Correct Answer - A
Given, p = 0.09 kg `m^(-3)`
Pressure, p (at STP) = `1.01xx10^(5)` Pa
According to kinetic theory og gas `p=(1//3)pv^(3)`
`rArr C=sqrt((3p)/(p))=sqrt((3xx1.01xx10^(5))/(0.09))=1837.5_(rms^(-1))`
Now, volume occupied by one mole of hydrogen at STP
`22.4L=22.4xx10^(-3)m^(3)`
`therefor` Mass of hydrogen = volume `xx` density
`=22.4xx10^(-3)xx0.09`
`=0.016xx10^(-3)kg`
`therefore` Average KE per mol `=1/2mc^(2)=1/2xx2.016xx10^(-3)xx(1837.5)^(2)`
= 3403.4J