Question:
If average KE of gas molecule at \( 27^\circ C \) is \( 3.3 \times 10^{-20} \, \text{J} \), find KE at \( 127^\circ C \):
If average KE of gas molecule at \( 27^\circ C \) is \( 3.3 \times 10^{-20} \, \text{J} \), find KE at \( 127^\circ C \):
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Kinetic energy is directly proportional to absolute temperature in ideal gases.
Updated On: May 13, 2025
- \( 15 \times 10^{-20} \, \text{J} \)
- \( 0.68 \times 10^{-20} \, \text{J} \)
- \( 4.4 \times 10^{-20} \, \text{J} \)
- \( 10.3 \times 10^{-21} \, \text{J} \)
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The Correct Option is C
Solution and Explanation
KE \( \propto T \). Convert temps: \( T_1 = 300 \, \text{K}, T_2 = 400 \, \text{K} \)
\[ \text{KE}_2 = \text{KE}_1 \cdot \frac{T_2}{T_1} = 3.3 \times 10^{-20} \cdot \frac{400}{300} = 4.4 \times 10^{-20} \, \text{J} \]
\[ \text{KE}_2 = \text{KE}_1 \cdot \frac{T_2}{T_1} = 3.3 \times 10^{-20} \cdot \frac{400}{300} = 4.4 \times 10^{-20} \, \text{J} \]
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