Answer : The rms speed of the molecules in a sample of $H_2$ gas at 300 K will be four times larger than the rms speed of $O_2$ molecules at the same temperature, and the ratio $\mu _{rms}(H_2)/\mu _{rms}(O_2)$ constant with increasing temperature.

Explanation :

Formula used for root mean square speed :

$\mu _{rms}=\sqrt{\frac{3RT}{M}}$

where,

$\mu _{rms}$ = rms speed of the molecule

R = gas constant

T = temperature

M = molar mass of the gas

At constant temperature, the formula becomes,

$\mu _{rms}=\sqrt{\frac{1}{M}}$

And the formula for two gases will be,

$\frac{\mu _{H_2}}{\mu _{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}$

Molar mass of $O_2$ = 32 g/mole

Molar mass of $H_2$ = 2 g/mole

Now put all the given values in the above formula, we get

$\frac{\mu _{H_2}}{\mu _{O_2}}=\sqrt{\frac{32g/mole}{M_{2g/mole}}}=4$

Therefore, the rms speed of the molecules in a sample of $H_2$ gas at 300 K will be four times larger than the rms speed of $O_2$ molecules at the same temperature.

And the ratio $\mu _{rms}(H_2)/\mu _{rms}(O_2)$ constant with increasing temperature because rms speed depends only on the molar mass of the gases at same temperature.

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2 years ago

Match these items.

dolphi86 [110]

g - Measurement of Mass

m² - Measurement of Area

m³ - Measurement of Volume

km- Measurement of Length

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2 years ago