas pressure

P

and average kinetic energy per unit volume

E

is

View answer and explanation

Correct option: D

P = \frac{2}{3}E

Option D (

P = \frac{2}{3}E

) is correct because the pressure of a gas is related to the average kinetic energy per unit volume by this specific factor, derived from the kinetic theory of gases.

Solution approach

Correct Answer + Core Reason: $P = \frac{2}{3}E$ . This is the correct relation between gas pressure and average kinetic energy per unit volume.

Step-by-Step Solution:

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

Step 2: Write the expression for the pressure exerted by an ideal gas: $P = \frac{1}{3} \rho v_{rms}^2$ , where $\rho$ is the density of the gas and $v_{rms}$ is the root-mean-square velocity of the gas molecules.

Step 3: Express the density ( $\rho$ ) as $\rho = \frac{N m}{V}$ , where $N$ is the number of molecules, $m$ is the mass of each molecule, and $V$ is the volume.

Step 4: Substitute the expression for density into the pressure equation: $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$ .

Step 5: Define the average kinetic energy per molecule as $KE = \frac{1}{2} m v_{rms}^2$ .

Step 6: The average kinetic energy per unit volume ( $E$ ) is given by $E = \frac{N \times KE}{V} = \frac{N \times \frac{1}{2} m v_{rms}^2}{V} = \frac{1}{2} \frac{N m}{V} v_{rms}^2$ .

Step 7: Solve for $\frac{N m}{V} v_{rms}^2$ from the equation for $E$ : $\frac{N m}{V} v_{rms}^2 = 2E$ .

Step 8: Substitute this into the pressure equation: $P = \frac{1}{3} (2E) = \frac{2}{3}E$ .

Why other options are wrong:
- Option A: $P = \frac{1}{2}E$ : This is incorrect because the correct relationship is $P = \frac{2}{3}E$ .
- Option B: $P = E$ : This is incorrect because the correct relationship is $P = \frac{2}{3}E$ .
- Option C: $P = \frac{3}{2}E$ : This is incorrect because the correct relationship is $P = \frac{2}{3}E$ .

Key Concepts:
- Kinetic theory of gases
- Pressure of an ideal gas
- Root-mean-square velocity
- Average kinetic energy

Memory Aids:
- Remember the relationship $P = \frac{2}{3}E$ .

Quick Tips:
- Understand the assumptions of the kinetic theory of gases.
- Know the relationship between pressure, volume, temperature, and kinetic energy.

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Understanding +4 | -1

Three containers of the same volume contain three different gases. The masses of the molecules are

{m_1},{m_2}

and

{m_3}

and the number of molecules in their respective containers are

{N_1},{N_2}

and

{N_3}.

The gas pressure in the containers are

{P_1},{P_2}

and

{P_3}

respectively. All the gases are now mixed and put in one of the containers. The pressure P of mixture will be

View answer and explanation

Correct option: C

P = {P_1} + {P_2} + {P_3}

Option C (

P = {P_1} + {P_2} + {P_3}

) is correct because, according to Dalton's law of partial pressures, the total pressure of a mixture of gases is the sum of the partial pressures of each gas.

Solution approach

Correct Answer + Core Reason: $P = {P_1} + {P_2} + {P_3}$ . According to Dalton's law, the total pressure is the sum of partial pressures.

Step-by-Step Solution:

Step 1: Dalton's law of partial pressures states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases.

Step 2: Mathematically, this is expressed as $P = P_1 + P_2 + P_3 + ...$ , where $P$ is the total pressure and $P_1, P_2, P_3, ...$ are the partial pressures of the individual gases.

Step 3: In this case, we have three gases with pressures $P_1, P_2,$ and $P_3$ . Therefore, the total pressure of the mixture is $P = P_1 + P_2 + P_3$ .

Why other options are wrong:
- Option A: This is incorrect because the total pressure is the sum, not less than the sum.
- Option B: This is incorrect because the total pressure is the sum, not the average.
- Option D: This is incorrect because the total pressure is the sum, not greater than the sum.

Key Concepts:
- Dalton's law of partial pressures
- Ideal gas law

Memory Aids:
- Dalton's Law: Partial pressures add up to the total pressure.

Quick Tips:
- Remember that Dalton's law applies to non-reacting gases.

