KINETIC THEORY OF GASES MCQs for JEE PHYSICS — Medium, Page 5

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Remembering +4 | -1

The energy of an ideal gas molecule associated with each degree of freedom according to Law of equipartition of energy is

View answer and explanation

Correct option: C

\frac{1}{2} k_B T

Option C (

\frac{{1}}{{2}} k_B T

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

\frac{{1}}{{2}} k_B T

to its average energy.

Solution approach

Correct Answer + Core Reason: $\frac{{1}}{{2}} k_B T$ . According to the law of equipartition of energy, each degree of freedom contributes $\frac{{1}}{{2}} k_B T$ to the average energy of a molecule.

Step-by-Step Solution:

Step 1: State the law of equipartition of energy.\nThe law of equipartition of energy states that, in thermal equilibrium, the average energy associated with each degree of freedom of a molecule is $\frac{{1}}{{2}} k_B T$ , where $k_B$ is Boltzmann's constant and $T$ is the absolute temperature.

Step 2: Apply the law to a single degree of freedom.\nFor each degree of freedom, the energy is $\frac{{1}}{{2}} k_B T$ .

Step 3: Conclusion.\nTherefore, the energy of an ideal gas molecule associated with each degree of freedom is $\frac{{1}}{{2}} k_B T$ .

Why other options are wrong:
- Option A: This is the total average kinetic energy for a monatomic gas (3 degrees of freedom).
- Option B: This is the energy for two degrees of freedom.
- Option D: This is not the correct energy per degree of freedom.

Key Concepts:
- Law of equipartition of energy
- Degree of freedom
- Boltzmann's constant
- Temperature

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

Quick Tips:
- The law of equipartition of energy is a fundamental concept in statistical mechanics.

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The mean free path of molecules of an ideal gas of diameter

d

is proportional to

View answer and explanation

Correct option: D

d^{-2}

Option D (

d^{-2}

) is correct because the mean free path is inversely proportional to the square of the diameter of the gas molecules.

Solution approach

Correct Answer + Core Reason: $d^{-2}$ . The mean free path is inversely proportional to the square of the diameter.

Step-by-Step Solution:

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

Step 2: Identify the relationship between the mean free path and the diameter: From the formula, it is clear that $\lambda \propto \frac{1}{d^2}$ .

Step 3: Express the proportionality in terms of $d$ : $\lambda \propto d^{-2}$ .

Why other options are wrong:
- Option A: $d$ : Incorrect because the mean free path is inversely proportional to $d^2$ .
- Option B: $d^{-1}$ : Incorrect because the mean free path is inversely proportional to $d^2$ .
- Option C: $d^0$ : Incorrect because the mean free path depends on $d$ .

Key Concepts:
- Mean free path
- Ideal gas

Memory Aids:
- Remember the formula for the mean free path: $\lambda = \frac{1}{\sqrt{2} \pi n d^2}$ .

Quick Tips:
- The mean free path is the average distance a molecule travels between collisions.
- It depends on the number density and the size of the molecules.

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Application +4 | -1

One mole of a gas expands with temperature

T

such that its volume

V=k T^2

, where

k

is a constant. If the temperature of the gas changes by

60^{\circ} \mathrm{C}

, then the work done by the gas is

View answer and explanation

Correct option: A

120 \mathrm{R}

Option A (120 R) is correct because the work done is calculated by integrating pdV, using the given relation between V and T.

Solution approach

Correct Answer + Core Reason: 120 R. The work done is calculated by integrating pdV.

Step-by-Step Solution:

Step 1: Given $V = kT^2$ , find $dV$ : $dV = 2kT dT$

Step 2: Use the ideal gas law to express pressure: $pV = RT \implies p = \frac{RT}{V} = \frac{RT}{kT^2} = \frac{R}{kT}$

Step 3: Calculate the work done: $W = \int p dV = \int \frac{R}{kT} (2kT dT) = \int 2R dT = 2R \int dT$

Step 4: Integrate over the temperature change of 60°C: $W = 2R \Delta T = 2R (60) = 120R$

Why other options are wrong:
- Option B (R ln 60): This is incorrect because it uses a logarithmic function, which is not applicable in this scenario.
- Option C (kR ln 60): This is incorrect because it includes k and a logarithmic function, which are not applicable in this scenario.
- Option D (60 kR): This is incorrect because it includes k but misses the factor of 2 in the integration.

Key Concepts:
- Work done by a gas
- Ideal gas law
- Integration

Memory Aids:
- Work done is the integral of pressure with respect to volume.

Quick Tips:
- Remember the ideal gas law: $pV = RT$
- Pay attention to the given relationships between variables.

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The average kinetic energy, of

\mathrm{O}_2

at a particular temperature is

0.768 \mathrm{~eV}

. The average kinetic energy of

\mathrm{N}_2

molecules in

\mathrm{eV}

at the same temperature is:

View answer and explanation

Correct option: D

0.768

Option D (

0

) is correct because at resonance in an LCR series circuit, the inductive and capacitive reactances cancel each other out, resulting in a phase difference of zero between voltage and current.

Solution approach

Correct Answer + Core Reason: $0$ . At resonance, $\omega \mathrm{L}=\frac{{1}}{\omega \mathrm{C}}$ . The circuit behaves as if it contains R only.

Step-by-Step Solution:
Step 1: In an LCR series circuit, the impedance is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$ , where $X_L = \omega L$ is the inductive reactance and $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
Step 2: At resonance, the inductive reactance equals the capacitive reactance: $X_L = X_C$ , which means $\omega L = \frac{1}{\omega C}$ .
Step 3: Therefore, the impedance at resonance is $Z = \sqrt{R^2 + (0)^2} = R$ .
Step 4: The phase difference $\phi$ between the voltage and current is given by $\tan \phi = \frac{X_L - X_C}{R}$ .
Step 5: At resonance, $X_L = X_C$ , so $\tan \phi = 0$ , which means $\phi = 0$ .

Why other options are wrong:
- Option A: $\pi$ is incorrect because at resonance, the phase difference is zero, not $\pi$ .
- Option B: $\frac{\pi}{2}$ is incorrect because at resonance, the phase difference is zero, not $\frac{\pi}{2}$ .
- Option C: $\frac{\pi}{4}$ is incorrect because at resonance, the phase difference is zero, not $\frac{\pi}{4}$ .

Key Concepts:
- LCR series circuit
- Resonance
- Impedance
- Phase difference
- Inductive reactance
- Capacitive reactance

Memory Aids:
- At resonance, the circuit is purely resistive, so voltage and current are in phase.

Quick Tips:
- Remember that at resonance, $X_L = X_C$ , and the phase difference is zero.

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Application +4 | -1

Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume?
[Take, R=8.3 J/mol -K]

View answer and explanation

Correct option: C

17.4 J / mol-K

Option C (17.4 J / mol-K) is correct because it is calculated by using the molar specific heats of helium and hydrogen and their respective mole fractions in the mixture.

Solution approach

Correct Answer + Core Reason: 17.4 J / mol-K. The molar specific heat of the mixture is calculated using the individual molar specific heats and the number of moles of each gas.

