At what temperature is the root mean square v | vedclass

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(A) The root mean square $(RMS)$ velocity of a gas is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The root mean square $(RMS)$ velocity of a gas is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature in Kelvin,and $M$ is the molar mass of the gas.Let $T_H$ be the temperature of hydrogen and $T_O$ be the temperature of oxygen. Given $T_O = 47^{\circ}C = 47 + 273 = 320 \; K$.The molar mass of hydrogen $(H_2)$ is $M_H = 2 \; g/mol$ and the molar mass of oxygen $(O_2)$ is $M_O = 32 \; g/mol$.According to the problem,the $RMS$ velocities are equal:$\sqrt{\frac{3RT_H}{M_H}} = \sqrt{\frac{3RT_O}{M_O}}$Squaring both sides and simplifying:$\frac{T_H}{M_H} = \frac{T_O}{M_O}$Substituting the values:$\frac{T_H}{2} = \frac{320}{32}$$\frac{T_H}{2} = 10$$T_H = 20 \; K$.

At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at $47^{\circ}C$ (in $; K$)?

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