Kinetic Theory of Gases JEE 2019: Why A, B and D Are Correct

What is the answer to the kinetic theory of gases JEE 2019 question?

A, B and D are correct in this kinetic theory of gases JEE 2019 question. Quadrupling absolute temperature doubles RMS speed but makes average translational kinetic energy four times larger. The option-by-option check below follows the official solution’s method.

This is the JEE Advanced 2019, Paper 1, Chemistry multiple-correct question. For a gas at equilibrium, decide which claims about molecular RMS speed and mean translational kinetic energy hold:

Why is the average translational kinetic energy independent of mass?

The official solution begins with the ideal-gas equilibrium relation:

E_{a v} = \frac{3}{2} k T

Option A is correct because this expression contains no molecular-mass term. Average translational kinetic energy per molecule depends only on absolute temperature, so changing molecular mass at fixed temperature does not change this average.

The symbols mean:

Average translational kinetic energy per molecule: the quantity on the left.
Boltzmann’s constant: lowercase k.
Absolute temperature in kelvin: uppercase T.

Following the official method, the corresponding translational energy for one mole is:

E_{t r a n s, 1 m o l} = \frac{3}{2} R T

The separate label matters: energy per molecule and energy for one mole are different quantities. Use Boltzmann’s constant for the former and the gas constant for the latter.

Keep translational in the definition. The question does not ask for total molecular internal energy, which can include rotational and vibrational contributions.

Why does quadrupling temperature make B correct but C incorrect?

RMS speed doubles, but mean translational kinetic energy quadruples. Speed depends on the square root of absolute temperature; energy depends directly on it. The gas stays the same during this comparison, so its molar mass remains fixed.

The RMS-speed expression is:

U_{r m s} = \sqrt{\frac{3 R T}{M}}

Here, uppercase R is the gas constant and uppercase M is molar mass. In SI calculations, use molar mass in kilograms per mole.

Set the new temperature to: $T^{'} = 4 T$

Substitute before simplifying:

U_{r m s}^{'} = \sqrt{\frac{3 R \times 4 T}{M}} = \sqrt{4} \sqrt{\frac{3 R T}{M}} = 2 U_{r m s}

Therefore, B is correct. Test energy independently:

E_{a v}^{'} = \frac{3}{2} k (4 T) = 4 [\frac{3}{2} k T] = 4 E_{a v}

Therefore, C is incorrect: it predicts a multiplier of two instead of four. The square root in the speed formula is why the multipliers differ.

Multiplying temperature means multiplying its kelvin value, not its Celsius reading. A change from 300 kelvin to 1200 kelvin gives precisely the speed and energy multipliers calculated above.

Why is option D correct, and what is the final selection?

Option D is correct: at fixed temperature, RMS speed varies inversely with the square root of molecular mass. The formula uses molar mass, but molar mass and the mass of one molecule differ only by the constant Avogadro factor.

First isolate the mass dependence:

U_{r m s} = \frac{\sqrt{3 R T}}{\sqrt{M}} \Rightarrow U_{r m s} \propto M^{- 1 / 2}

Connect molar mass with individual molecular mass:

M = N_{A} m \Rightarrow U_{r m s} \propto m^{- 1 / 2}

Here, lowercase m is one molecule’s mass, and the factor multiplying it is Avogadro’s constant. Temperature must remain fixed for this comparison.

Final selection: A, B and D.

What reasoning mistake produces option C?

The faulty chain is: absolute temperature quadruples, RMS speed doubles, therefore mean translational kinetic energy doubles. The last step is invalid because kinetic energy is quadratic in speed, not linear. The first two steps are correct; transferring the speed multiplier to energy creates the error.

For the same molecular mass:

E_{a v} = \frac{1}{2} m ⟨ u^{2} ⟩ = \frac{1}{2} m U_{r m s}^{2}

Consequently:

\frac{E_{a v}^{'}}{E_{a v}} = {(\frac{U_{r m s}^{'}}{U_{r m s}})}^{2} = 2^{2} = 4

Do not replace RMS speed with mean speed in this identity. RMS speed is the square root of the mean square of speed, not the mean speed itself:

U_{r m s}^{2} = ⟨ u^{2} ⟩ \neq ⟨ u ⟩^{2}

Write a separate ratio for each requested quantity. Never transfer a multiplier from speed to energy without checking the energy formula.

