Question

In a cubical vessel are enclosed n molecule of a gas each having mass 'm' and an average speed 'v'. if ${}^{\prime }{l}^{\prime }$ is the length of each edge on the cube, the pressure exerted by the gas will be

• A. $\frac{nm{v}^2}{l^3}$
• B. $\frac{n{m}^{2}v}{2l^3}$
• C. $\frac{mn{v}^2}{3l^3}$
• D. $\frac{nmv}{2l}$

Answer 1

The correct option is C $\frac{mn{v}^2}{3l^3}$

The momentum imparted per unit time to the wall by molecule will be

$\Delta F$=$n\Delta p$

so change in momentum will be$\Delta p$ $=2m{v}_x$

and molecule will collide form the two surfaces of wall so total time $\Delta t$ $=\frac{2L}{v_x}$

so number of collision by molecule in unit time will be n $=\frac{1}{\Delta t}$ $=\frac{v_x}{2L}$

so $F$ $=\frac{m{v}_x^2}{L}$

now for cube all the direction are equivalent,we have

$F=\frac{mn{v}^2}{3ln}$(i.e $\frac{v^2}{n}$=$v^2$ it is the sum of three velocity in x,y,z direction)

so pressure $p=\frac{F}{l^2}$

$p=\frac{mn{v}^2}{3l^3}$