The rms speed of oxygen molecules at 47∘C is equal to that of hydrogen molecules kept at ____ ∘ C (M(0)/M(h)=32/2)

Correct Option: -253

Step 1: Understand the Formula for RMS Speed
The root-mean-square (rms) speed of a gas molecule is given by the formula:

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

where:
• $R$ is the universal gas constant,
• $T$ is the absolute temperature in Kelvin (K),
• $M$ is the molar mass of the gas.

Step 2: Express the Given Values in Kelvin
The temperature of oxygen gas is given as $t_{{\text{O}}_2}=47\°\text{C}$.
Converting this to Kelvin:

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

We are given the ratio of molar masses of oxygen and hydrogen:

$\frac{M_{{\text{O}}_2}}{M_{{\text{H}}_2}}=\frac{32}{2}=16$

Step 3: Equate the RMS Speeds
According to the problem, the rms speed of oxygen molecules at $320\text{ K}$ is equal to the rms speed of hydrogen molecules at an unknown temperature $T_{{\text{H}}_2}$:

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

$\sqrt{\frac{3R{T}_{{\text{O}}_2}}{M_{{\text{O}}_2}}}=\sqrt{\frac{3R{T}_{{\text{H}}_2}}{M_{{\text{H}}_2}}}$

Squaring both sides and cancelling common factors ($3R$), we get:

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

Step 4: Solve for the Temperature of Hydrogen
Rearranging the equation to solve for $T_{{\text{H}}_2}$:

$T_{{\text{H}}_2}=T_{{\text{O}}_2}\×\left(\frac{M_{{\text{H}}_2}}{M_{{\text{O}}_2}}\right)$

$T_{{\text{H}}_2}=320\×\left(\frac{2}{32}\right)$

$T_{{\text{H}}_2}=\frac{320}{16}=20\text{ K}$

Step 5: Convert Temperature to Celsius
To express the temperature in degrees Celsius ($\°\text{C}$):

$t_{{\text{H}}_2}=20-273=-253\°\text{C}$

Thus, hydrogen molecules must be kept at -253°C to have the same rms speed as oxygen molecules at 47°C.