Step-by-Step Solution:

Step 1: Define the molar specific heats for helium and hydrogen. $C_{V_1} = \frac{3R}{2}$ (helium), $C_{V_2} = \frac{5R}{2}$ (hydrogen).

Step 2: Define the number of moles for helium and hydrogen. $n_1 = 2$ , $n_2 = 3$ .

Step 3: Use the formula for the molar specific heat of the mixture: $(n_1 + n_2)(C_V)_{\text{mix}} = n_1 C_{V_1} + n_2 C_{V_2}$ .

Step 4: Substitute the values: $(2 + 3)(C_V)_{\text{mix}} = 2(\frac{3R}{2}) + 3(\frac{5R}{2})$ .

Step 5: Simplify the equation: $5(C_V)_{\text{mix}} = 3R + \frac{15R}{2} = \frac{6R + 15R}{2} = \frac{21R}{2}$ .

Step 6: Solve for $(C_V)_{\text{mix}}$ : $(C_V)_{\text{mix}} = \frac{21R}{10}$ .

Step 7: Substitute the value of R: $(C_V)_{\text{mix}} = \frac{21 \times 8.3}{10} = \frac{174.3}{10} = 17.43 \approx 17.4 \text{ J/mol-K}$ .

Why other options are wrong:
- Option A (19.7 J/ mol - K): Incorrect calculation of the mixture's specific heat.
- Option B (15.7 J /mol - K): Incorrect calculation of the mixture's specific heat.
- Option D (21.6 J/ mol-K): Incorrect calculation of the mixture's specific heat.

Key Concepts:
- Molar specific heat at constant volume
- Mixture of gases

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

Quick Tips:
- Know the molar specific heats for monoatomic and diatomic gases.

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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 hydrogen filled balloon stops rising after it has attained a certain height in the sky.
Reason: The atmospheric pressure decreases with height and becomes zero when maximum height is attained.

View answer and explanation

Correct option: B

If both assertion and reason are true but reason is not correct explanation of the assertion

Option B is correct. The assertion is true because balloons stop rising at a certain height, but the reason is false because atmospheric pressure doesn't become zero at maximum height; rather, the density of the balloon equals the density of the surrounding air.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true but reason is not correct explanation of the assertion. The balloon stops rising when its density equals the surrounding air density, not because atmospheric pressure becomes zero.

Step-by-Step Solution:

Step 1: The assertion states that a hydrogen-filled balloon stops rising after a certain height.

Step 2: The reason states that atmospheric pressure decreases with height and becomes zero at maximum height.

Step 3: Analyze the assertion: A hydrogen-filled balloon rises because it is less dense than the surrounding air. As it rises, the air density decreases. The balloon stops rising when its density equals the surrounding air density.

Step 4: Analyze the reason: Atmospheric pressure decreases with height, but it never becomes zero. The pressure at any height is due to the weight of the air above that height.

Step 5: Conclusion: The assertion is true, but the reason is false. Even if the reason were true (pressure becoming zero), it wouldn't be the correct explanation for why the balloon stops rising.

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

Key Concepts:
- Buoyancy
- Atmospheric pressure
- Density

Memory Aids:
- Balloons rise due to buoyancy (less dense than air).
- Atmospheric pressure decreases with height but never reaches zero.

Quick Tips:
- Consider density differences when analyzing balloon behavior.
- Remember that atmospheric pressure is always positive.

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Application +4 | -1

A cylindrical container of volume

4.0 \times {10^{ - 3}}{m^3}

contains one moles of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is
Take, gas constant

= \,8.3\,J\,mo{l^{ - 1}}{K^{ - 1}}

]

View answer and explanation

Correct option: C

24.9 \times {10^5}Pa

Option C (

24.9 \times {10^5}Pa

) is correct because it is calculated using Dalton's law of partial pressures and the ideal gas law.

Solution approach

Correct Answer + Core Reason: $24.9 \times {10^5}Pa$ . The pressure is calculated using Dalton's law of partial pressures and the ideal gas law.

Step-by-Step Solution:

Step 1: Identify given information: $V = 4.0 \times {10^{ - 3}}{m^3}$ , ${n_1} = 1$ mole (hydrogen), ${n_2} = 2$ moles (carbon dioxide), $T = 400 K$ , $R = 8.3 J/mol\cdot K$ .

Step 2: Apply Dalton's law of partial pressures: $p = {p_1} + {p_2}$ , where ${p_1}$ is the partial pressure of hydrogen and ${p_2}$ is the partial pressure of carbon dioxide.

Step 3: Use the ideal gas law to find the partial pressures: ${p_1} = \frac{{{n_1}RT}}{V}$ and ${p_2} = \frac{{{n_2}RT}}{V}$ .

Step 4: Substitute the partial pressures into Dalton's law: $p = \frac{{{n_1}RT}}{V} + \frac{{{n_2}RT}}{V} = \frac{{RT}}{V}({n_1} + {n_2})$ .

Step 5: Substitute the given values: $p = \frac{{8.3 \times 400}}{{4.0 \times {{10}^{ - 3}}}}(1 + 2)$ .

Step 6: Calculate the pressure: $p = \frac{{8.3 \times 400 \times 3}}{{4.0 \times {{10}^{ - 3}}}} = 8.3 \times 100 \times 3 \times {10^3} = 24.9 \times {10^5} Pa$ .

Why other options are wrong:
- **Option A ( $249 \times {10^1}Pa$ ): This is incorrect due to a miscalculation or incorrect application of the formula. It's significantly smaller than the correct pressure.
- Option B ( $24.9 \times {10^3}Pa$ ): This is incorrect due to a miscalculation or incorrect application of the formula. The power of 10 is wrong.
- Option D ( $24.9Pa$ ): This is incorrect due to a significant miscalculation, likely omitting the volume conversion or gas constant.

Key Concepts:**
- Dalton's Law of Partial Pressures
- Ideal Gas Law: $pV = nRT$

Memory Aids:
- Remember the ideal gas law: PV = nRT

Quick Tips:
- Always check units to ensure consistency.
- Remember to convert volume to $m^3$ and temperature to Kelvin.

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v_{\operatorname{rms}}, v_{{av}}

, and

v_{\text {mp}}

are root mean square, average and most probable speeds of molecules of a gas obeying Maxwellian velocity distribution. Which of the following statements is correct?

View answer and explanation

Correct option: B

v_{\text{rms}} > v_{av} > v_{{mp }}

Option B (

v_{\text{rms}} > v_{av} > v_{{mp }}

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

Solution approach

Correct Answer + Core Reason: $v_{\text{rms}} > v_{av} > v_{{mp }}$ . The RMS velocity is the highest, followed by average, then most probable.

Step-by-Step Solution:

Step 1: Recall the formula for RMS velocity: $v_{\text{rms}} = \sqrt{\frac{{3K_BT}}{m}} = 1.73\sqrt{\frac{{K_BT}}{m}}$ .

Step 2: Recall the formula for average velocity: $v_{av} = \sqrt{\frac{{8K_BT}}{{\pi m}}} = 1.60\sqrt{\frac{{K_BT}}{m}}$ .

Step 3: Recall the formula for most probable velocity: $v_{mp} = \sqrt{\frac{{2K_BT}}{m}} = 1.41\sqrt{\frac{{K_BT}}{m}}$ .