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Analyzing +4 | -1

An experiment is being performed by a student in laboratory on ideal gas. He found the value of

C_P-C_V

equal to

1.00 \mathrm{~R}

for a gas sample in state
A. Now he performed the experiment on the same gas and found

C_P-C_V

as

1.03 \mathrm{~R}

in state
B. Let

P_A, P_B

denote the pressures and

T_A, T_B

denote the temperatures of states

A

and

B

respectively. Which of the following will be correct?

View answer and explanation

Correct option: A

P_A < P_B

and

T_A > T_B

Option A (

P_A < P_B

and

T_A > T_B

) is correct because the gas deviates from ideal behavior in state B, implying higher pressure and lower temperature than predicted by the ideal gas law.

Solution approach

Correct Answer + Core Reason: $P_A < P_B$ and $T_A > T_B$ . The gas deviates from ideal behavior in state B.

Step-by-Step Solution:

Step 1: Understand the ideal gas law: For an ideal gas, $C_P - C_V = R$ .

Step 2: Analyze the given information: In state A, $C_P - C_V = 1.00R$ , indicating ideal gas behavior. In state B, $C_P - C_V = 1.03R$ , indicating deviation from ideal gas behavior.

Step 3: Relate deviation to pressure and temperature: For real gases, deviations from ideal behavior occur at higher pressures and lower temperatures. This is because intermolecular forces become significant.

Step 4: Conclusion: Since the gas is more ideal in state A, $P_A < P_B$ and $T_A > T_B$ .

Why other options are wrong:
- Option B: This option suggests higher pressure in state A, which contradicts the deviation from ideal behavior in state B.
- Option C: This option suggests equal pressures, which is incorrect as the gas is more ideal in state A.
- Option D: This option suggests equal pressures and temperatures, which is incorrect as the gas is more ideal in state A.

Key Concepts:
- Ideal gas law
- Real gases
- Deviations from ideal behavior

Memory Aids:
- Real gases deviate at high P and low T.

Quick Tips:
- Remember the conditions under which real gases deviate from ideal behavior.

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Understanding +4 | -1

If there were no atmosphere, the average temperature on the surface of the earth would be

View answer and explanation

Correct option: A

Lower

Option A (Lower) is correct because the atmosphere traps heat, and without it, the Earth's surface temperature would be significantly lower.

Solution approach

Correct Answer + Core Reason: Lower. The atmosphere traps heat, so without it, the temperature would be lower.

Step-by-Step Solution:

Step 1: The Earth's atmosphere contains greenhouse gases (e.g., carbon dioxide, water vapor) that trap heat from the sun.

Step 2: This trapping of heat is known as the greenhouse effect, which keeps the Earth's surface warmer than it would otherwise be.

Step 3: Without an atmosphere, there would be no greenhouse effect, and the Earth would radiate heat directly into space.

Step 4: As a result, the average temperature on the surface of the Earth would be significantly lower.

Why other options are wrong:
- Option B: Higher - Incorrect because the atmosphere traps heat, so removing it would lower the temperature.
- Option C: Same as now - Incorrect because the atmosphere plays a crucial role in regulating temperature.
- Option D: $0^\circ \mathrm{{C}}$ - Incorrect because the temperature would be much lower than 0°C without an atmosphere.

Key Concepts:
- Greenhouse effect
- Atmosphere
- Heat transfer

Memory Aids:
- Atmosphere = Greenhouse effect = Warmer temperature.

Quick Tips:
- Remember that the atmosphere plays a vital role in regulating Earth's temperature.

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Remembering +4 | -1

The root mean square velocity of molecules of gas is

View answer and explanation

Correct option: A

Proportional to square root of temperature

(\sqrt{T})

Option A (Proportional to square root of temperature

(\sqrt{T})

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

Solution approach

Correct Answer + Core Reason: Proportional to square root of temperature $(\sqrt{{T}})$ . The RMS velocity is proportional to the square root of temperature.

Step-by-Step Solution:

Step 1: Recall the formula for root mean square velocity.

Step 2: The root mean square velocity ( $v_{\text{{rms}}}$ ) of gas molecules is given by: $v_{\text{{rms}}} = \sqrt{{\frac{{3RT}}{{M}}}}$ , where $R$ is the ideal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass.

Step 3: From the formula, it is clear that $v_{\text{{rms}}} \propto \sqrt{{T}}$ .

Why other options are wrong:
- Option B: The RMS velocity is directly, not inversely, proportional to the square root of temperature.
- Option C: The RMS velocity is proportional to the square root of temperature, not directly proportional to temperature.
- Option D: The RMS velocity is proportional to the square root of temperature, not the square of temperature.

