Kinetic Theory of Gases: Exam-Pattern Drill (6)

hard 2 min read

Question

A gas at 300 K has rms speed 500 m/s. If the temperature is doubled and the molar mass is halved (mixture changed), find the new rms speed.

Solution — Step by Step

vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}

So vrmsT/Mv_{rms} \propto \sqrt{T/M}.

v2v1=T2/M2T1/M1=T2T1M1M2\frac{v_2}{v_1} = \sqrt{\frac{T_2/M_2}{T_1/M_1}} = \sqrt{\frac{T_2}{T_1} \cdot \frac{M_1}{M_2}}

We have T2/T1=2T_2/T_1 = 2 and M1/M2=2M_1/M_2 = 2. So the ratio is 4=2\sqrt{4} = 2.

v2=2×500=1000 m/sv_2 = 2 \times 500 = 1000\text{ m/s}

Final answer: vrms,new=1000 m/sv_{rms,new} = 1000\text{ m/s}

Why This Works

The rms speed depends on the average kinetic energy per molecule (12mv2=32kBT\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_B T). Doubling TT doubles the kinetic energy, which would multiply vrmsv_{rms} by 2\sqrt{2}. Halving the mass at fixed kinetic energy multiplies vrmsv_{rms} by another 2\sqrt{2}. Combined: factor of 2.

This “ratio method” saves you from plugging actual values of RR, TT, MM — which is exactly what JEE wants you to skip in 90 seconds.

Alternative Method

Plug numbers brute force using vrms=3RT/Mv_{rms} = \sqrt{3RT/M}. You’d need to find MM first from the original data, then recompute. Slow and error-prone in an exam.

Common Mistake

Students confuse rms speed with mean speed and most probable speed. All three scale the same way with TT and MM, but their numerical values differ:

vmp:vavg:vrms=1:1.128:1.224v_{mp} : v_{avg} : v_{rms} = 1 : 1.128 : 1.224

Read the question carefully — if it says “average speed”, use vavg=8RT/πMv_{avg} = \sqrt{8RT/\pi M}, not vrmsv_{rms}.

For mixture problems, “molar mass halved” usually means the gas changed from O2\text{O}_2 (M=32M = 32) to something around CH4\text{CH}_4 (M=16M = 16). If the question gives specific gases, identify them first — easier than tracking MM ratios.

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