Verified Solution • 2025–26 Session moderate

QJEEM23Ph0194: JEE Main 2023 Physics - Class 11 JEEM23Ph0194 Solution: If the r. m.s speed of chlorine molecule is at , the r. m. s…

If the r. m.s speed of chlorine molecule is 490 m/s490 \mathrm{~m} / \mathrm{s} at 27∘C27^{\circ} \mathrm{C} , the r. m. s speed of argon molecules at the same temperature will be (Atomic mass of argon =39.9u=39.9 \mathrm{u} , molecular mass of chlorine =70.9u=70.9 \mathrm{u} )

Multiple Choice Options

  • A 451.7 m/s451.7 \mathrm{~m} / \mathrm{s}
  • B 751.7 m/s751.7 \mathrm{~m} / \mathrm{s}
  • C 551.7 m/s551.7 \mathrm{~m} / \mathrm{s}
  • D 651.7 m/s651.7 \mathrm{~m} / \mathrm{s}

Step-by-Step Verified Solution

The correct relationship between the rms speeds of the two gases is:

vArvCl=MClMAr\frac{v_{\mathrm{Ar}}}{v_{\mathrm{Cl}}} = \sqrt{\frac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}}

Given the molar masses for argon and chlorine:

MAr=39.9uM_{\mathrm{Ar}} = 39.9 \mathrm{u}

MCl2=70.9uM_{\mathrm{Cl}_2} = 70.9 \mathrm{u}

And the rms speed of chlorine molecules:

vCl=490 m/sv_{\mathrm{Cl}} = 490 \mathrm{~m} / \mathrm{s}

We can now solve for the rms speed of argon molecules:

vAr=70.939.9×490v_{\mathrm{Ar}} = \sqrt{\frac{70.9}{39.9}} \times 490

vAr≈651.7 m/sv_{\mathrm{Ar}} \approx 651.7 \mathrm{~m} / \mathrm{s}

The rms speed of argon molecules at the same temperature as the chlorine molecules is approximately 651.7 m/s651.7 \mathrm{~m} / \mathrm{s} .

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