Physics Fluids and Thermal - How to Solve: Kinetic Theory of Gases (RMS Speed, Degrees of Freedom, Equipartition) – IIT JEE Guide

How to Solve: Kinetic Theory of Gases (RMS Speed, Degrees of Freedom, Equipartition) – IIT JEE Guide

Introduction

Mastering Kinetic Theory of Gases unlocks 5-10 marks in IIT JEE (Main + Advanced) every year—questions on RMS speed, degrees of freedom, and equipartition appear in both numerical and conceptual forms. If you can solve these, you’re also ready for real-world applications like gas leak detection, rocket propulsion, and even weather forecasting!

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Ideal Gas Law – ( PV = nRT ) (and how to derive it from kinetic theory).
2. Basic Kinetic Energy – ( KE = \frac{1}{2}mv^2 ).
3. Molecular Speeds – Most probable, average, and RMS speeds (and how they differ).

If any of these are unclear, stop and review them first—this guide assumes you know them cold.

KEY TERMS & FORMULAS

1. Root Mean Square (RMS) Speed

Formula: [ v_{rms} = \sqrt{\frac{3RT}{M}} ] Variables: - ( v_{rms} ) = Root mean square speed (m/s) - ( R ) = Universal gas constant (8.314 J/mol·K) – MEMORISE THIS - ( T ) = Absolute temperature (K) - ( M ) = Molar mass of gas (kg/mol) – Convert to kg/mol if given in g/mol!

Alternative Form (using Boltzmann constant ( k_B )): [ v_{rms} = \sqrt{\frac{3k_B T}{m}} ] - ( k_B ) = Boltzmann constant (1.38 × 10⁻²³ J/K) – MEMORISE THIS - ( m ) = Mass of one molecule (kg)

When to use which? - Use ( R ) when given molar mass (M). - Use ( k_B ) when given mass of one molecule (m).

2. Degrees of Freedom (f)

Definition: The number of independent ways a molecule can store energy. Key Values (MEMORISE THESE): | Gas Type | Degrees of Freedom (f) | Example | |----------|----------------------|---------| | Monatomic | 3 (only translational) | He, Ar, Ne | | Diatomic (rigid) | 5 (3 translational + 2 rotational) | H₂, O₂, N₂ (at room temp) | | Diatomic (vibrating) | 7 (3 translational + 2 rotational + 2 vibrational) | H₂, O₂ (at high temps) | | Polyatomic | 6 (3 translational + 3 rotational) | CO₂, CH₄ |

Rule of Thumb: - At room temperature, diatomic gases have f = 5 (vibrations are "frozen out"). - At high temperatures, diatomic gases have f = 7 (vibrations contribute).

3. Equipartition Theorem

Statement: Energy is equally distributed among all active degrees of freedom. Formula: [ \text{Average energy per molecule} = \frac{f}{2} k_B T ] [ \text{Total internal energy (U)} = \frac{f}{2} nRT ]

Key Points: - Monatomic gas (f=3): ( U = \frac{3}{2} nRT ) - Diatomic gas (f=5): ( U = \frac{5}{2} nRT ) - Polyatomic gas (f=6): ( U = 3 nRT )

MEMORISE THESE! They appear directly in exams.

4. Most Probable Speed (( v_{mp} )) & Average Speed (( \bar{v} ))

Formulas: [ v_{mp} = \sqrt{\frac{2RT}{M}} ] [ \bar{v} = \sqrt{\frac{8RT}{\pi M}} ]

Relationship between speeds: [ v_{mp} : \bar{v} : v_{rms} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3} ] MEMORISE THE RATIO: ( 1 : 1.128 : 1.225 ) (approx.)

STEP-BY-STEP METHOD

Step 1: Identify What’s Given & What’s Asked

• Given: Temperature (T), molar mass (M), gas type (mono/diatomic), or internal energy (U).
• Asked: RMS speed, degrees of freedom, internal energy, or ratio of speeds.

Step 2: Choose the Right Formula

• RMS speed? → ( v_{rms} = \sqrt{\frac{3RT}{M}} ) or ( \sqrt{\frac{3k_B T}{m}} )
• Degrees of freedom? → Use the table (mono=3, diatomic=5/7, poly=6)
• Internal energy? → ( U = \frac{f}{2} nRT )
• Ratio of speeds? → ( v_{mp} : \bar{v} : v_{rms} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3} )

Step 3: Convert Units (CRUCIAL!)

• Temperature: Always in Kelvin (K). If given in °C, add 273.
• Molar mass (M): Must be in kg/mol. If given in g/mol, divide by 1000.
• Mass of molecule (m): Must be in kg. If given in amu, convert using ( 1 \text{ amu} = 1.66 × 10^{-27} \text{ kg} ).

