KINETIC THEORY OF GASES MCQs for JEE PHYSICS — Medium, Page 2

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Analyzing +4 | -1

Assertion: The change in air pressure effect the speed of sound.
Reason: The speed of sound in a gas is proportional to the square of pressure.

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 the speed of sound is independent of pressure changes when temperature is constant, and the speed of sound is proportional to the square root of pressure, not the square.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are false. Speed of sound is independent of pressure if temperature is constant, and it's proportional to the square root of pressure, not the square.

Step-by-Step Solution:

Step 1: Analyze the Assertion: The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma P}{\rho}}$ , where $\gamma$ is the adiabatic index, $P$ is the pressure, and $\rho$ is the density. If the temperature is constant, then $P/\rho$ remains constant (Boyle's Law). Therefore, the speed of sound is independent of pressure changes at constant temperature. The assertion is false.

Step 2: Analyze the Reason: The speed of sound is proportional to the square root of the pressure, $v \propto \sqrt{P}$ , not the square of the pressure. Therefore, the reason is false.

Step 3: Conclusion: Both the assertion and the reason are false.

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

Key Concepts:
- Speed of Sound in a Gas
- Boyle's Law
- Adiabatic Index

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

Quick Tips:
- Consider the effect of temperature when analyzing the speed of sound.

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Application +4 | -1

At what temperature, the rms velocity of gas molecules would be double of its value at NTP, if pressure is remaining constant?

View answer and explanation

Correct option: A

819^{\circ} \mathrm{C}

Option A (

819^{\circ} \mathrm{C}

) is correct because the rms velocity is proportional to the square root of the absolute temperature, and the temperature is calculated accordingly.

Solution approach

Correct Answer + Core Reason: $819^{\circ} \mathrm{C}$ . The temperature is calculated based on the relationship between rms velocity and temperature.

Step-by-Step Solution:

Step 1: The rms velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{{3RT}}{M}}$ , where $R$ is the gas constant, $T$ is the absolute temperature, and $M$ is the molar mass.

Step 2: Given that the rms velocity doubles, $v_{rms,2} = 2v_{rms,1}$ .

Step 3: Since the pressure is constant, we can write $\frac{{v_{rms,2}}}{{v_{rms,1}}} = \sqrt{\frac{{T_2}}{{T_1}}}$ .

Step 4: Substitute the given condition: $2 = \sqrt{\frac{{T_2}}{{T_1}}}$ .

Step 5: Square both sides: $4 = \frac{{T_2}}{{T_1}}$ .

Step 6: At NTP, $T_1 = 273 K$ . Therefore, $T_2 = 4 \times 273 = 1092 K$ .

Step 7: Convert back to Celsius: $T_2 = 1092 - 273 = 819^{\circ}C$ .

Why other options are wrong:
- **Option B ( $819 \mathrm{~K}$ ): This is incorrect because it gives the temperature in Kelvin, but the question asks for Celsius.
- Option C ( $546 \mathrm{~K}$ ): Incorrect due to miscalculation of the temperature.
- Option D ( $546^{\circ} \mathrm{C}$ ): Incorrect due to miscalculation of the temperature.

Key Concepts:
- RMS velocity of gas molecules
- Kinetic theory of gases
- Temperature conversion (Celsius to Kelvin)

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

Quick Tips:
- Always convert temperatures to Kelvin when using the rms velocity formula.

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Understanding +4 | -1

What is an ideal gas?

View answer and explanation

Correct option: B

A gas satisfying the assumptions of kinetic theory

The correct answer is B (A gas satisfying the assumptions of kinetic theory) because an ideal gas is defined as one that obeys the assumptions of the kinetic theory of gases.

Solution approach

Correct Answer + Core Reason: A gas satisfying the assumptions of kinetic theory. This is the definition of an ideal gas.

Step-by-Step Solution:

Step 1: Recall the assumptions of the kinetic theory of gases:
- The gas consists of a large number of identical molecules.
- The molecules are in random motion and obey Newton's laws of motion.
- The volume of the molecules is negligible compared to the volume of the container.
- No intermolecular forces act on the molecules except during collisions.
- Collisions between molecules and with the walls of the container are perfectly elastic.

Step 2: Define an ideal gas: An ideal gas is a gas that perfectly obeys these assumptions.

Step 3: Final Answer: An ideal gas is a gas satisfying the assumptions of kinetic theory.

Why other options are wrong:
- Option A: One that consists of molecules: All gases consist of molecules, but this doesn't define an ideal gas.
- Option C: A gas having Maxwellian distribution of speed: While ideal gases do have a Maxwellian distribution, this is a consequence of the kinetic theory, not the definition itself.
- Option D: A gas consisting of massless particles: Gas particles have mass.

Key Concepts:
- Ideal Gas
- Kinetic Theory of Gases

Memory Aids:
- Ideal Gas = Obeys Kinetic Theory.

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

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Application +4 | -1

A gas mixture consists of

4 \mathrm{~mol}

of oxygen and

6 \mathrm{~mol}

of argon at temperature

T

. Neglecting all vibrational modes, the total internal energy of the system is

View answer and explanation

Correct option: D

19 R T

Option D (19RT) is correct because it sums the internal energies of oxygen (diatomic) and argon (monoatomic) gases, considering their respective degrees of freedom.

Solution approach

Correct Answer + Core Reason: $19RT$ . The total internal energy is the sum of the internal energies of oxygen and argon.

Step-by-Step Solution:

Step 1: Identify the degrees of freedom for each gas. Oxygen (O2) is a diatomic gas, so it has 5 degrees of freedom (3 translational and 2 rotational, neglecting vibrational modes). Argon (Ar) is a monoatomic gas, so it has 3 degrees of freedom (3 translational).

Step 2: Calculate the internal energy of oxygen: $U_{O_2} = n_{O_2} \times \frac{f}{2} RT = 4 \times \frac{5}{2} RT = 10RT$ , where $n_{O_2}$ is the number of moles of oxygen, $f$ is the degrees of freedom, $R$ is the gas constant, and $T$ is the temperature.

Step 3: Calculate the internal energy of argon: $U_{Ar} = n_{Ar} \times \frac{f}{2} RT = 6 \times \frac{3}{2} RT = 9RT$ , where $n_{Ar}$ is the number of moles of argon.

Step 4: Calculate the total internal energy: $U_{total} = U_{O_2} + U_{Ar} = 10RT + 9RT = 19RT$ .

