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The temperature of a gas rises from 27^@...

The temperature of a gas rises from `27^@C` to `327^@C`. Show that the rms speed of the gas molecules would be `sqrt2` times its initial value at the final temperature.

Text Solution

Verified by Experts

`T_1=27^@C=(27+273)K=300K`,
`T_2=327^@C=(327+273)K=600K`,
As `c prop sqrtT`, we have `c_1/c_2=sqrt(T_1/T_2)`
or,`c_2=c_1sqrt(T_2/T_1)=c_1sqrt(600/300)=sqrt2c_1` (proved)
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