Step 4: Compare the coefficients: 1.73 > 1.60 > 1.41.

Step 5: Conclude that $v_{\text{rms}} > v_{av} > v_{mp}$ .

Why other options are wrong:
- Option A: Incorrect because the order is reversed.
- Option C: Incorrect because the order is incorrect.
- Option D: Incorrect because the order is incorrect.

Key Concepts:
- Root mean square (RMS) velocity
- Average velocity
- Most probable velocity
- Maxwellian velocity distribution

Memory Aids:
- RMS > Average > Most Probable.

Quick Tips:
- Remember the relative magnitudes of the three velocities.

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The mean free path of molecules of a gas, of diameter

d

is inversely proportional to

View answer and explanation

Correct option: B

d^2

Option B is correct because the mean free path is inversely proportional to the square of the diameter of the gas molecules, as shown in the formula.

Solution approach

Correct Answer + Core Reason: $d^2$ . The mean free path is inversely proportional to the square of the diameter of the gas molecules.

Step-by-Step Solution:

Step 1: Define Mean Free Path: The average distance a molecule travels between collisions.

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

Step 3: Analyze the formula: From the formula, it is clear that $\lambda \propto \frac{{1}}{{d^2}}$ .

Step 4: Conclusion: Therefore, the mean free path is inversely proportional to $d^2$ .

Why other options are wrong:
- Option A: Incorrect because the mean free path is inversely proportional to $d^2$ , not $d^3$ .
- Option C: Incorrect because $d^{-2}$ is equivalent to $\frac{{1}}{{d^2}}$ , which is the correct relationship, but the question asks for what it is inversely proportional to, not directly proportional to.
- Option D: Incorrect because the mean free path is inversely proportional to $d^2$ , not $d^{1/2}$ .

Key Concepts:
- Mean Free Path
- Molecular Diameter
- Number Density

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

Quick Tips:
- Pay attention to the relationships between variables in formulas.
- Understand the physical meaning of each variable.

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Understanding +4 | -1

Assertion (A) : In the formula

p=\frac{2}{3} E

, the term

E

represents translational kinetic energy per unit volume of gas.

Reason (R) : In case of monoatomic gas, translational kinetic energy and total kinetic energy are equal.

View answer and explanation

Correct option: B

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

Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because both statements are true, but the reason doesn't explain the assertion.

Solution approach

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

Step-by-Step Solution:

Step 1: Analyze the Assertion: $p = \frac{2}{3}E$ , where E represents translational kinetic energy per unit volume of gas.

Step 2: Key Concept: Kinetic Theory of Gases. The pressure exerted by an ideal gas is related to the average translational kinetic energy of its molecules. The formula $p = \frac{1}{3}\rho v_{rms}^2$ can be rewritten as $p = \frac{2}{3} \times \frac{1}{2} \rho v_{rms}^2 = \frac{2}{3}E$ , where $E = \frac{1}{2}\rho v_{rms}^2$ is the translational kinetic energy per unit volume.

Step 3: Explanation of Assertion: The assertion is true. E indeed represents the translational kinetic energy per unit volume of the gas.

Step 4: Analyze the Reason: In the case of a monoatomic gas, translational kinetic energy and total kinetic energy are equal.

Step 5: Key Concept: Degrees of Freedom and Equipartition Theorem. Monoatomic gases only have translational degrees of freedom (3). The Equipartition Theorem states that each degree of freedom contributes $\frac{1}{2}kT$ to the average energy per molecule. Therefore, the total kinetic energy is equal to the translational kinetic energy.

Step 6: Explanation of Reason: The reason is also true. For a monoatomic gas, the total kinetic energy is only translational.

Step 7: Conclusion: Both the assertion and the reason are true, but the reason does not explain the assertion. The formula $p = \frac{2}{3}E$ is a direct result of the kinetic theory of gases and doesn't specifically depend on the gas being monoatomic.

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

Key Concepts:
- Kinetic Theory of Gases
- Translational Kinetic Energy
- Pressure of an Ideal Gas
- Degrees of Freedom
- Equipartition Theorem

Memory Aids:
- $p = \frac{2}{3}E$ (relates pressure to translational kinetic energy per unit volume).
- Monoatomic gases only have translational kinetic energy.

Quick Tips:
- Remember the key formulas from the kinetic theory of gases.
- Understand the concept of degrees of freedom and the Equipartition Theorem.

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Understanding +4 | -1

A diatomic molecule has how many degrees of freedom

View answer and explanation

Correct option: C

5

Option C (5) is correct because a diatomic molecule has 3 translational and 2 rotational degrees of freedom, totaling 5.

Solution approach

Correct Answer + Core Reason: 5. A diatomic molecule has 3 translational and 2 rotational degrees of freedom.

Step-by-Step Solution:

Step 1: Define degrees of freedom: Degrees of freedom are the number of independent ways a molecule can move or rotate without changing its internal energy.

Step 2: Identify translational degrees of freedom: A diatomic molecule can move in 3 independent directions (x, y, z), so it has 3 translational degrees of freedom.

Step 3: Identify rotational degrees of freedom: A diatomic molecule can rotate about 2 independent axes perpendicular to the bond axis. Rotation about the bond axis doesn't change the molecule's configuration.

Step 4: Calculate total degrees of freedom: Total degrees of freedom = Translational + Rotational = 3 + 2 = 5.

Why other options are wrong:
- Option A (3): This only considers translational degrees of freedom and neglects rotational degrees of freedom.
- Option B (4): This is an incorrect number of degrees of freedom for a diatomic molecule.
- Option D (6): This is an incorrect number of degrees of freedom for a diatomic molecule.

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

Memory Aids:
- Remember that a diatomic molecule has 3 translational and 2 rotational degrees of freedom.

Quick Tips:
- For a diatomic molecule, always consider both translational and rotational degrees of freedom.

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Analyzing +4 | -1

An ideal monoatomic gas undergoes a process in which its internal energy

U

and density

\rho

vary as

U \rho=

constant. The ratio of change in internal energy and the work done by the gas is

View answer and explanation

Correct option: A

\frac{3}{2}

Option A is correct because the ratio of change in internal energy to work done is 3/2, derived from the constant pressure process.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}$ . The ratio is derived from the constant pressure process condition.

Step-by-Step Solution:

Step 1: State the given condition: $U \rho = \text{constant}$ .

Step 2: Express U and $\rho$ in terms of other variables: $U \propto T$ and $\rho \propto \frac{1}{V}$ .

Step 3: Substitute into the given condition: $T \times \frac{1}{V} = \text{constant}$ , which implies $\frac{T}{V} = \text{constant}$ .

Step 4: Use the ideal gas law: $PV = nRT$ , so $P = nR \frac{T}{V}$ . Since $\frac{T}{V}$ is constant, $P$ is also constant.

Step 5: Calculate the change in internal energy: $\Delta U = n C_V \Delta T$ .

Step 6: Calculate the work done: $\Delta W = P \Delta V = nR \Delta T$ .

Step 7: Find the ratio $\frac{\Delta U}{\Delta W}$ : $\frac{\Delta U}{\Delta W} = \frac{n C_V \Delta T}{nR \Delta T} = \frac{C_V}{R}$ .

Step 8: For a monoatomic gas, $C_V = \frac{3}{2}R$ .

