Question

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

• A. $-20$
• B. $-253$
• C. $-235$
• D. $-100$

Hint:

Always perform gas law calculations using absolute temperature (Kelvin). A common mistake is using Celsius directly in the ratios.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The root mean square (RMS) speed of gas molecules depends on the absolute temperature ($T$) and the molar mass ($M$) of the gas.

Step 2: Key Formula or Approach:
$v_{rms} = \sqrt{\frac{3RT}{M}}$.
If speeds are equal: $\frac{T_1}{M_1} = \frac{T_2}{M_2}$.

Step 3: Detailed Explanation:
For Oxygen (O$_2$): $T_O = 273 + 47 = 320$ K, $M_O = 32$.
For Hydrogen (H$_2$): $T_H = ?$, $M_H = 2$.
Equating the ratios:
\[ \frac{320}{32} = \frac{T_H}{2} \]
\[ 10 = \frac{T_H}{2} \implies T_H = 20 \text{ K} \]
Convert the absolute temperature back to Celsius:
\[ t_H = T_H - 273 = 20 - 273 = -253 \text{ } ^\circ \text{C} \]

Step 4: Final Answer:
The hydrogen molecules are kept at $-253 \text{ } ^\circ \text{C}$.