Why other options are wrong:
- Option A (4RT): This only considers the number of moles of oxygen and doesn't account for the degrees of freedom or the argon gas.
- Option B (14RT): This might be a result of incorrectly adding the number of moles or using the wrong degrees of freedom for either oxygen or argon.
- Option C (8RT): This is an incorrect calculation that doesn't properly account for the degrees of freedom and number of moles of each gas.

Key Concepts:
- Internal Energy of an Ideal Gas
- Degrees of Freedom (Diatomic vs. Monoatomic)
- Equipartition Theorem

Memory Aids:
- Monoatomic: 3 degrees of freedom
- Diatomic: 5 degrees of freedom (at moderate temperatures)

Quick Tips:
- Remember to consider the degrees of freedom for each gas when calculating internal energy.

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Application +4 | -1

Oxygen is filled in a closed metal jar of volume

1.0 \times 10^{-3} \mathrm{~m}^3

at a pressure of

1.5 \times 10^5 \mathrm{~Pa}

and temperature

400 \mathrm{~K}

. The jar has a small leak in it. The atmospheric pressure is

1.0 \times 10^5 \mathrm{~Pa}

and the atmospheric temperature is

300 \mathrm{~K}

. Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalise with the surrounding.

View answer and explanation

Correct option: C

0.16 g

Option C (0.16 g) is correct because the mass of gas leaked out is calculated by finding the difference in mass before and after the pressure and temperature equalize.

Solution approach

Correct Answer + Core Reason: 0.16 g. The mass of gas leaked out is calculated by finding the difference in mass before and after.

Step-by-Step Solution:

Step 1: Initial conditions: $V_1 = 1 \times 10^{-3} m^3$ , $P_1 = 1.5 \times 10^5 Pa$ , $T_1 = 400 K$ .

Step 2: Use the ideal gas law to find the initial number of moles: $P_1V_1 = n_1RT_1 \Rightarrow n_1 = \frac{P_1V_1}{RT_1} = \frac{1.5 \times 10^5 \times 1 \times 10^{-3}}{8.3 \times 400} = \frac{1.5}{8.3 \times 4}$ .

Step 3: Calculate the initial mass: $m_1 = n_1 \times M = \frac{1.5}{8.3 \times 4} \times 32 = 1.4457 \approx 1.446 g$ .

Step 4: Final conditions: $P_2 = 1 \times 10^5 Pa$ , $V_2 = 1 \times 10^{-3} m^3$ , $T_2 = 300 K$ .

Step 5: Use the ideal gas law to find the final number of moles: $P_2V_2 = n_2RT_2 \Rightarrow n_2 = \frac{P_2V_2}{RT_2} = \frac{10^5 \times 10^{-3}}{8.3 \times 300} = \frac{1}{3 \times 8.3} = 0.040$ .

Step 6: Calculate the final mass: $m_2 = n_2 \times M = 0.04 \times 32 = 1.285 g$ .

Step 7: Calculate the mass leaked out: $\Delta m = m_1 - m_2 = 1.446 - 1.285 = 0.161 g \approx 0.16 g$ .

Why other options are wrong:
- Option A: Incorrect because the mass leaked out is not 0.12 g.
- Option B: Incorrect because the mass leaked out is not 0.14 g.
- Option D: Incorrect because the mass leaked out is not 0.18 g.

Key Concepts:
- Ideal gas law
- Moles and mass relationship

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

Quick Tips:
- Use the ideal gas law to find the number of moles before and after the leak, then calculate the mass difference.

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Application +4 | -1

If the molar specific heat at constant pressure for a polyatomic non-linear
gas is

x

and the molar specific heat at constant volume for a diatomic gas
is

y,

find the value of

\frac{x}{y}.

View answer and explanation

Correct option: A

1.6

Option A (1.6) is correct because it calculates the ratio of molar specific heats at constant pressure and constant volume for polyatomic non-linear and diatomic gases respectively.

Solution approach

Correct Answer + Core Reason: 1.6. The ratio is calculated by dividing the molar specific heat at constant pressure for a polyatomic non-linear gas by the molar specific heat at constant volume for a diatomic gas.

Step-by-Step Solution:

Step 1: Determine the degrees of freedom for a polyatomic non-linear gas. It has 3 translational, 3 rotational, and vibrational degrees of freedom. Assuming each vibrational mode contributes R, the molar specific heat at constant volume is $C_V = 3R + 3R + R = 7R/2$ and $C_P = C_V + R = 4R$ .

Step 2: Determine the degrees of freedom for a diatomic gas. It has 3 translational and 2 rotational degrees of freedom. The molar specific heat at constant volume is $C_V = \frac{5}{2}R$ .

Step 3: Given that the molar specific heat at constant pressure for a polyatomic non-linear gas is $x$ , $x = 4R$ .

Step 4: Given that the molar specific heat at constant volume for a diatomic gas is $y$ , $y = \frac{5}{2}R$ .

Step 5: Calculate the ratio $\frac{x}{y} = \frac{4R}{\frac{5}{2}R} = \frac{4}{\frac{5}{2}} = \frac{8}{5} = 1.6$ .

Why other options are wrong:
- Option B: Incorrect ratio value.
- Option C: Incorrect ratio value.
- Option D: Incorrect ratio value.

Key Concepts:
- Degrees of Freedom
- Molar Specific Heat at Constant Pressure ( $C_P$ )
- Molar Specific Heat at Constant Volume ( $C_V$ )
- Relationship between $C_P$ and $C_V$ : $C_P = C_V + R$

Memory Aids:
- Monoatomic: $C_V = \frac{3}{2}R$ , $C_P = \frac{5}{2}R$
- Diatomic: $C_V = \frac{5}{2}R$ , $C_P = \frac{7}{2}R$
- Polyatomic Non-linear: $C_V = 3R$ , $C_P = 4R$ (without vibrational modes)

Quick Tips:
- Remember the degrees of freedom for different types of gases.
- Use the relationship $C_P = C_V + R$ to find $C_P$ if you know $C_V$ .

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Understanding +4 | -1

The temperature of Argon, kept in a vessel, is raised by

1^{\circ} \mathrm{C}

at a constant volume. The total heat supplied to the gas is a combination of translational and rotational energies. Their respective shares are:-

View answer and explanation

Correct option: D

100 \%

and

0 \%

Option D is correct because Argon is a monoatomic gas, possessing only translational kinetic energy. Therefore, all the heat supplied increases translational energy, and none increases rotational energy.

Solution approach

Correct Answer + Core Reason: $100 \\%$ and $0 \\%$ . Argon is a monoatomic gas and only has translational kinetic energy.

Step-by-Step Solution:

Step 1: Argon is a monoatomic gas.

