JEE MAIN - Physics (2023 - 12th April Morning Shift - No. 14)
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 $$=70.9 \mathrm{u}$$ )
$$451.7 \mathrm{~m} / \mathrm{s}$$
$$751.7 \mathrm{~m} / \mathrm{s}$$
$$551.7 \mathrm{~m} / \mathrm{s}$$
$$651.7 \mathrm{~m} / \mathrm{s}$$
Explanation
The correct relationship between the rms speeds of the two gases is:
$$\frac{v_{\mathrm{Ar}}}{v_{\mathrm{Cl}}} = \sqrt{\frac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}}$$
Given the molar masses for argon and chlorine:
$$M_{\mathrm{Ar}} = 39.9 \mathrm{u}$$
$$M_{\mathrm{Cl}_2} = 70.9 \mathrm{u}$$
And the rms speed of chlorine molecules:
$$v_{\mathrm{Cl}} = 490 \mathrm{~m} / \mathrm{s}$$
We can now solve for the rms speed of argon molecules:
$$v_{\mathrm{Ar}} = \sqrt{\frac{70.9}{39.9}} \times 490$$
$$v_{\mathrm{Ar}} \approx 651.7 \mathrm{~m} / \mathrm{s}$$
The rms speed of argon molecules at the same temperature as the chlorine molecules is approximately $$651.7 \mathrm{~m} / \mathrm{s}$$.
$$\frac{v_{\mathrm{Ar}}}{v_{\mathrm{Cl}}} = \sqrt{\frac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}}$$
Given the molar masses for argon and chlorine:
$$M_{\mathrm{Ar}} = 39.9 \mathrm{u}$$
$$M_{\mathrm{Cl}_2} = 70.9 \mathrm{u}$$
And the rms speed of chlorine molecules:
$$v_{\mathrm{Cl}} = 490 \mathrm{~m} / \mathrm{s}$$
We can now solve for the rms speed of argon molecules:
$$v_{\mathrm{Ar}} = \sqrt{\frac{70.9}{39.9}} \times 490$$
$$v_{\mathrm{Ar}} \approx 651.7 \mathrm{~m} / \mathrm{s}$$
The rms speed of argon molecules at the same temperature as the chlorine molecules is approximately $$651.7 \mathrm{~m} / \mathrm{s}$$.
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