Gas molecules each of mass `10^(-26)` kg are taken in a container of volume `1 dm^(3)` . The root mean square speed of gas molecules is 1 km `sec^(-1)` .What is the temperature fo gas molecules.
(Given : `N_(A) =6xx10^(23),R=8J//mol.K`)

Options

A. 298 K
B. 25 K
C. 250 K
D. 2500 K

Answer

c

Explanation

(c) `1000=sqrt((3RT)/(M)),10^(3)=[(3xx8.314xxT)/(10^(-26)xx6xx10^(23))]^(1//2)`
` T=(10^(6)xx10^(-26)xx6xx10^(23))/(3xx8)=250K`