How do you solve related temperature-and-mass questions?

Use the temperature-to-mass ratio for RMS speed and the temperature ratio alone for per-molecule translational energy. These are original related practice questions, not additional authenticated PYQs. Each answer keeps the two ratios separate: mass affects speed, but it does not enter the per-molecule energy ratio.

What happens when the same gas reaches nine times its initial temperature?

The final-to-initial RMS-speed ratio is three, and the energy ratio is nine. Since the gas is unchanged, molar mass cancels from the speed ratio: $T^{'} = 9 T$

\frac{U_{r m s}^{'}}{U_{r m s}} = \sqrt{\frac{T^{'}}{T}} = \sqrt{9} = 3

\frac{E_{a v}^{'}}{E_{a v}} = \frac{T^{'}}{T} = 9

How do gases X and Y compare if Y has four times the molecular mass?

At the same temperature, Y has half the RMS speed of X but equal average translational energy per molecule. Compare Y with X in both ratios:

m_{X} = m, m_{Y} = 4 m

\frac{U_{Y}}{U_{X}} = \sqrt{\frac{m_{X}}{m_{Y}}} = \sqrt{\frac{m}{4 m}} = \frac{1}{2}

\frac{E_{Y}}{E_{X}} = \frac{T_{Y}}{T_{X}} = 1

What temperature ratio gives equal RMS speeds when Y has four times the molar mass?

Y needs four times the absolute temperature of X. Equal RMS speeds require equal temperature-to-molar-mass ratios:

\frac{T_{Y}}{M_{Y}} = \frac{T_{X}}{M_{X}}, M_{Y} = 4 M_{X}

\frac{T_{Y}}{T_{X}} = \frac{M_{Y}}{M_{X}} = 4

The resulting per-molecule energy ratio is:

\frac{E_{Y}}{E_{X}} = \frac{T_{Y}}{T_{X}} = 4

Redo this last question without looking: write the equal-speed condition first, then calculate the energy ratio separately.

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Frequently asked questions

What is the answer to the kinetic theory of gases JEE 2019 question?

A, B and D are correct in the JEE Advanced 2019 Paper 1 Chemistry question. Option C is incorrect because quadrupling absolute temperature quadruples average translational kinetic energy, rather than doubling it.

What happens to RMS speed when temperature is quadrupled?

For the same gas, RMS speed doubles because it is proportional to the square root of absolute temperature. Average translational kinetic energy quadruples because it is directly proportional to absolute temperature. Use temperature in kelvin, not Celsius.

Why is average translational kinetic energy independent of molecular mass?

For an ideal gas at equilibrium, average translational kinetic energy per molecule is E = (3/2)kT. The expression contains no molecular-mass term, so gases at the same absolute temperature have equal average translational kinetic energy per molecule.

How does molecular mass affect RMS speed at the same temperature?

At fixed temperature, RMS speed is inversely proportional to the square root of molecular mass. A gas with four times the molecular mass therefore has half the RMS speed, while its average translational kinetic energy per molecule remains unchanged.

Kinetic Theory of Gases JEE 2019: Why A, B and D Are Correct

What is the answer to the kinetic theory of gases JEE 2019 question?

Why is the average translational kinetic energy independent of mass?

Why does quadrupling temperature make B correct but C incorrect?

Why is option D correct, and what is the final selection?

What reasoning mistake produces option C?

How do you solve related temperature-and-mass questions?

What happens when the same gas reaches nine times its initial temperature?

How do gases X and Y compare if Y has four times the molecular mass?

What temperature ratio gives equal RMS speeds when Y has four times the molar mass?

Frequently asked questions

What is the answer to the kinetic theory of gases JEE 2019 question?

What happens to RMS speed when temperature is quadrupled?

Why is average translational kinetic energy independent of molecular mass?

How does molecular mass affect RMS speed at the same temperature?

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