Step 2: Monoatomic gases have only translational degrees of freedom (3).

Step 3: They do not have rotational degrees of freedom at normal temperatures.

Step 4: Therefore, all the heat supplied goes into increasing the translational kinetic energy.

Why other options are wrong:
- Option A: Argon does not have rotational energy.
- Option B: Argon does not have rotational energy.
- Option C: Argon does not have rotational energy.

Key Concepts:
- Degrees of freedom
- Kinetic theory of gases
- Monoatomic gases

Memory Aids:
- Monoatomic gases only have translational energy.

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

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Application +4 | -1

At

11^{\circ} c

, the rms speed of a gas molecule is found to be

1000 \mathrm{~m} / \mathrm{s}

. Take

R=\frac{25}{3} \mathrm{SI}

units. The molecular mass (ingrams), of the gas molecule (in nearest integer) is

View answer and explanation

Correct option: A

7

Option A (7) is correct because by using the RMS speed formula and substituting the given values, the molecular mass is calculated to be approximately 7 grams.

Solution approach

Correct Answer + Core Reason: 7. The molecular mass is calculated using the RMS speed formula.

Step-by-Step Solution:

Step 1: Write down the formula for RMS speed: $v_{rms} = \sqrt{\frac{3RT}{M}}$

Step 2: Substitute the given values: $1000 = \sqrt{\frac{3 \times (25/3) \times 284}{M \times 10^{-3}}}$

Step 3: Square both sides: $10^6 = \frac{25 \times 284}{M \times 10^{-3}}$

Step 4: Solve for M: $M = \frac{25 \times 284}{10^6 \times 10^{-3}} = \frac{25 \times 284}{10^3} = \frac{7100}{1000} = 7.1 \approx 7$ grams

Why other options are wrong:
- Option B: 9 is incorrect as the calculation yields a value closer to 7.
- Option C: 11 is incorrect as the calculation yields a value closer to 7.
- Option D: 13 is incorrect as the calculation yields a value closer to 7.

Key Concepts:
- Root Mean Square (RMS) speed
- Kinetic theory of gases

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

Quick Tips:
- Ensure consistent units when using the RMS speed formula.

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Understanding +4 | -1

A box contains n molecules of a gas. How will the pressure of the gas be effected, if the number of molecules is made

2n

View answer and explanation

Correct option: C

Pressure will be doubled

Option C (Pressure will be doubled) is correct because the pressure of a gas is directly proportional to the number of molecules when volume and temperature are constant.

Solution approach

Correct Answer + Core Reason: Pressure will be doubled. Pressure is directly proportional to the number of molecules.

Step-by-Step Solution:

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

Step 2: Since the volume and temperature are constant, we can write $P \propto n$ .

Step 3: If the number of molecules is made $2n$ , then the new pressure $P'$ is proportional to $2n$ .

Step 4: Therefore, $P' = 2P$ , which means the pressure will be doubled.

Why other options are wrong:
- Option A: Pressure will decrease is incorrect because increasing the number of molecules at constant volume and temperature will increase the pressure.
- Option B: Pressure will remain unchanged is incorrect because increasing the number of molecules will increase the pressure.
- Option D: Pressure will become three times is incorrect because the pressure is directly proportional to the number of molecules, so doubling the molecules doubles the pressure, not triples it.

Key Concepts:
- Ideal gas law: $PV = nRT$
- Relationship between pressure and number of molecules

Memory Aids:
- Remember that pressure is directly proportional to the number of molecules at constant volume and temperature.

Quick Tips:
- Use the ideal gas law to analyze the relationship between pressure, volume, temperature, and the number of moles.
- If volume and temperature are constant, pressure is directly proportional to the number of molecules.

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Understanding +4 | -1

A gas is allowed to expand isothermally. The root mean square velocity of the molecules

View answer and explanation

Correct option: C

Will remain unchanged

Option C (Will remain unchanged) is correct because the root mean square velocity depends only on temperature, which is constant in an isothermal process.

Solution approach

Correct Answer + Core Reason: Will remain unchanged. RMS velocity depends only on temperature in isothermal process.

Step-by-Step Solution:

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

Step 2: Identify the process: The gas is allowed to expand isothermally. This means the temperature remains constant.

Step 3: Analyze the dependence of rms velocity on temperature: The formula shows that $v_{rms}$ is directly proportional to the square root of the temperature. Since the temperature is constant in an isothermal process, the rms velocity will also remain constant.

Why other options are wrong:
- Option A: Will increase. This would be true if the temperature increased.
- Option B: Will decrease. This would be true if the temperature decreased.
- Option D: Depends on the other factors. The rms velocity depends only on temperature and molar mass, and molar mass is constant.

Key Concepts:
- Root mean square (rms) velocity
- Isothermal process
- Kinetic theory of gases

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

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

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Application +4 | -1

A cubical box with porous walls containing an equal number of

{O_2}

and

{H_2}

molecules is placed in a large evacuated chamber. The entire system is maintained at constant temperature

T.

The ratio of

{v_{rms}}

of

{O_2}

molecules to that of the

{v_{rms}}

of

{H_2}

molecules, found in the chamber outside the box after a short interval is

View answer and explanation

Correct option: B

\frac{1}{4}

Option B (

\frac{1}{4}

) is correct because the ratio of RMS velocities is inversely proportional to the square root of the molar masses of the gases.

Solution approach

Correct Answer + Core Reason: $\frac{1}{4}$ . The ratio of RMS velocities is inversely proportional to the square root of the molar mass.

Step-by-Step Solution:

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

Step 2: Since the temperature is constant, $v_{rms} \propto \frac{1}{\sqrt{M}}$ .

Step 3: Calculate the ratio of RMS velocities: $\frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}$ .

Step 4: The molar mass of $O_2$ is 32 g/mol, and the molar mass of $H_2$ is 2 g/mol.

Step 5: Substitute the molar masses into the ratio: $\frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$ .

Why other options are wrong:
- Option A: $\frac{1}{{2\sqrt 2 }}$ This is an incorrect calculation, likely due to a misunderstanding of the molar masses or the square root relationship.
- Option C: $\frac{1}{{\sqrt 2 }}$ This is an incorrect calculation, possibly from not considering the correct molar masses of oxygen and hydrogen.
- Option D: $\sqrt 2$ This is incorrect because it inverts the ratio and doesn't take the square root of the molar mass ratio.

Key Concepts:
- Root-mean-square (RMS) velocity
- Kinetic theory of gases
- Molar mass

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

Quick Tips:
- Ensure you use the correct molar masses for the gases involved.
- Remember that RMS velocity is inversely proportional to the square root of the molar mass at constant temperature.

