Calculate the temperature at which r.m.s velocity of gas molecules is double its value at `27^(@)C`, pressure of the gas remaining the same.

Answer

927

Explanation

Let `t^@`C, be the temperature at which the r.m.s. velocity `(v_1)` of the gas molecules is double its value at `27^@C(v_(27))`
As `v prop sqrtT therefore (v_1)/(v_(27))=sqrt((273+t)/(273+27))`
But `v_t=2v_(27)` or `(v_t)/(v_(27))=2`
`therefore sqrt((273+t)/(300))=2` or `(273+t)/(300)=4`
or t=`927^@`C