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Kinetic Theory of Gases: JEE Main Physics Question with Solution

Q: Does molar mass change when gas temperature is raised?

No, molar mass is a characteristic property of the gas molecules and remains constant.

The temperature of an ideal gas is increased from

200\text{ K}

800\text{ K}

. If r.m.s. speed of gas at

200\text{K}

\mathrm{v}_{0}

. Then, r.m.s. speed of the gas at

800\text{ K}

will be:

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\mathrm{v}_{0}

4 \mathrm{v}_{0}

\frac{\mathrm{v}_{0}}{4}

2 \mathrm{v}_{0}

Step-by-step solutionView

Correct answer

Because the r.m.s. speed of an ideal gas is proportional to the square root of absolute temperature, quadrupling the temperature doubles the speed to $2\mathrm{v}_{0}$ .

Option analysis

Why each option works or fails

A ·

\mathrm{v}_{0}

Believing that r.m.s. speed is independent of temperature or confusing temperature with a parameter held constant. Recall that $v_{\text{rms}} = \sqrt{3RT/M}$ , meaning speed changes proportionally with $\sqrt{T}$ .

B ·

4 \mathrm{v}_{0}

Assuming that r.m.s. speed is directly proportional to temperature ( $v \propto T$ ) rather than its square root. Apply the square root relationship: when temperature is multiplied by $4$ , the speed is multiplied by $\sqrt{4} = 2$ .

C ·

\frac{\mathrm{v}_{0}}{4}

Inverting the ratio or confusing heating with cooling, alongside missing the square root. Recognize that increasing temperature increases kinetic energy and speed, so the factor must be greater than 1.

D ·

2 \mathrm{v}_{0}

None. This option correctly applies $v_{\text{rms}} \propto \sqrt{T}$ . Quadrupling the absolute temperature from $200\text{ K}$ to $800\text{ K}$ increases the r.m.s. speed by a factor of $\sqrt{800/200} = 2$ , giving $2\mathrm{v}_{0}$ .

Reviewed route

Solution

StepWorking

01given

$T_1 = 200\text{ K}$ , $T_2 = 800\text{ K}$ , $v_{\text{rms},1} = v_0$ .

02find

Find $v_{\text{rms},2}$ at $T_2 = 800\text{ K}$ .

03strategise

Use $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ , which gives $v_{\text{rms}} \propto \sqrt{T}$ . Therefore, $\frac{v_{\text{rms},2}}{v_{\text{rms},1}} = \sqrt{\frac{T_2}{T_1}}$ .

04execute

$v_{\text{rms},2} = v_0 \sqrt{\frac{800}{200}} = v_0 \sqrt{4} = 2v_0$ .

✓verify

Since temperature quadrupled, the root-mean-square speed must increase by $\sqrt{4} = 2$ . Magnitude and direct variation are verified.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Does molar mass

M

change when gas temperature is raised?

No, molar mass $M$ is a characteristic property of the gas molecules and remains constant.

Answer

Because the r.m.s. speed of an ideal gas is proportional to the square root of absolute temperature, quadrupling the temperature doubles the speed to $2\mathrm{v}_{0}$ .

Why each option works or fails

A: $\mathrm{v}_{0}$ - Believing that r.m.s. speed is independent of temperature or confusing temperature with a parameter held constant. Recall that $v_{\text{rms}} = \sqrt{3RT/M}$ , meaning speed changes proportionally with $\sqrt{T}$ .
B: $4 \mathrm{v}_{0}$ - Assuming that r.m.s. speed is directly proportional to temperature ( $v \propto T$ ) rather than its square root. Apply the square root relationship: when temperature is multiplied by $4$ , the speed is multiplied by $\sqrt{4} = 2$ .
C: $\frac{\mathrm{v}_{0}}{4}$ - Inverting the ratio or confusing heating with cooling, alongside missing the square root. Recognize that increasing temperature increases kinetic energy and speed, so the factor must be greater than 1.
D · correct: $2 \mathrm{v}_{0}$ - None. This option correctly applies $v_{\text{rms}} \propto \sqrt{T}$ . Quadrupling the absolute temperature from $200\text{ K}$ to $800\text{ K}$ increases the r.m.s. speed by a factor of $\sqrt{800/200} = 2$ , giving $2\mathrm{v}_{0}$ .

Step-by-step solution

given: $T_1 = 200\text{ K}$ , $T_2 = 800\text{ K}$ , $v_{\text{rms},1} = v_0$ .
find: Find $v_{\text{rms},2}$ at $T_2 = 800\text{ K}$ .
strategise: Use $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ , which gives $200\text{ K}$ . Therefore, $\frac{v_{\text{rms},2}}{v_{\text{rms},1}} = \sqrt{\frac{T_2}{T_1}}$ .
execute: $v_{\text{rms},2} = v_0 \sqrt{\frac{800}{200}} = v_0 \sqrt{4} = 2v_0$ .
verify: Since temperature quadrupled, the root-mean-square speed must increase by $\sqrt{4} = 2$ . Magnitude and direct variation are verified.

Shortcut:

When to use it: Direct inspection using $v \propto \sqrt{T}$ .

quick_kill: $v_{\text{rms}} \propto \sqrt{T} \implies v_2 = v_0 \sqrt{800/200} = 2v_0$ , which corresponds directly to Option (3).

Kinetic Theory of Gases: JEE Main Physics Question with Solution

What feels right?

Because the r.m.s. speed of an ideal gas is proportional to the square root of absolute temperature, quadrupling the temperature doubles the speed to $2\mathrm{v}_{0}$ .

Why each option works or fails

Solution

01given

02find

03strategise

04execute

✓verify

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