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Understanding +4 | -1

Degree of freedom for polyatomic gas

View answer and explanation

Correct option: C

\geq 6

Option C is correct because a polyatomic gas has at least 6 degrees of freedom (3 translational and 3 rotational).

Solution approach

Correct Answer + Core Reason: $\geq 6$ . Polyatomic gases have translational, rotational, and vibrational degrees of freedom, resulting in at least 6.

Step-by-Step Solution:

Step 1: Recall the degrees of freedom for different types of gases:
- Monoatomic: 3 (translational)
- Diatomic: 5 (3 translational, 2 rotational)
- Polyatomic: 6 or more (3 translational, 3 rotational, and vibrational)

Step 2: Polyatomic gases have 3 translational and 3 rotational degrees of freedom as a minimum. They can also have vibrational modes, increasing the total.

Why other options are wrong:
- Option A: This is too low; diatomic gases have 5 degrees of freedom.
- Option B: This is too low; diatomic gases can have 5 degrees of freedom.
- Option D: This is too high; polyatomic gases start at 6 degrees of freedom.

Key Concepts:
- Degrees of Freedom
- Translational, Rotational, and Vibrational Motion

Memory Aids:
- Remember the degrees of freedom for monoatomic, diatomic, and polyatomic gases.

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

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Remembering +4 | -1

Which of the following options may be the correct estimate of the mean free path of gas particles?

[n:

Number of gas particle per unit volume,

d:

diameter]

View answer and explanation

Correct option: D

\lambda=\frac{1}{\sqrt{2} n d^2}

Option D (

\lambda=\frac{1}{\sqrt{2} n d^2}

) is correct because it represents the correct formula for the mean free path of gas particles, where n is the number of particles per unit volume and d is the diameter.

Solution approach

Correct Answer + Core Reason: $\lambda=\frac{1}{\sqrt{2} n d^2}$ . This is the correct formula for the mean free path.

Step-by-Step Solution:

Step 1: Define mean free path: The average distance a gas particle travels before colliding with another particle.

Step 2: Recall the formula for mean free path: $\lambda = \frac{1}{\sqrt{2} n \pi d^2}$ , where $n$ is the number density of particles and $d$ is the diameter of the particles. The $\pi$ is often omitted for simplicity in estimations.

Step 3: Compare the given options with the formula.

Step 4: Option D matches the general form of the mean free path formula.

Why other options are wrong:
- Option A: This option has $n^2$ in the denominator, which is incorrect. The correct formula has $n$ to the power of 1.
- Option B: This option has $n^2 d$ in the numerator, which is incorrect. The correct formula has $n$ and $d$ in the denominator.
- Option C: This option has $n d^2$ in the numerator, which is incorrect. The correct formula has $n$ and $d$ in the denominator.

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

Memory Aids:
- Remember the inverse relationship between mean free path and number density and diameter.

Quick Tips:
- Always remember the correct formula for mean free path.

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Application +4 | -1

An ideal gas filled in closed vessel at

27^{\circ} \mathrm{C}

temperature if gas is heated then its pressure is made 2 times. Then find out the final temperature of gas.

View answer and explanation

Correct option: A

600 \mathrm{~K}

Option A (

600 \mathrm{~K}

) is correct because using Gay-Lussac's Law (

\frac{P_1}{T_1} = \frac{P_2}{T_2}

) with constant volume, the final temperature is twice the initial temperature in Kelvin.

Solution approach

Correct Answer + Core Reason: $600 \mathrm{~K}$ . Using Gay-Lussac's Law, the final temperature is twice the initial temperature in Kelvin.

Step-by-Step Solution:

Step 1: Identify the given information: Initial temperature $T_1 = 27^{\circ} \mathrm{C} = 300 \mathrm{K}$ , final pressure $P_2 = 2P_1$ , and volume is constant.

Step 2: Apply Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Step 3: Substitute the given values: $\frac{P_1}{300} = \frac{2P_1}{T_2}$

Step 4: Solve for $T_2$ : $T_2 = 2 \times 300 = 600 \mathrm{K}$

Step 5: Final Answer: $T_2 = 600 \mathrm{K}$

Why other options are wrong:
- **Option B ( $500 \mathrm{~K}$ ): This is incorrect because it doesn't correctly apply the pressure ratio to the initial temperature.
- Option C ( $400 \mathrm{~K}$ ): This is incorrect because it doesn't correctly apply the pressure ratio to the initial temperature.
- Option D ( $700 \mathrm{~K}$ ): This is incorrect because it doesn't correctly apply the pressure ratio to the initial temperature.

Key Concepts:
- Gay-Lussac's Law
- Ideal Gas Law

Memory Aids:
- PV = nRT (Ideal Gas Law)
- Gay-Lussac's Law: Pressure is directly proportional to temperature at constant volume.

Quick Tips:**
- Always convert Celsius to Kelvin before using gas laws.

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Application +4 | -1

The average translational kinetic energy of air molecules is

0.040 \mathrm{eV}\,\, \left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)

. Calculate the temperature of the air. Boltzmann constant

k=1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}

.

View answer and explanation

Correct option: C

310 K

Option C (310 K) is correct because the average translational kinetic energy is directly proportional to the temperature, and the calculation yields approximately 310 K.

Solution approach

Correct Answer + Core Reason: 310 K. The temperature is calculated from the average translational kinetic energy.

Step-by-Step Solution:

Step 1: Identify the given values: Average kinetic energy $KE = 0.040 \text{{ eV}} = 0.040 \times 1.6 \times 10^{-19} \text{{ J}}$ and Boltzmann constant $k = 1.38 \times 10^{-23} \text{{ J K}}^{-1}$ .

Step 2: Recall the formula for average translational kinetic energy: $KE = \frac{3}{2}kT$ , where $T$ is the temperature in Kelvin.

Step 3: Substitute the given values into the formula: $0.040 \times 1.6 \times 10^{-19} = \frac{3}{2} \times 1.38 \times 10^{-23} \times T$ .

Step 4: Solve for $T$ : $T = \frac{{2 \times 0.040 \times 1.6 \times 10^{-19}}}{{3 \times 1.38 \times 10^{-23}}} = \frac{{0.128 \times 10^{-19}}}{{4.14 \times 10^{-23}}} = \frac{{0.128}}{{4.14}} \times 10^4 \approx 0.0309 \times 10^4 \approx 309 \text{{ K}}$ .