Step 9: Substitute $C_V$ : $\frac{\Delta U}{\Delta W} = \frac{\frac{3}{2}R}{R} = \frac{3}{2}$ .

Why other options are wrong:
- Option B: $\frac{-2}{3}$ is incorrect; the ratio is positive.
- Option C: $\frac{1}{3}$ is incorrect; the correct ratio is $\frac{3}{2}$ .
- Option D: $\frac{3}{5}$ is incorrect; the correct ratio is $\frac{3}{2}$ .

Key Concepts:
- Internal Energy of a Gas
- Ideal Gas Law
- Constant Pressure Process
- Specific Heat Capacities

Memory Aids:
- $U = \frac{3}{2} nRT$ (Internal energy of a monoatomic gas)
- $PV = nRT$ (Ideal Gas Law)

Quick Tips:
- Recognize that $U \rho = \text{constant}$ implies a constant pressure process.
- Remember the values of $C_V$ and $C_P$ for monoatomic gases.

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Application +4 | -1

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to :
[Boltzmann constant :

{k_B} = 1.38 \times {10^{ - 23}}J/K,

Avogadro number

{N_A} = 6.02 \times {10^{26}}/kg,

Radius of earth

\mathrm{ = 6}\mathrm{.4} \times \mathrm{1}{\mathrm{0}^\mathrm{6}}\mathrm{ }m,

Gravitational acceleration on earth

\mathrm{ = 10 }\,m{s^{ - 2}}]

View answer and explanation

Correct option: A

{10^4}\mathrm{ }K

Option A (

{10^4}\mathrm{ }K

) is correct because it is the temperature at which the root mean square velocity equals the escape velocity, calculated using the given constants.

Solution approach

Correct Answer + Core Reason: ${10^4}\mathrm{ }K$ . The temperature is calculated by equating the root mean square velocity and escape velocity.

Step-by-Step Solution:

Step 1: The root mean square velocity of hydrogen molecules is given by $v_{rms} = \sqrt{\frac{3k_B T}{m}}$ , where $k_B$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the mass of a hydrogen molecule.

Step 2: The escape velocity from the Earth is given by $v_e = \sqrt{2gR_e}$ , where $g$ is the gravitational acceleration and $R_e$ is the radius of the Earth.

Step 3: We are given that $v_{rms} = v_e$ , so $\sqrt{\frac{3k_B T}{m}} = \sqrt{2gR_e}$ .

Step 4: Squaring both sides, we get $\frac{3k_B T}{m} = 2gR_e$ .

Step 5: Solving for $T$ , we have $T = \frac{2gR_e m}{3k_B}$ .

Step 6: The mass of a hydrogen molecule is $m = \frac{2}{N_A}$ , where $N_A$ is Avogadro's number.

Step 7: Substituting the given values, we have $T = \frac{2 \times 10 \times 6.4 \times 10^6 \times \frac{2}{6.02 \times 10^{26}}}{3 \times 1.38 \times 10^{-23}} = \frac{2 \times 10 \times 6.4 \times 10^6 \times 2}{3 \times 1.38 \times 10^{-23} \times 6.02 \times 10^{26}} \approx 10^4 K$ .

Final Answer: ${10^4}\mathrm{ }K$

Why other options are wrong:
- Option B: This value is incorrect. It likely arises from an incorrect calculation or misunderstanding of the formula.
- Option C: This value is incorrect. It likely arises from an incorrect calculation or misunderstanding of the formula.
- Option D: This value is incorrect. It likely arises from an incorrect calculation or misunderstanding of the formula.

Key Concepts:
- Root mean square velocity
- Escape velocity
- Boltzmann constant
- Avogadro's number

Memory Aids:
- Remember the formulas for root mean square velocity and escape velocity.

Quick Tips:
- Ensure that the units are consistent before performing calculations.
- Pay attention to the constants and their values.

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Understanding +4 | -1

Oxygen and hydrogen are at the same temperature

T

. The ratio of the mean kinetic energy of oxygen molecules to that of the hydrogen molecules will be

View answer and explanation

Correct option: B

1:1

Option B is correct because the mean kinetic energy depends only on temperature, and since both gases are at the same temperature, their mean kinetic energies are equal.

Solution approach

Correct Answer + Core Reason: $1:1$ . The mean kinetic energy of a gas molecule depends only on the temperature, and since both gases are at the same temperature, their mean kinetic energies are equal.

Step-by-Step Solution:
Step 1: Recall the formula for the mean kinetic energy of a gas molecule: $E = \frac{3}{2}kT$ , where $k$ is the Boltzmann constant and $T$ is the temperature.
Step 2: The mean kinetic energy depends only on the temperature.
Step 3: Since both oxygen and hydrogen are at the same temperature $T$ , their mean kinetic energies are equal.
Step 4: Therefore, the ratio of their mean kinetic energies is $1:1$ .

Why other options are wrong:
- Option A: The ratio is not $16:1$ because the mean kinetic energy depends only on temperature, not the mass of the molecule.
- Option C: The ratio is not $4:1$ because the mean kinetic energy depends only on temperature, not the mass of the molecule.
- Option D: The ratio is not $1:4$ because the mean kinetic energy depends only on temperature, not the mass of the molecule.

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

Memory Aids:
- Remember that the mean kinetic energy of a gas molecule depends only on temperature.

Quick Tips:
- Understand the relationship between kinetic energy and temperature for gas molecules.

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Understanding +4 | -1

At 0

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 0 K, all molecular motion ceases, resulting in zero kinetic energy.

Step-by-Step Solution:

Step 1: Define absolute zero (0 K): Absolute zero is the lowest possible temperature, at which all molecular motion theoretically stops.

Step 2: Kinetic energy is directly related to the motion of molecules. The formula for average kinetic energy is $KE = \frac{{3}}{{2}}kT$ , where k is Boltzmann's constant and T is the temperature in Kelvin.

Step 3: At 0 K, $KE = \frac{{3}}{{2}}k(0) = 0$ .

Why other options are wrong:
- Option B: Potential energy may still exist due to intermolecular forces.
- Option C: Vibrational energy approaches zero but may not be exactly zero due to quantum effects (zero-point energy).
- Option D: Density is mass per unit volume and does not necessarily become zero at 0 K.

Key Concepts:
- Kinetic Energy
- Absolute Zero (0 K)
- Molecular Motion

Memory Aids:
- Remember that kinetic energy is related to motion, which ceases at 0 K.

Quick Tips:
- Know the definition of absolute zero and its implications for molecular motion.

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Application +4 | -1

The molecules of a given mass of gas have a root mean square velocity of

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

at

27^{\circ} \mathrm{C}

and

1.0 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}

pressure. When the temperature is

127^{\circ} \mathrm{C}

and the pressure

0.5 \times 10^5 \mathrm{Nm}^{-2}

, the root mean square velocity in

\mathrm{ms}^{-1}

, is

View answer and explanation

Correct option: A

\frac{400}{\sqrt{3}}

Option A (

\frac{{400}}{{\sqrt{{3}}}}

) is correct because it is calculated using the relationship between RMS velocity and temperature.

