Question 951: Thermodynamics & KTG

Question

The root mean square velocity of molecules of a gas is $200 \mathrm{~m} / \mathrm{s}$. What will be the root mean square velocity of the molecules, if the molecular weight is doubled and the absolute temperature is halved?

Accepted Answer

The root mean square (rms) velocity of gas molecules is given by the formula:
$ v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}} $
This shows that:
$ v_{\mathrm{rms}} \propto \sqrt{\frac{T}{M}} $
If we have two scenarios with different conditions and we are comparing their rms velocities, the relationship can be expressed as:
$ \frac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2} = \sqrt{\frac{T_1}{M_1}} 	imes \sqrt{\frac{M_2}{T_2}} $
Here, when the molecular weight is doubled, $ M_2 = 2M_1 $, and when the absolute temperature is halved, $ T_2 = \frac{T_1}{2} $.
Substituting these values into the equation, we get:
$ \frac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2} = \sqrt{\frac{T_1}{M_1}} 	imes \sqrt{\frac{4M_1}{T_1}} = 2 $
This implies that:
$ (v_{\mathrm{rms}})_2 = \frac{(v_{\mathrm{rms}})_1}{2} = \frac{200}{2} = 100 \, 	ext{m/s} $

Answer

$50 \mathrm{~m} / \mathrm{s}$

Answer

$200 \mathrm{~m} / \mathrm{s}$

Answer

$100 \mathrm{~m} / \mathrm{s}$

Answer

$\frac{100}{\sqrt{2}} \mathrm{~m} / \mathrm{s}$

Thermodynamics & KTG

The root mean square velocity of molecules of a gas is 200 \mathrm{~m} / \mathrm{s}. What will be the root mean square velocity of the molecules, if the molecular weight is doubled and the absolute temperature is halved?

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