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Question

The average kinetic energy of a gas molecule at $${27}^{o}C$$ is $$6.21\times {10}^{-21}J$$, then its average kinetic energy at $${227}^{o}C$$ is:


A
10.35×1021J
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B
11.35×1021J
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C
52.2×1021J
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D
5.22×1021J
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Solution

The correct option is B $$10.35\times {10}^{-21}J$$
Average kinetic energy of gas molecules $$\propto$$ Temperature (Absolute)
$$\cfrac{K.E(at\quad {227}^{o}C)}{K.E (at\quad {27}^{o}C)}=\cfrac{273+227}{273+27}=\cfrac{500}{300}=\cfrac{5}{3}$$
$$K.E({227}^{o})=\cfrac{5}{3}\times 6.21\times {10}^{-21}J=10.35\times {10}^{-21}J$$

Physics

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