Step 5: Final Answer: 310 K

Why other options are wrong:
- Option A (315 K): This is incorrect due to a calculation error.
- Option B (305 K): This is incorrect due to a calculation error.
- Option D (320 K): This is incorrect due to a calculation error.

Key Concepts:
- Average Translational Kinetic Energy
- Boltzmann Constant

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

Quick Tips:
- Remember to convert electron volts to joules before calculation.

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Application +4 | -1

An oxygen cylinder of volume

30 {~L}

has an initial gauge pressure of

15 {~atm}

and a temperature of

27^{\circ} {C}

. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to

11 {~atm}

and its temperature drops to

17^{\circ} {C}

. Estimate the mass of oxygen taken out of the cylinder.
(Take,

R=8.31 \,{Jmol}^{-1} {~K}^{-1}

and molecular mass of

\left.{O}_2=32 {u}\right)

View answer and explanation

Correct option: C

0.14 {~kg}

Option C (0.14 kg) is correct because it is the mass of oxygen calculated using the ideal gas law.

Solution approach

Correct Answer + Core Reason: 0.14 kg. The mass of oxygen is calculated using the ideal gas law before and after withdrawal.

Step-by-Step Solution:

Step 1: Convert gauge pressures to absolute pressures: $P_1 = 15 + 1 = 16 \text{{ atm}}$ , $P_2 = 11 + 1 = 12 \text{{ atm}}$ .

Step 2: Convert temperatures to Kelvin: $T_1 = 27 + 273.15 = 300.15 \text{{ K}}$ , $T_2 = 17 + 273.15 = 290.15 \text{{ K}}$ .

Step 3: Use the ideal gas law to find the initial number of moles: $n_1 = \frac{{P_1V}}{{RT_1}} = \frac{{16 \times 1.013 \times 10^5 \times 30 \times 10^{-3}}}{{8.314 \times 300.15}} = 19.48 \text{{ moles}}$

Step 4: Use the ideal gas law to find the final number of moles: $n_2 = \frac{{P_2V}}{{RT_2}} = \frac{{12 \times 1.013 \times 10^5 \times 30 \times 10^{-3}}}{{8.314 \times 290.15}} = 15.12 \text{{ moles}}$

Step 5: Calculate the number of moles removed: $n = n_1 - n_2 = 19.48 - 15.12 = 4.36 \text{{ moles}}$

Step 6: Calculate the mass of oxygen removed: $m = n \times M = 4.36 \times 32 = 139.52 \text{{ g}} = 0.13952 \text{{ kg}} \approx 0.14 \text{{ kg}}$

Why other options are wrong:
- Option A (0.14 g): This is incorrect because the unit is wrong. The mass is in kg, not g.
- Option B (0.02 g): This is incorrect due to a miscalculation or using wrong values in the formula.
- Option D (0.014 kg): This is incorrect due to a miscalculation or using wrong values in the formula.

Key Concepts:
- Ideal gas law
- Absolute pressure
- Moles and mass

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

Quick Tips:
- Always convert gauge pressure to absolute pressure before using the ideal gas law.
- Ensure consistent units throughout the calculation.

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Understanding +4 | -1

In kinetic theory of gases, which of the following statements regarding elastic collisions of the molecules is wrong

View answer and explanation

Correct option: A

Kinetic energy is lost in collisions

Option A is correct because in the kinetic theory of gases, collisions are assumed to be perfectly elastic, meaning no kinetic energy is lost.

Solution approach

Correct Answer + Core Reason: Kinetic energy is lost in collisions. In the kinetic theory of gases, collisions are assumed to be perfectly elastic, meaning kinetic energy is conserved.

Step-by-Step Solution:

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

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

Step 3: In an elastic collision, kinetic energy is conserved.

Step 4: Therefore, the statement "Kinetic energy is lost in collisions" is wrong.

Why other options are wrong:
- Option B: Kinetic energy remains constant in collision is correct.
- Option C: Momentum is conserved in collision is correct.
- Option D: Pressure of the gas remains constant in collisions is correct (on average).

Key Concepts:
- Kinetic theory of gases
- Elastic collisions
- Conservation of kinetic energy
- Conservation of momentum

Memory Aids:
- Elastic collisions conserve kinetic energy.

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

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Understanding +4 | -1

For a gas, if the ratio of specific heats at constant pressure and constant volume is

\gamma

, then the value of degrees of freedom is

View answer and explanation

Correct option: D

\frac{2}{\gamma-1}

Option D (

\frac{2}{\gamma-1}

) is correct because the degrees of freedom

n

are related to the ratio of specific heats

\gamma

by the formula

\gamma = 1 + \frac{2}{n}

, which can be rearranged to

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

.

Solution approach

Correct Answer + Core Reason: $\frac{2}{\gamma-1}$ . The degrees of freedom are related to the ratio of specific heats by the formula $\gamma = 1 + \frac{2}{n}$ .

Step-by-Step Solution:

Step 1: Recall the relationship between the ratio of specific heats ( $\gamma$ ) and the degrees of freedom ( $n$ ): $\gamma = 1 + \frac{2}{n}$

Step 2: Rearrange the formula to solve for $n$ :
$\gamma - 1 = \frac{2}{n}$
$n = \frac{2}{\gamma - 1}$

Step 3: Final Answer: $n = \frac{2}{\gamma - 1}$

Why other options are wrong:
- **Option A ( $\frac{\gamma+1}{\gamma-1}$ ): This is not the correct formula for degrees of freedom.
- Option B ( $\frac{\gamma-1}{\gamma+1}$ ): This is not the correct formula for degrees of freedom.
- Option C ( $\frac{(\gamma-1)}{2}$ ): This is the inverse of the correct formula, multiplied by 1/4.

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

Memory Aids:**
- $\gamma = 1 + \frac{2}{n}$

Quick Tips:
- Remember the formula relating $\gamma$ and $n$ .

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Remembering +4 | -1

The translational kinetic energy of gas molecule for one mole of the gas is equal to

View answer and explanation

Correct option: A

\frac{3}{2}RT

Option A (

\frac{3}{2}RT

) is correct because the translational kinetic energy of a gas molecule for one mole of the gas is equal to

\frac{3}{2}RT

, where R is the gas constant and T is the temperature.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}RT$ . The translational kinetic energy of a gas molecule for one mole of the gas is equal to $\frac{3}{2}RT$ .

Step-by-Step Solution:

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

Step 2: The translational kinetic energy for one mole of gas is given by $E = \frac{f}{2}RT$ , where f is the degrees of freedom.

Step 3: For translational motion, $f = 3$ .

Step 4: Therefore, $E = \frac{3}{2}RT$ .

