The r.m.s. speed of oxygen molecules at $47^\circ\text{C}$ is equal to that of the hydrogen molecules kept at __________ $^\circ\text{C}$. (Mass of oxygen molecule/mass of hydrogen molecule $= 32/2$)

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The r.m.s. speed of oxygen molecules at $47^\circ\text{C}$ is equal to that of the hydrogen molecules kept at __________ $^\circ\text{C}$. (Mass of oxygen molecule/mass of hydrogen molecule $= 32/2$)

$-20$

$-253$

$-235$

$-100$

The correct answer is

$-253$

R.M.S. Speed Equivalence Calculation

The root-mean-square (r.m.s.) speed of gas molecules is given by the formula:

$ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} $

where $R$ is the ideal gas constant, $T$ is the absolute temperature (in Kelvin), and $M$ is the molar mass of the gas.

Equating Speeds for Oxygen and Hydrogen

We are given that the r.m.s. speed of oxygen molecules ($v_{\text{rms, O}_2}$) at $47^\circ\text{C}$ is equal to the r.m.s. speed of hydrogen molecules ($v_{\text{rms, H}_2}$) at an unknown temperature $T_{\text{H}_2}$.

First, convert the temperature of oxygen to Kelvin:

$ T_{\text{O}_2} = 47 + 273 = 320 \text{ K} $

The molar mass ratio is given as $\frac{M_{\text{O}_2}}{M_{\text{H}_2}} = \frac{32}{2} = 16$. This implies $\frac{M_{\text{H}_2}}{M_{\text{O}_2}} = \frac{1}{16}$.

Equating the speeds:

$ v_{\text{rms, O}_2} = v_{\text{rms, H}_2} $

$ \sqrt{\frac{3RT_{\text{O}_2}}{M_{\text{O}_2}}} = \sqrt{\frac{3RT_{\text{H}_2}}{M_{\text{H}_2}}} $

Squaring both sides and cancelling constants:

$ \frac{T_{\text{O}_2}}{M_{\text{O}_2}} = \frac{T_{\text{H}_2}}{M_{\text{H}_2}} $

Calculating Hydrogen Temperature

Rearrange the formula to solve for $T_{\text{H}_2}$:

$ T_{\text{H}_2} = T_{\text{O}_2} \times \frac{M_{\text{H}_2}}{M_{\text{O}_2}} $

Substitute the known values:

$ T_{\text{H}_2} = 320 \text{ K} \times \frac{1}{16} $

$ T_{\text{H}_2} = 20 \text{ K} $

Converting to Celsius

Convert the absolute temperature of hydrogen back to Celsius:

$ T_{\text{H}_2} (\text{in } ^\circ\text{C}) = T_{\text{H}_2} (\text{in K}) - 273 $

$ T_{\text{H}_2} (\text{in } ^\circ\text{C}) = 20 - 273 $

$ T_{\text{H}_2} (\text{in } ^\circ\text{C}) = -253 ^\circ\text{C} $

Therefore, the r.m.s. speed of oxygen molecules at $47^\circ\text{C}$ is equal to that of hydrogen molecules at $-253^\circ\text{C}$.

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