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

We are required to compare the root mean square (r.m.s.) speeds of oxygen and hydrogen molecules at different temperatures.

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Given Formula:

The r.m.s. speed of gas molecules is:

\[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \]

where:

• \(k\) = Boltzmann constant
• \(T\) = absolute temperature (K)
• \(m\) = mass of a molecule

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Condition Given:

The r.m.s. speed of oxygen molecules at \(47^\circ\text{C}\) is equal to that of hydrogen molecules at some temperature \(T_H\).

\[ \sqrt{\frac{3kT_O}{m_O}} = \sqrt{\frac{3kT_H}{m_H}} \]

Squaring both sides:

\[ \frac{T_O}{m_O} = \frac{T_H}{m_H} \]

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Step 1: Mass Ratio

Molecular masses:

• Oxygen molecule \(O_2 = 32\)
• Hydrogen molecule \(H_2 = 2\)

\[ \frac{m_O}{m_H} = \frac{32}{2} = 16 \]

Thus,

\[ T_H = \frac{T_O}{16} \]

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Step 2: Temperature Conversion

Convert oxygen temperature to Kelvin:

\[ T_O = 47^\circ\text{C} + 273 = 320~\text{K} \]

Now calculate hydrogen temperature:

\[ T_H = \frac{320}{16} = 20~\text{K} \]

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Step 3: Convert Back to Celsius

\[ T_H = 20 - 273 = -253^\circ\text{C} \]

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Final Answer:

\(\boxed{-253^\circ\text{C}}\)