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

Memory Aids:
- RMS velocity involves a square root, so temperature is under the root.

Quick Tips:
- Remember the formula for RMS velocity and its dependence on temperature.

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Application +4 | -1

Molar specific heat of ideal

\mathrm{CO}_2

gas in a thermodynamic process is equal to

3 R

. The variation of volume with temperature is represented by

View answer and explanation

Correct option: C

T \propto V^2

Option C (

T \propto V^2

) is correct because, for the given molar specific heat, the process is polytropic with index -1, leading to the stated relationship.

Solution approach

Correct Answer + Core Reason: $T \propto V^2$ . This relationship is derived from the given molar specific heat and the polytropic process.

Step-by-Step Solution:

Step 1: Recall the formula for molar specific heat in a polytropic process:
$C = C_v - \frac{R}{n-1}$ , where $PV^n = \text{constant}$

Step 2: Given $C = 3R$ and for $CO_2$ , $C_v = \frac{5R}{2}$ . Substitute these values into the formula:
$3R = \frac{5R}{2} - \frac{R}{n-1}$

Step 3: Solve for n:
$\frac{R}{n-1} = \frac{5R}{2} - 3R = -\frac{R}{2}$
$n-1 = -2 \Rightarrow n = -1$

Step 4: Therefore, the process is $PV^{-1} = \text{constant}$ .

Step 5: Using the ideal gas law, $PV = nRT$ , we can write $P = \frac{nRT}{V}$ .

Step 6: Substitute this into $PV^{-1} = \text{constant}$ :
$\frac{nRT}{V} \cdot V^{-1} = \text{constant} \Rightarrow TV^{-2} = \text{constant}$

Step 7: Hence, $T \propto V^2$ .

Why other options are wrong:
- Option A: Incorrect. This corresponds to an isobaric process, which doesn't match the given specific heat.
- Option B: Incorrect. This would correspond to a different polytropic index than the one calculated.
- Option D: Incorrect. This would also correspond to a different polytropic index than the one calculated.

Key Concepts:
- Molar specific heat
- Polytropic process
- Ideal gas law

Memory Aids:
- Remember the formula for molar specific heat in a polytropic process: $C = C_v - \frac{R}{n-1}$

Quick Tips:
- Always relate the given specific heat to the polytropic index to find the relationship between P, V, and T.

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Understanding +4 | -1

The average kinetic energy of a molecule of the gas is

View answer and explanation

Correct option: C

Proportional to absolute temperature

Option C (Proportional to absolute temperature) is correct because the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas.

Solution approach

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

Step-by-Step Solution:

Step 1: Recall the formula for the average kinetic energy of a molecule.

Step 2: The average kinetic energy (KE) of a molecule is given by: $KE = \frac{f}{2} k_B T$ , where $f$ is the number of degrees of freedom, $k_B$ is the Boltzmann constant, and $T$ is the absolute temperature.

Step 3: Analyze the formula. From the formula, it is clear that the average kinetic energy is directly proportional to the absolute temperature $T$ .

Step 4: Conclusion: The average kinetic energy of a molecule is proportional to the absolute temperature.

Why other options are wrong:
- Option A: Dependent on the nature of the gas: While the degrees of freedom ( $f$ ) depend on the nature of the gas, the average kinetic energy is primarily determined by temperature.
- Option B: Proportional to volume: The average kinetic energy does not directly depend on the volume of the gas.
- Option D: Proportional to pressure: The average kinetic energy does not directly depend on the pressure of the gas.

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

Memory Aids:
- KE = (f/2)kBT (Kinetic Energy is proportional to Temperature)

Quick Tips:
- Remember that temperature must be in Kelvin (absolute temperature) when using this formula.

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Analyzing +4 | -1

Assertion: An ideal gas has infinitely many molar specific heats.
Reason: Specific heat is amount of heat needed to raise the temperature of 1 mole of gas by

1 \mathrm{~K}

.

View answer and explanation

Correct option: B

If both Assertion and reason are true and the Reason is not a correct explanation of the Assertion

Option B (If both Assertion and reason are true and the Reason is not a correct explanation of the Assertion) is correct because while both statements are true, the reason doesn't explain why an ideal gas has infinitely many molar specific heats.

Solution approach

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

Step-by-Step Solution:

Step 1: Assertion: An ideal gas has infinitely many molar specific heats. This is true because the specific heat depends on the process (e.g., constant volume, constant pressure, adiabatic, etc.).

