MHT CET · Physics · Kinetic Theory of Gases
The temperature at which r.m.s. velocity of hydrogen molecules is 4.5 times that of an oxygen molecule at \(47^{\circ} \mathrm{C}\) is (Molecular weight of hydrogen and oxygen molecules are 2 and 32 respectively)
- A \(47^{\circ} \mathrm{C}\)
- B \(132^{\circ} \mathrm{C}\)
- C \(320^{\circ} \mathrm{C}\)
- D \(405^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(B) \(132^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
We know,
\(\begin{aligned}
& \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\
& \text {Given } \mathrm{V}_{\mathrm{H}_{\mathrm{mms}}}=4.5 \mathrm{~V}_{\mathrm{O}_{\mathrm{rms}}} \\
& \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=4.5 \sqrt{\frac{3 \mathrm{R}(320)}{32}} \\
& \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=4.5 \sqrt{30 \mathrm{R}} \\
& 3 \mathrm{RT}=20.25 \times 30 \mathrm{R} \times 2 \\
& \therefore \quad \mathrm{~T}=20.25 \times 2 \times 10=405 \mathrm{~K}=132^{\circ} \mathrm{C}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\
& \text {Given } \mathrm{V}_{\mathrm{H}_{\mathrm{mms}}}=4.5 \mathrm{~V}_{\mathrm{O}_{\mathrm{rms}}} \\
& \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=4.5 \sqrt{\frac{3 \mathrm{R}(320)}{32}} \\
& \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=4.5 \sqrt{30 \mathrm{R}} \\
& 3 \mathrm{RT}=20.25 \times 30 \mathrm{R} \times 2 \\
& \therefore \quad \mathrm{~T}=20.25 \times 2 \times 10=405 \mathrm{~K}=132^{\circ} \mathrm{C}
\end{aligned}\)
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