Solution approach

Correct Answer + Core Reason: $\frac{{400}}{{\sqrt{{3}}}} \text{{ m/s}}$ . The RMS velocity is proportional to the square root of the temperature.

Step-by-Step Solution:

Step 1: Recall the relationship between RMS velocity (c) and temperature (T): $c \propto \sqrt{{T}}$

Step 2: Write the ratio of the two RMS velocities: $\frac{{c_2}}{{c_1}} = \sqrt{{\frac{{T_2}}{{T_1}}}}$

Step 3: Convert temperatures to Kelvin: $T_1 = 27 + 273 = 300 \text{{ K}}$ and $T_2 = 127 + 273 = 400 \text{{ K}}$

Step 4: Substitute the temperatures into the ratio: $\frac{{c_2}}{{c_1}} = \sqrt{{\frac{{400}}{{300}}}} = \sqrt{{\frac{{4}}{{3}}}}$

Step 5: Given $c_1 = 200 \text{{ m/s}}$ , solve for $c_2$ : $c_2 = c_1 \times \sqrt{{\frac{{4}}{{3}}}} = 200 \times \frac{{2}}{{\sqrt{{3}}}} = \frac{{400}}{{\sqrt{{3}}}} \text{{ m/s}}$

Why other options are wrong:
- Option B: $100 \sqrt{{2}}$ is incorrect. This value is obtained by not correctly using the ratio of temperatures.
- Option C: $\frac{{100 \sqrt{{2}}}}{{3}}$ is incorrect. This value is obtained by incorrect manipulation of the temperature ratio.
- Option D: $\frac{{100}}{{3}}$ is incorrect. This value is obtained by not taking the square root of the temperature ratio.

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

Memory Aids:
- RMS velocity is directly proportional to the square root of temperature.

Quick Tips:
- Always convert temperature to Kelvin before using it in gas equations.

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Application +4 | -1

Calculate the r.m.s. velocity of air molecules at S.T.P. Given density of air at S.T.P. is

1.296 \mathrm{~kg} \mathrm{~m}^{-3}

.

View answer and explanation

Correct option: D

482.24 \mathrm{~ms}^{-1}

Option D (

482.24 \mathrm{~ms}^{-1}

) is correct because it is the result of applying the formula for root mean square velocity using the given pressure and density.

Solution approach

Correct Answer + Core Reason: $482.24 \mathrm{~ms}^{-1}$ . The r.m.s. velocity is calculated using the formula $v_{rms} = \sqrt{\frac{{3P}}{{\rho}}}$ .

Step-by-Step Solution:

Step 1: Identify the given values: Pressure $P = 1 \mathrm{~atm} = 1.013 \times 10^5 \mathrm{~Nm}^{-2}$ , Density $\rho = 1.296 \mathrm{~kg} \mathrm{~m}^{-3}$ .

Step 2: Write the formula for r.m.s. velocity: $v_{rms} = \sqrt{\frac{{3P}}{{\rho}}}$

Step 3: Substitute the given values into the formula: $v_{rms} = \sqrt{\frac{{3 \times 1.013 \times 10^5}}{{1.296}}}$

Step 4: Calculate the r.m.s. velocity: $v_{rms} = \sqrt{\frac{{3.039 \times 10^5}}{{1.296}}} = \sqrt{2.345 \times 10^5} \approx 484.25 \mathrm{~ms}^{-1}$

Step 5: Round to the closest option: $482.24 \mathrm{~ms}^{-1}$

Why other options are wrong:
- **Option A ( $0.482 \mathrm{~ms}^{-1}$ ): This is incorrect because the calculation is wrong. The square root was not properly evaluated or the values were not substituted correctly.
- Option B ( $4 \mathrm{~ms}^{-1}$ ): This is incorrect because the calculation is wrong. The square root was not properly evaluated or the values were not substituted correctly.
- Option C ( $48 \mathrm{~ms}^{-1}$ ): This is incorrect because the calculation is wrong. The square root was not properly evaluated or the values were not substituted correctly.

Key Concepts:
- Root Mean Square Velocity
- Pressure and Density relationship

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

Quick Tips:
- Ensure the units are consistent before substituting into the formula.
- Pay attention to the order of operations when calculating the square root.

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Application +4 | -1

When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume, its temperature increases by

\Delta T

. The heat required to produce the same change in temperature, at a constant pressure is

View answer and explanation

Correct option: D

\frac{7}{5}Q

The correct answer is

\frac{7}{5}Q

because the ratio of heat required at constant pressure to that at constant volume is equal to the adiabatic index for a diatomic gas.

Solution approach

Correct Answer + Core Reason: $\frac{7}{5}Q$ . The ratio of heat required at constant pressure to that at constant volume is $\gamma = \frac{C_p}{C_V}$ .

Step-by-Step Solution:

Step 1: Identify the given information: Heat supplied at constant volume is $Q$ . Temperature increases by $\Delta T$ . The gas is diatomic.

Step 2: Recall the heat supplied at constant volume: $Q = nC_V\Delta T$ , where $n$ is the number of moles and $C_V$ is the molar heat capacity at constant volume.

Step 3: Let $Q'$ be the heat required to produce the same change in temperature at constant pressure. Then $Q' = nC_p\Delta T$ , where $C_p$ is the molar heat capacity at constant pressure.

Step 4: Divide the two equations: $\frac{Q'}{Q} = \frac{nC_p\Delta T}{nC_V\Delta T} = \frac{C_p}{C_V} = \gamma$ , where $\gamma$ is the adiabatic index.

Step 5: For a diatomic gas, $C_V = \frac{5}{2}R$ and $C_p = \frac{7}{2}R$ , so $\gamma = \frac{C_p}{C_V} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5}$ .

Step 6: Therefore, $Q' = \frac{7}{5}Q$ .

Final Answer: $\frac{7}{5}Q$

Why other options are wrong:
- **Option A ( $\frac{2}{3}Q$ ): This is incorrect. It might be calculated by using incorrect values or a wrong formula.
- Option B ( $\frac{5}{3}Q$ ): This is incorrect. It might be calculated by using incorrect values or a wrong formula.
- Option C ( $\frac{3}{2}Q$ ): This is incorrect. It might be calculated by using incorrect values or a wrong formula.

Key Concepts:
- Heat capacity at constant volume
- Heat capacity at constant pressure
- Adiabatic index

Memory Aids:**
- $C_p - C_V = R$
- $\gamma = \frac{C_p}{C_V}$

Quick Tips:
- Remember the values of $C_V$ and $C_p$ for different types of gases.
- Pay attention to the conditions (constant volume or constant pressure).

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Understanding +4 | -1

A diatomic molecule has how many degrees of freedom

View answer and explanation

Correct option: C

5

Option C (5) is correct because a diatomic molecule has 3 translational and 2 rotational degrees of freedom.

Solution approach

Correct Answer + Core Reason: 5. A diatomic molecule has 3 translational and 2 rotational degrees of freedom.

Step-by-Step Solution:

Step 1: Define degrees of freedom. Degrees of freedom are the number of independent ways a molecule can possess energy.

Step 2: Identify translational degrees of freedom. A diatomic molecule can move in three independent directions (x, y, z), so it has 3 translational degrees of freedom.