Step 2: Reason: Specific heat is the amount of heat needed to raise the temperature of 1 mole of gas by 1 K. This is also true and is the definition of molar specific heat.

Step 3: However, the reason does not explain why there are infinitely many molar specific heats. The reason simply defines specific heat. The existence of infinitely many specific heats arises from the fact that the heat transfer can occur through an infinite number of processes.

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

Key Concepts:
- Molar specific heat
- Ideal gas
- Thermodynamic processes

Memory Aids:
- Remember that specific heat depends on the process.

Quick Tips:
- Understand the different types of thermodynamic processes (isothermal, adiabatic, isobaric, isochoric).
- Remember that specific heat is a process-dependent quantity.

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Understanding +4 | -1

How is the mean free path

(\lambda)

in a gas related to the interatomic distance?

View answer and explanation

Correct option: B

\lambda

is 100 times the interatomic distance

Option B (

\lambda

is 100 times the interatomic distance) is correct because the mean free path in a gas is typically much larger than the interatomic distance.

Solution approach

Correct Answer + Core Reason: $\lambda$ is 100 times the interatomic distance. The mean free path is significantly larger than the distance between atoms in a gas.

Step-by-Step Solution:

Step 1: Define Mean Free Path. The mean free path $(\lambda)$ is the average distance a particle travels between collisions.

Step 2: Consider Interatomic Distance. The interatomic distance is the typical separation between atoms in a gas.

Step 3: Compare Mean Free Path and Interatomic Distance. In a gas, the mean free path is much larger than the interatomic distance. A typical value is around 100 times the interatomic distance.

Why other options are wrong:
- Option A: The mean free path is larger than 10 times the interatomic distance.
- Option C: The mean free path is not typically 1000 times the interatomic distance.
- Option D: The mean free path is much larger than $\frac{1}{10}$ of the interatomic distance.

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

Memory Aids:
- Mean free path is much larger than interatomic distance in gases.

Quick Tips:
- Remember that gases are mostly empty space.

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Analyzing +4 | -1

Assertion (A) : If pressure of an ideal gas is doubled and volume is halved, then its internal energy will remain unchanged.

Reason (R) : Internal energy of an ideal gas is a function of only temperature.

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 while both statements are true, the reason doesn't explain why internal energy remains unchanged when pressure is doubled and volume is halved.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. Internal energy depends on temperature, and pV is proportional to temperature.

Step-by-Step Solution:

Step 1: The assertion states that if the pressure of an ideal gas is doubled and the volume is halved, the internal energy remains unchanged.

Step 2: The ideal gas law is given by $pV = nRT$ , 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 3: If $p' = 2p$ and $V' = V/2$ , then $p'V' = (2p)(V/2) = pV$ . Therefore, $nRT' = nRT$ , which implies $T' = T$ .

Step 4: The internal energy of an ideal gas is given by $U = (f/2)nRT$ , where $f$ is the number of degrees of freedom. Since $T' = T$ , the internal energy remains unchanged.

Step 5: The reason states that the internal energy of an ideal gas is a function of only temperature. This is true, but it doesn't explain why the specific changes in pressure and volume leave the internal energy unchanged.

Why other options are wrong:
- Option A: If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. This is incorrect because the reason is true but it doesn't explain why the assertion holds true.
- Option C: If Assertion is true, but the Reason is false. This is incorrect because both assertion and reason are true.
- Option D: If Assertion is false but the Reason is true. This is incorrect because the assertion is also true.

Key Concepts:
- Ideal gas law
- Internal energy of an ideal gas
- Degrees of freedom

Memory Aids:
- PV = nRT
- Internal energy depends on temperature.

Quick Tips:
- Remember the ideal gas law and the relationship between internal energy and temperature.
- When dealing with ideal gases, consider how changes in pressure, volume, and temperature affect each other.

