Question:
The temperature at which the rms speed of hydrogen molecules is equal to that of oxygen molecules at \(47^\circ C\) is
The temperature at which the rms speed of hydrogen molecules is equal to that of oxygen molecules at \(47^\circ C\) is
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For equal rms speeds:
\[
\frac{T}{M}=\text{constant}
\]
Lighter gases require lower temperature to have same rms speed.
Updated On: Jun 17, 2026
- \(20\,K\)
- \(80\,K\)
- \(73\,K\)
- \(3\,K\)
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The Correct Option is A
Solution and Explanation
Concept:
Root mean square speed:
\[
v_{rms}=\sqrt{\frac{3RT}{M}}
\]
For equal rms speeds:
\[
\frac{T_1}{M_1}=\frac{T_2}{M_2}
\]
Step 1: Convert temperature of oxygen into Kelvin. \[ 47^\circ C=320\,K \]
Step 2: Apply equality of rms speeds. For hydrogen: \[ M_H=2 \] For oxygen: \[ M_O=32 \] Let required temperature be \(T\). \[ \frac{T}{2}=\frac{320}{32} \] \[ \frac{T}{2}=10 \] \[ T=20\,K \] Hence, \[ \boxed{20\,K} \]
Step 1: Convert temperature of oxygen into Kelvin. \[ 47^\circ C=320\,K \]
Step 2: Apply equality of rms speeds. For hydrogen: \[ M_H=2 \] For oxygen: \[ M_O=32 \] Let required temperature be \(T\). \[ \frac{T}{2}=\frac{320}{32} \] \[ \frac{T}{2}=10 \] \[ T=20\,K \] Hence, \[ \boxed{20\,K} \]
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