Why other options are wrong:
- Option B: $\frac{2}{3}RT$ is incorrect. The correct factor is $\frac{3}{2}$ .
- Option C: $\frac{1}{2}RT$ is incorrect. The correct factor is $\frac{3}{2}$ .
- Option D: $\frac{2}{3}KT$ is incorrect. It uses Boltzmann constant K instead of the gas constant R and has the wrong factor.

Key Concepts:
- Kinetic Theory of Gases
- Translational Kinetic Energy
- Degrees of Freedom

Memory Aids:
- Translational KE is $\frac{3}{2}RT$ for one mole.

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

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Application +4 | -1

The velocities of three molecules are

3{v}, 4 {v}

and

5 {v}

, respectively. Their rms speed will be

View answer and explanation

Correct option: A

\sqrt{\frac{50}{3}} v

Option A (

\sqrt{\frac{50}{3}} v

) is correct because it is the root mean square speed calculated using the given velocities of the molecules.

Solution approach

Correct Answer + Core Reason: $\sqrt{\frac{50}{3}} v$ . The RMS speed is calculated by taking the square root of the average of the squares of the velocities.

Step-by-Step Solution:

Step 1: The velocities of the three molecules are $3v$ , $4v$ , and $5v$ .

Step 2: Calculate the squares of the velocities: $(3v)^2 = 9v^2$ , $(4v)^2 = 16v^2$ , and $(5v)^2 = 25v^2$ .

Step 3: Calculate the average of the squares of the velocities: $\frac{9v^2 + 16v^2 + 25v^2}{3} = \frac{50v^2}{3}$ .

Step 4: Calculate the RMS speed: $v_{rms} = \sqrt{\frac{50v^2}{3}} = \sqrt{\frac{50}{3}}v$ .

Why other options are wrong:
- **Option B ( $\sqrt{\frac{5}{2}}v$ ): This is incorrect. It might be obtained by incorrectly calculating the average of the squares of the velocities.
- Option C ( $\frac{7}{2} v$ ): This is incorrect. It might be obtained by calculating the average velocity instead of the RMS velocity.
- Option D ( $\frac{5}{2} v$ ): This is incorrect. It might be obtained by calculating the average velocity instead of the RMS velocity.

Key Concepts:
- Root Mean Square (RMS) Speed

Memory Aids:**
- $v_{rms} = \sqrt{\frac{v_1^2 + v_2^2 + ... + v_n^2}{n}}$

Quick Tips:
- Remember to square the velocities before averaging them when calculating the RMS speed.

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Understanding +4 | -1

Assertion(A): Translational kinetic energy of any gas at temperature

T

would be

\frac{1}{2} m v_{\mathrm{rms}}^2

.
Reason(R): Translational kinetic energy of any type of gas at temperature

T

would be

\frac{3}{2} R T

for one mole.

View answer and explanation

Correct option: B

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

Option B (If both Assertion and Reason are correct 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 correct but Reason is not the correct explanation of Assertion. Both statements are true, but unrelated.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): Translational kinetic energy of any gas at temperature $T$ would be $\frac{1}{2} m v_{\mathrm{rms}}^2$ . This is a correct statement.

Step 2: Analyze Reason (R): Translational kinetic energy of any type of gas at temperature $T$ would be $\frac{3}{2} R T$ for one mole. This is also a correct statement.

Step 3: Determine if Reason (R) is the correct explanation of Assertion (A). While both are true, the reason doesn't directly explain why the assertion is true. The assertion is a general statement about kinetic energy, while the reason is a specific case for one mole of gas.

Why other options are wrong:
- Option A: Reason is not the correct explanation of Assertion.
- Option C: Reason is correct.
- Option D: Assertion is correct.

Key Concepts:
- Translational kinetic energy
- Root mean square velocity
- Ideal gas law

Memory Aids:
- Remember the formula for translational kinetic energy: $\frac{1}{2} m v_{\mathrm{rms}}^2$
- Remember the kinetic energy of one mole of gas: $\frac{3}{2} R T$

Quick Tips:
- Distinguish between general formulas and specific cases.

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Understanding +4 | -1

The vapour of a substance behaves as a gas

View answer and explanation

Correct option: B

Above critical temperature

Option B is correct because a vapor behaves as a gas above its critical temperature.

Solution approach

Correct Answer + Core Reason: Above critical temperature. Above this temperature, a substance cannot be liquefied by pressure alone.

Step-by-Step Solution:

Step 1: Define critical temperature: The temperature above which a gas cannot be liquefied by pressure alone.

Step 2: Explain the behavior of a substance below the critical temperature: Below the critical temperature, a substance can exist in either the liquid or vapor phase, and it can be liquefied by applying pressure.

Step 3: Explain the behavior of a substance above the critical temperature: Above the critical temperature, the substance exists only as a gas, and no amount of pressure can liquefy it. It behaves as a gas.

Step 4: Conclude that the vapor of a substance behaves as a gas above the critical temperature.

Why other options are wrong:
- Option A: Below the critical temperature, it can be liquefied.
- Option C: 100°C is not a universal critical temperature.
- Option D: 1000°C is not a universal critical temperature.

Key Concepts:
- Critical Temperature
- Vapor
- Gas
- Liquefaction

Memory Aids:
- Remember that above the critical temperature, a substance is always a gas.

Quick Tips:
- Understand the concept of critical temperature and its significance.

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Application +4 | -1

What is the ratio of specific heats of constant pressure and constant volume of

N{H_3}

is

View answer and explanation

Correct option: A

1.33

Option A (1.33) is correct because it calculates the ratio of specific heats using the degrees of freedom for

NH_3

.

Solution approach

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

Step-by-Step Solution:

Step 1: $NH_3$ has 3 translational and 3 rotational degrees of freedom, so $f = 6$ .

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

Step 3: Substitute $f = 6$ into the equation: $\gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} = 1.33$ .

Why other options are wrong:
- Option B: 1.44 is incorrect due to using a wrong degree of freedom.
- Option C: 1.28 is incorrect due to using a wrong degree of freedom.
- Option D: 1.67 is incorrect due to using a wrong degree of freedom.

Key Concepts:
- Specific heat
- Degrees of freedom

Memory Aids:
- $\gamma = 1 + \frac{2}{f}$

Quick Tips:
- Remember the degrees of freedom for different types of molecules.
- Use the correct formula for the ratio of specific heats.

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Remembering +4 | -1

Mean kinetic energy per degree of freedom of gas molecules is

View answer and explanation

Correct option: C

\frac{1}{2}kT

Option C (

\frac{1}{2}kT

) is correct because, according to the law of equipartition of energy, the mean kinetic energy per degree of freedom is

\frac{1}{2}kT

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}kT$ . According to the law of equipartition of energy.