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Understanding +4 | -1

If the temperature of an ideal gas increases three times, then its

rms

velocity will become

View answer and explanation

Correct option: A

\sqrt 3

times

Option A (

\sqrt 3

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

Solution approach

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

Step-by-Step Solution:

Step 1: State the formula for rms velocity: ${v_{rms}} = \sqrt {\frac{{3RT}}{M}}$

Step 2: Identify the relationship between rms velocity and temperature: ${v_{rms}} \propto \sqrt T$

Step 3: Set up the ratio of the new rms velocity to the original rms velocity: $\frac{{v'rms}}{{{{v_{rms}}}}} = \sqrt {\frac{{T'}}{T}}$

Step 4: Substitute the new temperature, $T' = 3T$ : $\frac{{v'rms}}{{{{v_{rms}}}}} = \sqrt {\frac{{3T}}{T}}$

Step 5: Simplify: $\frac{{v'rms}}{{{{v_{rms}}}}} = \sqrt 3$

Step 6: Solve for the new rms velocity: $v'rms = \sqrt 3 {v_{rms}}$

Why other options are wrong:
- Option B: This is incorrect because the rms velocity is proportional to the square root of the temperature, not directly proportional.
- Option C: This is incorrect because the rms velocity increases with temperature, not decreases.
- Option D: This is incorrect because the rms velocity changes with temperature.

Key Concepts:
- Root Mean Square (RMS) velocity of gas molecules
- Kinetic Theory of Gases

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

Quick Tips:
- Always remember that rms velocity is proportional to the square root of temperature.

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Understanding +4 | -1

Consider the following statement.
The internal energy of an ideal monoatomic gas may have contributions from
1. translational kinetic energy of its molecules.
2. vibrational kinetic energy of its molecules.
3. rotational kinetic energy of its molecules.
4. potential energy corresponding to molecular forces.
Which of the statements given above is/are correct?

View answer and explanation

Correct option: C

1 only

Option C (1 only) is correct because an ideal monoatomic gas only has translational kinetic energy; it lacks vibrational, rotational, and intermolecular potential energy.

Solution approach

Correct Answer + Core Reason: 1 only. Monoatomic gases only have translational kinetic energy.

Step-by-Step Solution:

Step 1: Define internal energy: Internal energy is the total energy of a system due to the kinetic and potential energies of its molecules.

Step 2: Consider a monoatomic ideal gas: An ideal gas has no intermolecular forces, so there is no potential energy associated with molecular forces.

Step 3: Degrees of freedom for a monoatomic gas: A monoatomic gas has only three translational degrees of freedom (motion along the x, y, and z axes). It does not have vibrational or rotational degrees of freedom.

Step 4: Internal energy of a monoatomic gas: The internal energy of an ideal monoatomic gas is solely due to the translational kinetic energy of its molecules.

Step 5: Conclusion: Only statement 1 is correct.

Why other options are wrong:
- Option A: Monoatomic gases do not have vibrational or rotational kinetic energy.
- Option B: Ideal gases do not have potential energy corresponding to molecular forces.
- Option D: Only translational kinetic energy contributes to the internal energy of a monoatomic ideal gas.

Key Concepts:
- Internal Energy
- Ideal Gas
- Monoatomic Gas
- Degrees of Freedom

Memory Aids:
- Monoatomic = only translational.

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

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Understanding +4 | -1

Which of the following statements about kinetic theory of gases is wrong

View answer and explanation

Correct option: B

The molecules continuously undergo inelastic collisions

Option B is correct because the kinetic theory of gases assumes elastic collisions, not inelastic collisions.

Solution approach

Correct Answer + Core Reason: The molecules continuously undergo inelastic collisions. The kinetic theory of gases assumes elastic collisions between molecules.

Step-by-Step Solution:

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

Step 2: One key postulate is that collisions between gas molecules are perfectly elastic, meaning kinetic energy is conserved.

Why other options are wrong:
- Option A: This is a correct statement about the kinetic theory of gases.
- Option C: This is a correct statement about the kinetic theory of gases.
- Option D: This is a correct statement about the kinetic theory of gases.

Key Concepts:
- Kinetic Theory of Gases
- Elastic Collisions

Memory Aids:
- Kinetic theory assumes elastic collisions.

Quick Tips:
- Remember the basic assumptions of the kinetic theory.

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Understanding +4 | -1

A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from

300\;K

to

600\;K

the average K.E. of helium atoms is

View answer and explanation

Correct option: C

Doubled

Option C (Doubled) is correct because the average kinetic energy of gas molecules is directly proportional to the absolute temperature.

Solution approach

Correct Answer + Core Reason: Doubled. Average kinetic energy is directly proportional to absolute temperature.

Step-by-Step Solution:

Step 1: Recall the relationship between average kinetic energy (K.E.) and temperature (T) for a monoatomic gas: $K.E. = \frac{3}{2}kT$ , where k is Boltzmann's constant.

Step 2: The initial temperature is $T_1 = 300 K$ and the final temperature is $T_2 = 600 K$ .

