Easy MCQ +4 / -1 PYQ · JEE Mains 2023

If the r. m.s speed of chlorine molecule is $490 \mathrm{~m} / \mathrm{s}$ at $27^{\circ} \mathrm{C}$, the r. m. s speed of argon molecules at the same temperature will be (Atomic mass of argon $=39.9 \mathrm{u}$, molecular mass of chlorine $=70.9 \mathrm{u}$ )

  1. A $451.7 \mathrm{~m} / \mathrm{s}$
  2. B $751.7 \mathrm{~m} / \mathrm{s}$
  3. C $551.7 \mathrm{~m} / \mathrm{s}$
  4. D $651.7 \mathrm{~m} / \mathrm{s}$ Correct answer

Solution

The correct relationship between the rms speeds of the two gases is: <br/><br/> $$\frac{v_{\mathrm{Ar}}}{v_{\mathrm{Cl}}} = \sqrt{\frac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}}$$ <br/><br/> Given the molar masses for argon and chlorine: <br/><br/> $M_{\mathrm{Ar}} = 39.9 \mathrm{u}$<br/><br/> $M_{\mathrm{Cl}_2} = 70.9 \mathrm{u}$ <br/><br/> And the rms speed of chlorine molecules: <br/><br/> $v_{\mathrm{Cl}} = 490 \mathrm{~m} / \mathrm{s}$ <br/><br/> We can now solve for the rms speed of argon molecules: <br/><br/> $v_{\mathrm{Ar}} = \sqrt{\frac{70.9}{39.9}} \times 490$ <br/><br/> $v_{\mathrm{Ar}} \approx 651.7 \mathrm{~m} / \mathrm{s}$ <br/><br/> The rms speed of argon molecules at the same temperature as the chlorine molecules is approximately $651.7 \mathrm{~m} / \mathrm{s}$.

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law

This question is part of PrepWiser's free JEE Main question bank. 271 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →