JEE MainPhysics

Answer verified against the official NTA key

+4 marks−1 if incorrectSingle correctPrevious-year question

Dependence of Gas RMS Speed on Absolute Temperature

Q: 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.

The temperature of an ideal gas is increased from 200K to 800K. If r.m.s. speed of gas at 200K is v_0. Then, r.m.s. speed of the gas at 800K 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.

Av_0B4 v_0C(v_0)/4D2 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 $2 v_{0}$ .

Option analysis

Why each option works or fails

A · v_0

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

B · 4 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 · (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 v_0

This option correctly applies $v_{rms} \propto \sqrt{T}$ . Quadrupling the absolute temperature from $200 K$ to $800 K$ increases the r.m.s. speed by a factor of $\sqrt{800 / 200} = 2$ , giving $2 v_{0}$ .

Reviewed route

Solution

StepWorking

01Given

$T_{1} = 200 K$ , $T_{2} = 800 K$ , $v_{rms, 1} = v_{0}$ .

02Find

Find $v_{rms, 2}$ at $T_{2} = 800 K$ .

03Strategise

Use $v_{rms} = \sqrt{\frac{3 R T}{M}}$ , which gives $v_{rms} \propto \sqrt{T}$ . Therefore, $\frac{v_{rms, 2}}{v_{rms, 1}} = \sqrt{\frac{T_{2}}{T_{1}}}$ .

04Execute

$v_{rms, 2} = v_{0} \sqrt{\frac{800}{200}} = v_{0} \sqrt{4} = 2 v_{0}$ .

✓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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $2 v_{0}$ .

Why each option works or fails

A: v_0 - Believing that r.m.s. speed is independent of temperature or confusing temperature with a parameter held constant. Recall that $v_{rms} = \sqrt{3 R T / M}$ , meaning speed changes proportionally with $\sqrt{T}$ .
B: 4 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: (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 v_0 - This option correctly applies $v_{rms} \propto \sqrt{T}$ . Quadrupling the absolute temperature from $200 K$ to $800 K$ increases the r.m.s. speed by a factor of $\sqrt{800 / 200} = 2$ , giving $2 v_{0}$ .

Step-by-step solution

Given: $T_{1} = 200 K$ , $T_{2} = 800 K$ , $v_{rms, 1} = v_{0}$ .
Find: Find $v_{rms, 2}$ at $T_{2} = 800 K$ .
Strategise: Use $v_{rms} = \sqrt{\frac{3 R T}{M}}$ , which gives $v_{rms} \propto \sqrt{T}$ . Therefore, $\frac{v_{rms, 2}}{v_{rms, 1}} = \sqrt{\frac{T_{2}}{T_{1}}}$ .
Execute: $v_{rms, 2} = v_{0} \sqrt{\frac{800}{200}} = v_{0} \sqrt{4} = 2 v_{0}$ .
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: $v_{rms} \propto \sqrt{T} ⟹ v_{2} = v_{0} \sqrt{800 / 200} = 2 v_{0}$ , which corresponds directly to Option D.

Dependence of Gas RMS Speed on Absolute Temperature

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 $2 v_{0}$ .

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

✓Verify

We think you should solve this next.

Match List I with List II…

The width of fringe is $2 mm$ on the screen in a double slits experiment for the l…

The number of colloidal systems from the following, which will have 'liquid' as the dispe…

Students also ask

Dependence of Gas RMS Speed on Absolute Temperature

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 2v02v0​.

01Given

02Find

03Strategise

04Execute

✓Verify

Match List I with List II…

The width of fringe is 2 mm2 mm on the screen in a double slits experiment for the l…

The number of colloidal systems from the following, which will have 'liquid' as the dispe…

Students also ask

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 $2 v_{0}$ .

The width of fringe is $2 mm$ on the screen in a double slits experiment for the l…