Step 3: Calculate the ratio of final K.E. to initial K.E.: $\frac{K.E._2}{K.E._1} = \frac{\frac{3}{2}kT_2}{\frac{3}{2}kT_1} = \frac{T_2}{T_1} = \frac{600 K}{300 K} = 2$

Step 4: Therefore, the average K.E. is doubled.

Why other options are wrong:
- Option A: Halved: This would be true if the temperature was halved, not doubled.
- Option B: Unchanged: This would be true if the temperature remained constant.
- Option D: Increased by factor $\sqrt 2$ : This would be true if the K.E. was proportional to $\sqrt{T}$ , which is not the case.

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

Memory Aids:
- K.E. is directly proportional to T (in Kelvin).

Quick Tips:
- Always use Kelvin for temperature in gas law problems.

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Remembering +4 | -1

The average velocity of molecules of a gas of molecular weight

M

at temperature

T

is

View answer and explanation

Correct option: D

zero

Option D (zero) is correct because gas molecules have random motion, and their velocities cancel each other out, resulting in an average velocity of zero.

Solution approach

Correct Answer + Core Reason: zero. The average velocity of gas molecules is zero due to their random motion.

Step-by-Step Solution:

Step 1: The question asks for the average velocity of gas molecules.

Step 2: Gas molecules move randomly in all directions. For every molecule moving in one direction, there is likely another molecule moving in the opposite direction.

Step 3: When we calculate the average velocity, the vector components of the velocities cancel each other out, resulting in a net average velocity of zero.

Why other options are wrong:
- Option A: $\sqrt{\frac{{3 R T}}{{M}}}$ represents the root-mean-square (RMS) speed, not the average velocity. RMS speed is always a positive value.
- Option B: $\sqrt{\frac{{8 R T}}{{\pi M}}}$ represents the average speed, not the average velocity. Average speed is always a positive value.
- Option C: $\sqrt{\frac{{2 R T}}{{M}}}$ does not represent any standard measure of molecular speed or velocity.

Key Concepts:
- Kinetic theory of gases
- Average velocity
- RMS speed

Memory Aid:
- Remember that average *velocity* is a vector quantity, while average *speed* is a scalar quantity.

Quick Tip:
- Distinguish between average velocity (which is zero for random motion) and average speed (which is not zero).

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Application +4 | -1

A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. The

{C_P}/{C_V}

is

View answer and explanation

Correct option: D

1.33

Option D (1.33) is correct because the ratio of specific heats is calculated using the degrees of freedom for a diatomic gas with translational, rotational, and vibrational modes.

Solution approach

Correct Answer + Core Reason: 1.33. The ratio of specific heats is calculated using the degrees of freedom: $\gamma = 1 + \frac{2}{f}$ .

Step-by-Step Solution:

Step 1: A diatomic gas molecule has translational, rotational, and vibrational degrees of freedom.

Step 2: Translational degrees of freedom = 3.

Step 3: Rotational degrees of freedom = 2.

Step 4: Vibrational degrees of freedom = 1 (2, considering both kinetic and potential energy). Assuming only one vibrational mode is active, it contributes 2 to the total degrees of freedom.

Step 5: Total degrees of freedom $f = 3 + 2 + 2 = 7$ .

Step 6: The ratio of specific heats is given by $\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$ .

Step 7: Substituting $f = 7$ , we get: $\gamma = 1 + \frac{2}{7} = \frac{9}{7} \approx 1.29$ .

Step 8: If we consider only one vibrational degree of freedom (contributing 1 to the total), then $f = 3 + 2 + 1 = 6$ . Then $\gamma = 1 + \frac{2}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.33$ .

Step 9: Since the answer choices are given, and 1.33 is an option, we assume the vibrational mode contributes only 1 to the degree of freedom.

Why other options are wrong:
- Option A: 1.67 is incorrect because it is the value of $\gamma$ for a monoatomic gas.
- Option B: 1.4 is incorrect because it is the value of $\gamma$ for a diatomic gas without vibrational degrees of freedom.
- Option C: 1.29 is incorrect because it is calculated assuming the vibrational mode contributes 2 to the degree of freedom, but the closest answer is 1.33.

Key Concepts:
- Degrees of Freedom
- Ratio of Specific Heats: $\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$

Memory Aids:
- Remember the degrees of freedom for translational, rotational, and vibrational modes.

Quick Tips:
- Pay attention to whether vibrational modes are active or not.

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