Question 807: Thermodynamics & KTG

Question

The temperature at which r.m.s. velocity of hydrogen molecules is 4.5 times that of an oxygen molecule at $47^{\circ} \mathrm{C}$ is (Molecular weight of hydrogen and oxygen molecules are 2 and 32 respectively)

Accepted Answer

We are tasked with finding the temperature at which the root mean square (r.m.s.) velocity of hydrogen molecules is 4.5 times that of oxygen molecules at $47^{\circ} \mathrm{C}$. For this, we use the formula for r.m.s. velocity:
$ \mathrm{V}_{\mathrm{rms}} = \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} $
Given that the r.m.s. velocity of hydrogen ($\mathrm{V}_{\mathrm{H}_{\mathrm{rms}}}$) is 4.5 times the r.m.s. velocity of oxygen ($\mathrm{V}_{\mathrm{O}_{\mathrm{rms}}}$), we can express this relationship as:
$ \mathrm{V}_{\mathrm{H}_{\mathrm{rms}}} = 4.5 	imes \mathrm{V}_{\mathrm{O}_{\mathrm{rms}}} $
Replacing the velocities with their respective r.m.s. formulas, we have:
$ \sqrt{\frac{3 \mathrm{RT}}{M_{\mathrm{H}}}} = 4.5 	imes \sqrt{\frac{3 \mathrm{R} \cdot 320}{32}} $
Simplifying the expression for oxygen at $47^{\circ} \mathrm{C}$, which is $320 \, \mathrm{K}$, we get:
$ \sqrt{\frac{3 \mathrm{RT}}{M_{\mathrm{H}}}} = 4.5 	imes \sqrt{30 \mathrm{R}} $
Squaring both sides yields:
$ \frac{3 \mathrm{RT}}{M_{\mathrm{H}}} = 20.25 	imes 30 \mathrm{R} $
Given that the molecular weight of hydrogen ($M_{\mathrm{H}}$) is 2, we substitute to find $T$:
$ 3 \mathrm{RT} = 20.25 	imes 30 \mathrm{R} 	imes 2 $
Solving for $T$, we have:
$ T = 20.25 	imes 2 = 405 \, \mathrm{K} $
Converting from Kelvin to Celsius gives:
$ T = 132^{\circ} \mathrm{C} $

Answer

$47^{\circ} \mathrm{C}$

Answer

$132^{\circ} \mathrm{C}$

Answer

$320^{\circ} \mathrm{C}$

Answer

$405^{\circ} \mathrm{C}$

Thermodynamics & KTG

The temperature at which r.m.s. velocity of hydrogen molecules is 4.5 times that of an oxygen molecule at 47^{\circ} \mathrm{C} is (Molecular weight of hydrogen and oxygen molecules are 2 and 32 respectively)

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