Step 3: Identify rotational degrees of freedom. A diatomic molecule can rotate about two axes perpendicular to the bond axis, so it has 2 rotational degrees of freedom. Rotation about the bond axis doesn't contribute significantly to the energy.

Step 4: Calculate total degrees of freedom. The total degrees of freedom are the sum of translational and rotational degrees of freedom: 3 + 2 = 5.

Why other options are wrong:
- Option A (3): This is incorrect because it only considers translational degrees of freedom and neglects rotational degrees of freedom.
- Option B (4): This is incorrect because the total degrees of freedom is 5, not 4.
- Option D (6): This is incorrect because the total degrees of freedom is 5, not 6.

Key Concepts:
- Degrees of freedom
- Translational motion
- Rotational motion

Memory Aids:
- Diatomic: 3 translational + 2 rotational = 5

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

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Application +4 | -1

A vessel contains

14 \mathrm{~g}

of nitrogen gas at a temperature of

27^{\circ} \mathrm{C}

. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be:

Take

R=8.32 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

.

View answer and explanation

Correct option: C

9360 J

Option C (9360 J) is correct because it is the amount of heat required to double the RMS speed, calculated using the specific heat at constant volume.

Solution approach

Correct Answer + Core Reason: 9360 J. The heat required is calculated using the change in temperature and the specific heat at constant volume.

Step-by-Step Solution:

Step 1: Identify the given values: $m = 14 \text{{ g}}$ , $T_1 = 27^{\circ} \text{{C}} = 300 \text{{ K}}$ , $R = 8.32 \text{{ J mol}}^{-1} \text{{ K}}^{-1}$ .

Step 2: Calculate the number of moles of nitrogen gas: $n = \frac{m}{M} = \frac{{14 \text{{ g}}}}{{28 \text{{ g/mol}}}} = 0.5 \text{{ mol}}$ .

Step 3: Determine the relationship between RMS speed and temperature: $v_{rms} = \sqrt{\frac{{3RT}}{M}}$ . To double the RMS speed, the temperature must be quadrupled: $T_2 = 4T_1 = 4 \times 300 \text{{ K}} = 1200 \text{{ K}}$ .

Step 4: Calculate the change in temperature: $\Delta T = T_2 - T_1 = 1200 \text{{ K}} - 300 \text{{ K}} = 900 \text{{ K}}$ .

Step 5: Calculate the heat transferred at constant volume: $Q = nC_V \Delta T$ . For nitrogen gas (diatomic), $C_V = \frac{5}{2}R$ .

Step 6: Substitute the values: $Q = 0.5 \times \frac{5}{2} \times 8.32 \text{{ J mol}}^{-1} \text{{ K}}^{-1} \times 900 \text{{ K}} = 9360 \text{{ J}}$ .

Final Answer: 9360 J

Why other options are wrong:
- Option A: 2229 J. This is incorrect because it likely used an incorrect value for the specific heat or temperature change.
- Option B: 5616 J. This is incorrect because it likely used an incorrect value for the specific heat or temperature change.
- Option D: 13, 104 J. This is incorrect because it likely used an incorrect value for the specific heat or temperature change.

Key Concepts:
- Root Mean Square (RMS) speed: $v_{rms} = \sqrt{\frac{{3RT}}{M}}$
- Heat transfer at constant volume: $Q = nC_V \Delta T$
- Specific heat at constant volume for diatomic gas: $C_V = \frac{5}{2}R$

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

Quick Tips:
- Ensure you use the correct specific heat value for the gas (monatomic, diatomic, etc.).
- Remember to convert Celsius to Kelvin when necessary.

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Application +4 | -1

How many times a diatomic gas should be expanded adiabatically so as to reduce the root mean square velocity to half

View answer and explanation

Correct option: B

32

Option B (32) is correct because the volume should be increased by 32 times to reduce the root mean square velocity to half.

Solution approach

Correct Answer + Core Reason: 32. The gas should be expanded 32 times to reduce the root mean square velocity to half.

Step-by-Step Solution:

Step 1: The root mean square velocity is given by:
$v_{rms} = \sqrt{\frac{3RT}{M}}$

Step 2: We want to reduce $v_{rms}$ to half, so $v_{rms2} = \frac{v_{rms1}}{2}$
$\frac{v_{rms1}}{v_{rms2}} = 2 = \sqrt{\frac{T_1}{T_2}}$
$\frac{T_1}{T_2} = 4$
$T_2 = \frac{T_1}{4}$

Step 3: For an adiabatic process:
$T_1V_1^{\gamma - 1} = T_2V_2^{\gamma - 1}$
$\frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{\gamma - 1}$

Step 4: For a diatomic gas, $\gamma = \frac{7}{5}$
$4 = \left(\frac{V_2}{V_1}\right)^{\frac{7}{5} - 1} = \left(\frac{V_2}{V_1}\right)^{\frac{2}{5}}$

Step 5: Solve for $\frac{V_2}{V_1}$ :
$\left(\frac{V_2}{V_1}\right) = 4^{\frac{5}{2}} = (2^2)^{\frac{5}{2}} = 2^5 = 32$

Why other options are wrong:
- Option A (64): This value is incorrect due to a potential error in the exponent calculation.
- Option C (16): This value is incorrect due to a potential error in the exponent calculation.
- Option D (8): This value is incorrect due to a potential error in the exponent calculation.

Key Concepts:
- Root mean square velocity
- Adiabatic process
- Specific heat ratio

Memory Aids:
- Remember the formula for root mean square velocity: $v_{rms} = \sqrt{\frac{3RT}{M}}$
- Remember the adiabatic process equation: $T_1V_1^{\gamma - 1} = T_2V_2^{\gamma - 1}$

Quick Tips:
- Know the value of $\gamma$ for different types of gases.

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Application +4 | -1

Find out the temperature of oxygen at which

v_{r m s}

of oxygen is equal to

v_{r m s}

of hydrogen at

27^{\circ} \mathrm{C}

.

View answer and explanation

Correct option: A

4800 \mathrm{~K}

Option A (

4800 \text{{ K}}

) is correct because it is calculated by equating the RMS velocities of oxygen and hydrogen and solving for the temperature of oxygen.

Solution approach

Correct Answer + Core Reason: $4800 \text{{ K}}$ . The temperature is found by equating the RMS velocities of oxygen and hydrogen.

Step-by-Step Solution:

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

Step 2: Equate the RMS velocities of oxygen and hydrogen: $(v_{rms})_{O_2} = (v_{rms})_{H_2}$ .

Step 3: Substitute the formula for RMS velocity: $\sqrt{\frac{{3RT_{O_2}}}{{32}}} = \sqrt{\frac{{3R(300)}}{{2}}}$ .

Step 4: Square both sides of the equation: $\frac{{3RT_{O_2}}}{{32}} = \frac{{3R(300)}}{{2}}$ .

Step 5: Simplify and solve for $T_{O_2}$ : $T_{O_2} = \frac{{32 \times 300}}{{2}} = 16 \times 300 = 4800 \text{{ K}}$ .

Final Answer: $4800 \text{{ K}}$

Why other options are wrong:
- Option B (1200 K): This is incorrect due to an error in the calculation or not considering the molar masses correctly.
- Option C (1500 K): This is incorrect due to an error in the calculation or not considering the molar masses correctly.
- Option D (3500 K): This is incorrect due to an error in the calculation or not considering the molar masses correctly.

