Question

The temperature at which oxygen molecules will have same r.m.s. speed as helium molecules at 57$^\circ$C is (molecular masses of oxygen and helium are 32 and 4 respectively.) ______.

• A. 1320 K
• B. 2240 K
• C. 2640 K
• D. 3230 K

Hint:

Heavier molecules are sluggish! To make a massive Oxygen molecule ($32$ g/mol) move as fast as a tiny Helium atom ($4$ g/mol), you must heat it up by the exact ratio of their masses ($32/4 = 8$ times hotter in Kelvin). $330 \text{ K} \times 8 = 2640 \text{ K}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We must calculate the required temperature for oxygen gas so that its molecules travel at the exact same root-mean-square (r.m.s.) speed as helium molecules at a given temperature.

Step 2: Detailed Explanation:
The formula for the r.m.s. speed of a gas molecule is:
$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$
where:
$R$ = Universal gas constant
$T$ = Absolute temperature (in Kelvin)
$M$ = Molar mass of the gas
The problem states that the r.m.s. speed of Oxygen ($O_2$) equals the r.m.s. speed of Helium ($He$):
$(v_{\text{rms}})_{O_2} = (v_{\text{rms}})_{He}$
$\sqrt{\frac{3R T_{O_2}}{M_{O_2}}} = \sqrt{\frac{3R T_{He}}{M_{He}}}$
Square both sides and cancel out the common $3R$ term:
$\frac{T_{O_2}}{M_{O_2}} = \frac{T_{He}}{M_{He}}$
We are given:
$M_{O_2} = 32$
$M_{He} = 4$
$T_{He} = 57^\circ\text{C}$
Crucial Step: Convert the given Celsius temperature to absolute Kelvin!
$T_{He} = 57 + 273 = 330 \text{ K}$.
Substitute the values into the equality:
$\frac{T_{O_2}}{32} = \frac{330}{4}$
Solve for $T_{O_2}$:
$T_{O_2} = 32 \times \left( \frac{330}{4} \right)$
$T_{O_2} = 8 \times 330$
$T_{O_2} = 2640 \text{ K}$

Step 3: Final Answer:
The temperature is 2640 K, matching option (c).