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 of gas molecules?(Given: `N_(A)= 6xx 10^(23), R= 8J"/"mol.K`).

Options

A. 298 K
B. 25 K
C. 240 K
D. 2500 K

Answer

C

Explanation

`1000 = sqrt((3RT)/(M)) , 10^(3) = [(3 xx 8.314 xx T)/(10^(-26) xx 6 xx 10^(23))]^(1//2)`
` T = (10^(6) xx 10^(-26) xx 6 xx 10^(23))/(3 xx 8.314) = 240 K`