If the mean kinetic energy of the molecules of a gas is `(1/3)^(rd)` of its value at `27^(@)C`, then the temperature of the gas will be
If the mean kinetic energy of the molecules of a gas is `(1/3)^(rd)` of its value at `27^(@)C`, then the temperature of the gas will be
A. `100^(@)C`
B. `-173^(@)C`
C. `900^(@)C`
D. `627^(@)C`
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Correct Answer - B
`KE_(2)=(1)/(3)KE_(1)`
`(KE_(2))/(KE_(1))=(T_(2))/(T_(1))" "therefore(1)/(3)=(T_(2))/(T_(1))`
`therefore T_(2)=(T_(1))/(3)=(300)/(3)=100K`
`T_(2)=100-273=-173^(@)C`
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