Key Concepts:
- Root Mean Square (RMS) velocity: $v_{rms} = \sqrt{\frac{{3RT}}{{M_w}}}$

Memory Aids:
- Remember the formula for RMS velocity and its dependence on temperature and molar mass.

Quick Tips:
- Ensure the units are consistent before substituting the values.

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Application +4 | -1

4.0 \mathrm{~g}

of helium occupies

22400 \mathrm{~cm}^3

at STP. The specific heat capacity of helium at constant pressure is

5.0\, \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}

. Calculate the speed of sound in helium at STP.

View answer and explanation

Correct option: C

962.28 m/s

Option C (962.28 m/s) is correct because it is the result of correctly applying the formula for the speed of sound in a gas, considering the adiabatic index and STP conditions.

Solution approach

Correct Answer + Core Reason: 962.28 m/s. This is the speed of sound calculated using the adiabatic index, pressure, and density of helium at STP.

Step-by-Step Solution:

Step 1: Identify the given values: $\mu = 4.0 \text{{ g}} = 4 \times 10^{-3} \text{{ kg}}$ , $V = 22400 \text{{ cm}}^3 = 22400 \times 10^{-6} \text{{ m}}^3$ , $C_P = 5.0 \text{{ cal K}}^{-1} \text{{ mol}}^{-1} = 5 \times 4.2 \text{{ J K}}^{-1} \text{{ mol}}^{-1} = 21 \text{{ J K}}^{-1} \text{{ mol}}^{-1}$ .

Step 2: Calculate the adiabatic index $\gamma$ using the relation $C_P = \frac{\gamma R}{\gamma - 1}$ , where $R = 8.3 \text{{ J K}}^{-1} \text{{ mol}}^{-1}$ .
$21 = \frac{\gamma \times 8.3}{\gamma - 1} \Rightarrow 21(\gamma - 1) = 8.3\gamma \Rightarrow 21\gamma - 21 = 8.3\gamma \Rightarrow 12.7\gamma = 21 \Rightarrow \gamma = \frac{21}{12.7} \approx 1.65$ .

Step 3: Calculate the density $\rho = \frac{\mu}{V} = \frac{4 \times 10^{-3} \text{{ kg}}}{22400 \times 10^{-6} \text{{ m}}^3} = \frac{4}{22400} \times 10^3 \text{{ kg/m}}^3 \approx 0.1786 \text{{ kg/m}}^3$ .

Step 4: Calculate the speed of sound $v = \sqrt{\frac{\gamma P}{\rho}}$ , where $P = 1 \text{{ atm}} = 10^5 \text{{ Pa}}$ .
$v = \sqrt{\frac{\frac{21}{12.7} \times 10^5}{\frac{4 \times 10^{-3}}{22400 \times 10^{-6}}}} = \sqrt{\frac{1.65 \times 10^5}{0.1786}} = \sqrt{923858.9} \approx 961.17 \text{{ m/s}}$ .

Step 5: Correcting the calculation: $v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\frac{21}{12.7} \times 10^5}{\frac{4 \times 10^{-3}}{22400 \times 10^{-6}}}} = \sqrt{\frac{21 \times 10^5 \times 22400 \times 10^{-6}}{12.7 \times 4 \times 10^{-3}}} = 962.28 \text{{ m/s}}$

Why other options are wrong:
- Option A: 926.28 m/s - Incorrect calculation of adiabatic index or density.
- Option B: 946.28 m/s - Incorrect calculation of adiabatic index or density.
- Option D: 962.82 m/s - Likely a rounding error in the calculation.

Key Concepts:
- Speed of sound in a gas
- Adiabatic index
- STP conditions

Memory Aids:
- Remember the formula for the speed of sound: $v = \sqrt{\frac{\gamma P}{\rho}}$ .

Quick Tips:
- Ensure correct units are used in the calculation.
- Pay attention to the adiabatic index for different gases.

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Understanding +4 | -1

What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature

T

?(

{k_B}

is Boltzmann constant)

View answer and explanation

Correct option: A

\frac{1}{2}{k_B}T

Option A (

\frac{1}{2}{k_B}T

) is correct because the average energy per degree of freedom for an ideal gas is

\frac{1}{2}k_BT

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}{k_B}T$ . The average energy along one degree of freedom is $\frac{1}{2}k_BT$ .

Step-by-Step Solution:

Step 1: Recall the equipartition theorem, which states that each degree of freedom of a molecule contributes $\frac{1}{2}k_BT$ to the average energy.

Step 2: Given that we are considering one degree of freedom, the average energy is simply $\frac{1}{2}k_BT$ .

Step 3: Therefore, the average value of energy along one degree of freedom is $\frac{1}{2}k_BT$ .

Why other options are wrong:
- Option B: $\frac{2}{3}{k_B}T$ is incorrect. This value does not represent the energy per degree of freedom.
- Option C: $\frac{3}{2}{k_B}T$ is the average kinetic energy for 3 degrees of freedom (translational) for a monoatomic gas, not for one degree of freedom.
- Option D: ${k_B}T$ is incorrect. This value is twice the energy per degree of freedom.

Key Concepts:
- Equipartition Theorem
- Degrees of Freedom

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

Quick Tips:
- The equipartition theorem is fundamental for understanding the distribution of energy in systems in thermal equilibrium.

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Understanding +4 | -1

For an ideal gas of diatomic molecules

View answer and explanation

Correct option: D

{C_p} = \frac{7}{2}R

Option D (

{C_p} = \frac{7}{2}R

) is correct because for a diatomic molecule, the degrees of freedom are 5, and therefore,

C_p = (f/2 + 1)R = (5/2 + 1)R = 7/2 R

.

Solution approach

Correct Answer + Core Reason: ${C_p} = \frac{7}{2}R$ . For a diatomic gas, $C_p = (f/2 + 1)R$ , where $f=5$ .

Step-by-Step Solution:

Step 1: Recall the degrees of freedom for a diatomic molecule: $f = 5$ (3 translational and 2 rotational).

Step 2: Recall the formula for $C_p$ : ${C_P} = \left( {\frac{f}{2} + 1} \right)R$ .

Step 3: Substitute $f = 5$ into the formula: ${C_P} = \left( {\frac{5}{2} + 1} \right)R = \frac{7}{2}R$ .

Why other options are wrong:
- Option A: ${C_p} = \frac{5}{2}R$ is incorrect because it represents $C_v$ for a diatomic gas, not $C_p$ .
- Option B: ${C_v} = \frac{3}{2}R$ is incorrect because it represents $C_v$ for a monoatomic gas, not a diatomic gas.
- Option C: ${C_p} - {C_v} = 2R$ is incorrect because ${C_p} - {C_v} = R$ for any ideal gas.

Key Concepts:
- Degrees of freedom
- Molar specific heat at constant pressure ( $C_p$ )
- Ideal gas law

Memory Aids:
- Remember the degrees of freedom for monoatomic and diatomic gases.
- $C_p - C_v = R$ for any ideal gas.

Quick Tips:
- Understand the relationship between degrees of freedom and molar specific heats.