Step 4: Plug & Solve

• Substitute values into the formula.
• Double-check units before calculating.

Step 5: Interpret the Answer

• If asked for RMS speed, ensure the answer is in m/s.
• If asked for internal energy, ensure it’s in Joules (J).
• If asked for degrees of freedom, justify your choice (e.g., "At room temp, diatomic gases have f=5").

WORKED EXAMPLES

Example 1 – Basic (RMS Speed)

Question: Calculate the RMS speed of oxygen (O₂) molecules at 300 K. (Molar mass of O₂ = 32 g/mol)

Step-by-Step Solution:
1. Given: - T = 300 K - M = 32 g/mol = 0.032 kg/mol (converted to kg/mol) - R = 8.314 J/mol·K

1. Formula: ( v_{rms} = \sqrt{\frac{3RT}{M}} )

2. Plug in values: ( v_{rms} = \sqrt{\frac{3 × 8.314 × 300}{0.032}} )

3. Calculate:

4. Numerator: ( 3 × 8.314 × 300 = 7482.6 )
5. Denominator: 0.032
6. Inside sqrt: ( 7482.6 / 0.032 = 233,831.25 )

7. ( v_{rms} = \sqrt{233,831.25} ≈ 483.6 \text{ m/s} )

8. Answer: 484 m/s (rounded to 3 significant figures)

What we did and why: - We used the RMS speed formula because the question asked for it. - We converted molar mass to kg/mol to match SI units. - The answer is realistic (O₂ molecules at 300 K move at ~500 m/s).

Example 2 – Medium (Degrees of Freedom & Internal Energy)

Question: A diatomic gas at 500 K has an internal energy of 10 kJ. Find the number of moles of the gas. (Assume f=5)

Step-by-Step Solution:
1. Given: - T = 500 K - U = 10 kJ = 10,000 J (converted to Joules) - f = 5 (diatomic at moderate temps)

1. Formula: ( U = \frac{f}{2} nRT )

2. Rearrange for n: ( n = \frac{2U}{fRT} )

3. Plug in values: ( n = \frac{2 × 10,000}{5 × 8.314 × 500} )

4. Calculate:

5. Denominator: ( 5 × 8.314 × 500 = 20,785 )

6. ( n = \frac{20,000}{20,785} ≈ 0.962 \text{ moles} )

7. Answer: 0.962 moles

What we did and why: - We used equipartition theorem to relate internal energy to moles. - We assumed f=5 because the gas is diatomic and temperature is moderate (not extremely high). - Unit conversion (kJ → J) was crucial to avoid errors.

Example 3 – Exam-Style (Disguised Question)

Question: The ratio of the RMS speeds of two gases A and B at the same temperature is 2:1. If the molar mass of A is 4 g/mol, what is the molar mass of B?

Step-by-Step Solution:
1. Given: - ( \frac{v_{rms,A}}{v_{rms,B}} = \frac{2}{1} ) - ( M_A = 4 \text{ g/mol} ) - Same temperature (T is constant)

1. Formula: ( v_{rms} = \sqrt{\frac{3RT}{M}} ) So, ( \frac{v_{rms,A}}{v_{rms,B}} = \sqrt{\frac{M_B}{M_A}} )

2. Square both sides: ( \left( \frac{2}{1} \right)^2 = \frac{M_B}{M_A} ) ( 4 = \frac{M_B}{4} )

3. Solve for ( M_B ): ( M_B = 4 × 4 = 16 \text{ g/mol} )

4. Answer: 16 g/mol

What we did and why: - We recognized the ratio and used the RMS speed formula to relate speeds to molar masses. - Squaring the ratio was key to eliminating the square root. - The question disguised the concept (ratio of speeds → molar mass), a common IIT JEE trick.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is all you need to remember for Kinetic Theory of Gases in IIT JEE:

1. RMS speed formula: ( v_{rms} = \sqrt{\frac{3RT}{M}} ). M must be in kg/mol!
2. Degrees of freedom:
3. Monatomic = 3
4. Diatomic (room temp) = 5
5. Diatomic (high temp) = 7
6. Polyatomic = 6
7. Internal energy: ( U = \frac{f}{2} nRT ). Memorise the values for f=3,5,6,7.
8. Speed ratios: ( v_{mp} : \bar{v} : v_{rms} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3} ). Approx 1 : 1.128 : 1.225.
9. Unit conversions are everything! Kelvin for T, kg/mol for M, kg for m.

If you see a ratio question, square it. If you see internal energy, check degrees of freedom. If you see RMS speed, convert units first. You’ve got this!