Question:
The temperature at which the r.m.s. velocity of a gas triples to its r.m.s. velocity at \(0^\circ\text{C}\) is
The temperature at which the r.m.s. velocity of a gas triples to its r.m.s. velocity at \(0^\circ\text{C}\) is
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The r.m.s. speed of a gas varies as the square root of absolute temperature:
\[
v_{\text{rms}}\propto \sqrt{T}.
\]
Always convert Celsius temperature into Kelvin before using this relation.
Updated On: Jun 18, 2026
- \(2184\,\text{K}\)
- \(2184^\circ\text{C}\)
- \(2100^\circ\text{C}\)
- \(2100\,\text{K}\)
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The Correct Option is B
Solution and Explanation
Step 1: Use the relation between r.m.s. speed and temperature.
For an ideal gas, \[ v_{\text{rms}}\propto \sqrt{T} \] where \(T\) is the absolute temperature in Kelvin.
Step 2: Apply the given condition.
Let the initial temperature be \[ T_1=0^\circ\text{C}=273\,\text{K} \] The r.m.s. speed becomes three times its initial value. Therefore, \[ \frac{v_2}{v_1}=3 \] Using \[ \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}, \] we get \[ 3=\sqrt{\frac{T_2}{273}} \] Squaring both sides, \[ 9=\frac{T_2}{273} \] \[ T_2=2457\,\text{K} \]
Step 3: Convert Kelvin into Celsius.
\[ T_2=2457-273 \] \[ T_2=2184^\circ\text{C} \]
Step 4: Final conclusion.
Hence, the required temperature is \[ \boxed{2184^\circ\text{C}} \]
For an ideal gas, \[ v_{\text{rms}}\propto \sqrt{T} \] where \(T\) is the absolute temperature in Kelvin.
Step 2: Apply the given condition.
Let the initial temperature be \[ T_1=0^\circ\text{C}=273\,\text{K} \] The r.m.s. speed becomes three times its initial value. Therefore, \[ \frac{v_2}{v_1}=3 \] Using \[ \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}, \] we get \[ 3=\sqrt{\frac{T_2}{273}} \] Squaring both sides, \[ 9=\frac{T_2}{273} \] \[ T_2=2457\,\text{K} \]
Step 3: Convert Kelvin into Celsius.
\[ T_2=2457-273 \] \[ T_2=2184^\circ\text{C} \]
Step 4: Final conclusion.
Hence, the required temperature is \[ \boxed{2184^\circ\text{C}} \]
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