Graph shows a hypothetical speed distribution for a sample of `N` gas particle :- (for`VgtV_(0), (dN)/(dV)= 0`)

Options

A. The value of a`V_(0)` is `2N`.
B. The ratio `V_("avg")//V_(0)` is equal to `2//3`.
C. The ratio `V_("avg")//V_(0)` is equal to `1//sqrt(2)`.
D. Three fourth of the total particle has a speed between `0.5 V_(0)` and `V_(0)`.

Answer

A, B, C, D

Explanation

Area under the curves is equal to number of molecules of the gas sample.
Hence `N= (1)/(2)aV_(0)implies aV_(0) = 2N`
`V_(avg) = (1)/(N)underset(0)overset(oo)intvN(V)dV = (1)/(N)underset(0)overset(V_(0))intC .((a)/(V_(0)V))dV = (2)/(3)V_(0)`
`(V_(avg))/(V_(0)) = (2)/(3)`
`V_(rms)^(2) = (1)/(N)underset(0)overset(oo)intV^(2)N(V)dV = (1)/(N)underset(0)overset(V_(0))intV^(2) .((a)/(V_(0)V))dV = (V_(0)^(2))/(2)`
`implies (V_(rms))/(V_(0)) = (1)/(sqrt(2))`
Area under the curve from `0.5 V_(0)` to `V_(0)` is `3//4` of total area