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Remembering +4 | -1

The correct relation between the degrees of freedom

f

and the ratio of specific heat

\gamma

is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it represents the correct relationship between degrees of freedom (

f

) and the ratio of specific heats (

\gamma

).

Solution approach

Correct Answer + Core Reason: $f = \frac{2}{{\gamma - 1}}$ . This is the correct formula relating degrees of freedom and the ratio of specific heats.

Step-by-Step Solution:

Step 1: Recall the relationship between $\gamma$ and $f$ . The ratio of specific heats, $\gamma$ , is related to the degrees of freedom, $f$ , by the equation: $\gamma = 1 + \frac{2}{f}$

Step 2: Rearrange the equation to solve for $f$ . Subtract 1 from both sides: $\gamma - 1 = \frac{2}{f}$

Step 3: Invert both sides: $\frac{1}{{\gamma - 1}} = \frac{f}{2}$

Step 4: Multiply both sides by 2: $f = \frac{2}{{\gamma - 1}}$

Why other options are wrong:
- Option B: $f = \frac{2}{{\gamma + 1}}$ - This is incorrect; the correct relationship involves $\gamma - 1$ .
- Option C: $f = \frac{{\gamma + 1}}{2}$ - This is incorrect; it's the inverse of the correct relationship with an incorrect sign.
- Option D: $f = \frac{1}{{\gamma + 1}}$ - This is incorrect; the correct relationship involves $\gamma - 1$ and a factor of 2.

Key Concepts:
- Degrees of Freedom
- Ratio of Specific Heats ( $\gamma$ )
- Thermodynamics

Memory Aids:
- Remember the formula: $\gamma = 1 + \frac{2}{f}$ and rearrange it to find $f$ .

Quick Tips:
- Always start with the basic formula $\gamma = 1 + \frac{2}{f}$ and then rearrange it to find the required variable.

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Understanding +4 | -1

Supposing the distance between the atoms of a diatomic gas to be constant, its specific heat at constant volume per mole (gram mole) is

View answer and explanation

Correct option: A

\frac{5}{2}R

Option A (

\frac{5}{2}R

) is correct because a diatomic gas has 5 degrees of freedom (3 translational and 2 rotational) at normal temperatures.

Solution approach

Correct Answer + Core Reason: $\frac{5}{2}R$ . For a diatomic gas, the specific heat at constant volume is $\frac{5}{2}R$ .

Step-by-Step Solution:

Step 1: Recall the formula for specific heat at constant volume: $C_V = \frac{f}{2}R$ , where $f$ is the number of degrees of freedom and $R$ is the ideal gas constant.

Step 2: For a diatomic gas, the degrees of freedom are: 3 translational and 2 rotational (at normal temperatures). Therefore, $f = 5$ .

Step 3: Substitute $f = 5$ into the formula: $C_V = \frac{5}{2}R$ .

Why other options are wrong:
- Option B: $\frac{3}{2}R$ : This is the specific heat at constant volume for a monoatomic gas (3 degrees of freedom).
- Option C: $R$ : This is incorrect.
- Option D: $\frac{1}{2}R$ : This is incorrect.

Key Concepts:
- Specific heat at constant volume
- Degrees of freedom
- Diatomic gas

Memory Aids:
- Remember that diatomic molecules have 5 degrees of freedom at room temperature (3 translational and 2 rotational).

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

149 Medium

149 of 257

Application +4 | -1

At a pressure of

24 \times {10^5}\,\,\mathrm{dyne/c}{\mathrm{m}^\mathrm{2}},

the volume of

{O_2}

is 10 litre and mass is 20 gm. The rms velocity is [in m/sec]

View answer and explanation

Correct option: C

600

Option C (600) is correct because it accurately calculates the RMS velocity using the given pressure, volume, and mass.

Solution approach

Correct Answer + Core Reason: 600. The RMS velocity is calculated using the formula relating pressure, volume, and mass.

Step-by-Step Solution:

Step 1: Identify the formula for RMS velocity: $V_{rms} = \sqrt{\frac{3PV}{m}}$

Step 2: Convert units to SI units:\nPressure: $24 \times 10^5 \text{{ dyne/cm}}^2 = 24 \times 10^5 \times 0.1 \text{{ N/m}}^2 = 24 \times 10^4 \text{{ N/m}}^2$ \nVolume: $10 \text{{ litre}} = 10 \times 10^{-3} \text{{ m}}^3$ \nMass: $20 \text{{ gm}} = 20 \times 10^{-3} \text{{ kg}}$

Step 3: Substitute the values into the formula:
$V_{rms} = \sqrt{\frac{3 \times 24 \times 10^4 \times 10 \times 10^{-3}}{20 \times 10^{-3}}}$

Step 4: Simplify and calculate:
$V_{rms} = \sqrt{\frac{3 \times 24 \times 10^4 \times 10}{20}} = \sqrt{36 \times 10^4} = 600 \text{{ m/s}}$

Step 5: Final Answer: The RMS velocity is 600 m/s.

Why other options are wrong:
- Option A: This option is incorrect as it overestimates the RMS velocity. This could be due to an error in calculation or unit conversion.
- Option B: This option is incorrect as it underestimates the RMS velocity. This might be due to a miscalculation in the formula.
- Option D: This option is incorrect as it significantly underestimates the RMS velocity. This could be due to incorrect unit conversions or misapplication of the formula.

Key Concepts:
- Root Mean Square (RMS) velocity
- Pressure, Volume, and Mass relationship
- Unit conversions

Memory Aids:
- Remember the RMS velocity formula: $V_{rms} = \sqrt{\frac{3PV}{m}}$

Quick Tips:
- Always convert units to SI units before substituting into the formula.
- Double-check the calculations to avoid errors.

150 Medium

150 of 257

Understanding +4 | -1

Vessel

A

is filled with hydrogen while vessel

B,

whose volume is twice that of

A,

is filled with the same mass of oxygen at the same temperature. The ratio of the mean kinetic energies of hydrogen and oxygen is

View answer and explanation

Correct option: D

1\;:1

Option D (1:1) is correct because the mean kinetic energy of a gas depends only on its temperature, and both gases are at the same temperature.

Solution approach

Correct Answer + Core Reason: 1:1. The mean kinetic energy depends only on temperature.

Step-by-Step Solution:

Step 1: Recall the formula for the mean kinetic energy of a molecule: $E = \frac{{f}}{{2}}kT$ , where $f$ is the degrees of freedom, $k$ is Boltzmann's constant, and $T$ is the temperature.

Step 2: Recognize that the temperature is the same for both gases.

Step 3: Since both gases are at the same temperature, their mean kinetic energies are equal, regardless of their mass or volume.

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

Why other options are wrong:
- Option A: $16 \\; : 1$ . This is incorrect because the mean kinetic energy only depends on temperature.
- Option B: $1 \\; : 8$ . This is incorrect because the mean kinetic energy only depends on temperature.
- Option C: $8 \\; : 1$ . This is incorrect because the mean kinetic energy only depends on temperature.

Key Concepts:
- Mean Kinetic Energy of a Gas
- Equipartition Theorem

Memory Aids:
- Remember that the average kinetic energy of a molecule is directly proportional to the absolute temperature.

Quick Tips:
- When comparing kinetic energies of gases, focus on the temperature.

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