Step-by-Step Solution:

Step 1: Recall the law of equipartition of energy.

Step 2: The law of equipartition of energy states that for a system in thermal equilibrium, the average energy per degree of freedom is $\frac{1}{2}kT$ , where k is the Boltzmann constant and T is the absolute temperature.

Step 3: Therefore, the mean kinetic energy per degree of freedom is $\frac{1}{2}kT$ .

Why other options are wrong:
- Option A: $\frac{3}{2}kT$ is the total kinetic energy for 3 degrees of freedom (translational) for a monoatomic gas, not per degree of freedom.
- Option B: $kT$ is not the correct expression for kinetic energy per degree of freedom.
- Option D: $\frac{3}{2}RT$ represents the total kinetic energy of one mole of an ideal gas, not per degree of freedom and uses the gas constant R instead of the Boltzmann constant k.

Key Concepts:
- Law of Equipartition of Energy
- Boltzmann Constant
- Degrees of Freedom

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

Quick Tips:
- Be careful to distinguish between energy per degree of freedom and total energy for a molecule or a mole of gas.

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Analyzing +4 | -1

Assertion (A) : According to the law of equipartition of energy, internal energy of an ideal gas at a given temperature, is equally distributed in translational and rotational kinetic energies.

Reason (R) : Rotational kinetic energy of a monoatomic gas is zero

View answer and explanation

Correct option: D

If Assertion is false but the Reason is true.

Option D is correct because the assertion is false, as equipartition of energy distributes energy equally among all degrees of freedom, not just translational and rotational, and the reason is true, as monoatomic gases have no rotational kinetic energy.

Solution approach

Correct Answer + Core Reason: If Assertion is false but the Reason is true.

Step-by-Step Solution:

Step 1: Analyze the Assertion: According to the law of equipartition of energy, internal energy is equally distributed among all degrees of freedom, not just translational and rotational.

Step 2: Analyze the Reason: Rotational kinetic energy of a monoatomic gas is zero because monoatomic gases are considered point masses and do not have rotational degrees of freedom.

Step 3: Combine the analysis: The assertion is incorrect because it limits the distribution to only translational and rotational kinetic energies. The reason is correct.

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

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

Memory Aids:
- Equipartition: Energy equally distributed among all degrees of freedom.

Quick Tips:
- Remember to consider all degrees of freedom when applying the equipartition theorem.

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Application +4 | -1

Find the loss in the mass of 1 mole of an ideal monatomic gas kept in a rigid container as it cools down by

10^{\circ} \mathrm{C}

. The gas constant

R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

.

View answer and explanation

Correct option: A

1.38 \times 10^{-15} \mathrm{~kg}

Option A (

1.38 \times 10^{-15} \mathrm{~kg}

) is correct because it represents the mass loss calculated from the energy change due to cooling, using

E=mc^2

.

Solution approach

Correct Answer + Core Reason: $1.38 \times 10^{-15} \mathrm{~kg}$ . The mass loss is calculated using the energy change and $E=mc^2$ .

Step-by-Step Solution:

Step 1: Calculate the energy change of the gas.

Step 2: The energy possessed by a monoatomic gas is given by $E = \frac{3}{2} nR\Delta T$ , where $n$ is the number of moles, $R$ is the gas constant, and $\Delta T$ is the change in temperature.

Step 3: Substitute the given values: $n = 1 \mathrm{~mole}$ , $R = 8.3 \mathrm{~J \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}$ , and $\Delta T = 10 \mathrm{~^{\circ}C}$ .

Step 4: Calculate the energy change: $E = \frac{3}{2} \times 1 \times 8.3 \times 10 = 124.5 \mathrm{~J}$ .

Step 5: Use the mass-energy equivalence formula $E = mc^2$ to find the mass loss, where $c$ is the speed of light ( $3 \times 10^8 \mathrm{~m/s}$ ).

Step 6: Rearrange the formula to solve for mass: $m = \frac{E}{c^2}$ .

Step 7: Substitute the values: $m = \frac{124.5}{(3 \times 10^8)^2} = \frac{124.5}{9 \times 10^{16}} = 1.383 \times 10^{-15} \mathrm{~kg}$ .

Step 8: Round to two significant figures: $m = 1.38 \times 10^{-15} \mathrm{~kg}$ .

Why other options are wrong:
- **Option B ( $3.18 \times 10^{-15} \mathrm{~kg}$ ): This value is incorrect. It might arise from using an incorrect factor in the energy calculation or a mistake in the division.
- Option C ( $8.31 \times 10^{-15} \mathrm{~kg}$ ): This value is incorrect. It might arise from using the gas constant directly without considering the factor of 3/2 in the energy calculation.
- Option D ( $13.8 \times 10^{-15} \mathrm{~kg}$ ): This value is incorrect. It might arise from an error in the exponent during the calculation.

Key Concepts:**
- Kinetic theory of gases
- Mass-energy equivalence ( $E=mc^2$ )
- Ideal gas law

Memory Aids:
- Remember the formula for the kinetic energy of a monoatomic gas: $E = \frac{3}{2} nRT$ .

Quick Tips:
- Always use consistent units when performing calculations.
- Remember the value of the speed of light, $c = 3 \times 10^8 \mathrm{~m/s}$ .
- Be careful with exponents when dividing by large numbers.

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Understanding +4 | -1

Mean kinetic energy (or average energy) per

g

molecule of a monoatomic gas is given by

View answer and explanation

Correct option: A

\frac{3}{2}RT

Option A (

\frac{{3}}{{2}}RT

) is correct because it represents the average kinetic energy per mole of a monoatomic gas.

Solution approach

Correct Answer + Core Reason: $\frac{{3}}{{2}}RT$ . This is the average kinetic energy per mole of a monoatomic gas.

Step-by-Step Solution:

Step 1: Recall the formula for the average kinetic energy per mole of a monoatomic gas.

Step 2: Key Concept: The average kinetic energy per mole is given by $\frac{{3}}{{2}}RT$ , where R is the gas constant and T is the temperature.

Step 3: Apply the formula: Mean kinetic energy = $\frac{{3}}{{2}}RT$

Why other options are wrong:
- Option B: Incorrect because $\frac{{1}}{{2}}kT$ is the average kinetic energy per molecule, not per mole.
- Option C: Incorrect because it is missing the factor of 3.
- Option D: Incorrect because $\frac{{3}}{{2}}kT$ is the average kinetic energy per molecule, not per mole.

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

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

Quick Tips:
- Distinguish between kinetic energy per molecule and per mole.

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Understanding +4 | -1

At the same temperature and pressure and volume of two gases, which of the following quantities is constant

View answer and explanation

Correct option: B

Average kinetic energy

Option B (Average kinetic energy) is correct because at the same temperature, the average kinetic energy of gas molecules is the same, regardless of the type of gas.

Solution approach

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

Step-by-Step Solution:

Step 1: Recall the equipartition theorem: The average kinetic energy of a molecule is proportional to the temperature: $KE_{avg} = \frac{{f}}{{2}}kT$ , where f is the degrees of freedom, k is Boltzmann's constant, and T is the temperature.

Step 2: Since the temperature is the same for both gases, their average kinetic energies are the same.

Why other options are wrong:
- Option A: Total number of molecules. The number of molecules depends on the volume and the type of gas.
- Option C: Root mean square velocity. RMS velocity depends on both temperature and molar mass.
- Option D: Mean free path. Mean free path depends on the number density and the size of the molecules.

Key Concepts:
- Average Kinetic Energy
- Equipartition Theorem
- Temperature

Memory Aids:
- Kinetic energy is directly proportional to temperature.

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

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Understanding +4 | -1

If the volume of the gas containing

n

number of molecules is

V,

then the pressure will decrease due to force of intermolecular attraction in the proportion

View answer and explanation

Correct option: C

{\left( {n/V} \right)^2}

Option C (

{\left( {n/V} \right)^2}

) is correct because the pressure decrease due to intermolecular attraction is proportional to the square of the number density of the gas molecules.

Solution approach

Correct Answer + Core Reason: ${\left( {n/V} \right)^2}$ . The pressure correction term in the real gas equation is proportional to the square of the density.

Step-by-Step Solution:

Step 1: Recall the van der Waals equation of state for real gases: $(P + a(\frac{{n}}{{V}})^2)(V - nb) = nRT$ .

Step 2: Identify the pressure correction term: $a(\frac{{n}}{{V}})^2$ , where $a$ is a constant that accounts for intermolecular attraction.

Step 3: Recognize that the pressure decrease is due to the term $a(\frac{{n}}{{V}})^2$ .

Step 4: Therefore, the pressure decreases in proportion to $(\frac{{n}}{{V}})^2$ .

Why other options are wrong:
- Option A: Incorrect because the pressure decrease is proportional to the square of the number density, not the number density itself.
- Option B: Incorrect because the pressure decrease is proportional to the square of the number density, not $n/V^2$ .
- Option C: Correct because the pressure decrease is proportional to the square of the number density.
- Option D: Incorrect because the pressure decrease depends on the number of molecules, $n$ , not just the volume.

Key Concepts:
- van der Waals Equation of State
- Intermolecular Forces
- Real Gases

Memory Aids:
- Remember the van der Waals equation and the significance of the $a$ and $b$ constants.

Quick Tips:
- When dealing with real gases, always consider the effects of intermolecular forces and finite molecular size.

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Remembering +4 | -1

In Vander Wal's equation the critical pressure

P_c

is given by

View answer and explanation

Correct option: B

\frac{\mathrm{a}}{27 \mathrm{~b}^2}

Option B (

\frac{\mathrm{a}}{27 \mathrm{~b}^2}

) is correct because it is the formula for critical pressure derived from Vander Wal's equation of state.

Solution approach

Correct Answer + Core Reason: $\frac{\mathrm{a}}{27 \mathrm{~b}^2}$ . The critical pressure is derived from the Vander Waals equation by setting the first and second derivatives of pressure with respect to volume to zero at the critical point.

Step-by-Step Solution:

Step 1: Write the Vander Waals equation of state: $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}$ , or $\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}-\mathrm{b}}-\frac{\mathrm{a}}{\mathrm{V}^2}$

Step 2: At the critical point, $\mathrm{P}=\mathrm{P}_{\mathrm{c}}, \mathrm{V}=\mathrm{V}_{\mathrm{c}}$ , and $\mathrm{T}=\mathrm{T}_{\mathrm{c}}$ ; $\therefore \quad \mathrm{P}_{\mathrm{c}}=\frac{\mathrm{RT}_{\mathrm{c}}}{\mathrm{V}_{\mathrm{c}}-\mathrm{b}}-\frac{\mathrm{a}}{\mathrm{V}_{\mathrm{c}}^2}$

Step 3: Apply the conditions for the critical point: $\frac{\mathrm{dP}_{\mathrm{c}}}{\mathrm{dV}_{\mathrm{c}}}=0$ and $\frac{\mathrm{d}^2 \mathrm{P}_{\mathrm{c}}}{\mathrm{dV}_{\mathrm{c}}^2}=0$

Step 4: Calculate the first derivative and set it to zero: $0=\frac{-\mathrm{RT}_\mathrm{c}}{\left(\mathrm{V}_\mathrm{c}-\mathrm{b}\right)^2}+\frac{2 \mathrm{a}}{\mathrm{V}_\mathrm{c}^3}$

Step 5: Calculate the second derivative and set it to zero: $0=\frac{2 \mathrm{RT}_{\mathrm{c}}}{\left(\mathrm{V}_{\mathrm{c}}-\mathrm{b}\right)^3}-\frac{6 \mathrm{a}}{\mathrm{V}_{\mathrm{c}}^4}$

Step 6: Solve the equations to find $\mathrm{V}_{\mathrm{c}}=3 \mathrm{b}$ and $\mathrm{T}_{\mathrm{c}}=\frac{8 \mathrm{a}}{27 \mathrm{bR}}$

Step 7: Substitute these values into the equation for $\mathrm{P}_{\mathrm{c}}$ : $P_c=\frac{\mathrm{R}}{2 \mathrm{b}}\left(\frac{8\mathrm{a}}{27 \mathrm{bR}}\right)-\frac{\mathrm{a}}{9\mathrm{~b}^2}=\frac{\mathrm{a}}{27 \mathrm{~b}^2}$

Why other options are wrong:
- Option A: This option is incorrect because it does not represent the critical pressure derived from the Vander Waals equation.
- Option C: This option is incorrect because it has the inverse relationship with b² and an incorrect constant.
- Option D: This option is incorrect because it has an inverse relationship and the variables are swapped.

Key Concepts:
- Vander Waals equation of state
- Critical point
- Derivatives

Memory Aids:
- Remember the conditions for the critical point: first and second derivatives are zero.

Quick Tips:
- Understand the derivation of the critical constants from the Vander Waals equation.

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