Question

Under 3 $atm$, 12.5 litres of a certain gas weighs 15 $g$. The average speed of gaseous molecules is $x\times {10}^{4}cmse{c}^{-1}$.
Value of $x$ is......(only integer part)

Answer 1

$u_{AV}=\sqrt{\left[\right(8RT\left)/\pi M\right]}$ ...(i)
From gas equation $PV=\left(w/N\right)RT$
Given, $P=3atm,V=12.5litres,w=15g$
$\therefore 3\times 12.5=\left(15/M\right)\times 0.0821\times T$
$\therefore T/M=30.45$ ...(ii)
By Eqs. (i) and (ii)
$u_{AV}=2.05\times {10}^{4}cmse{c}^{-1}$
So, value of $x$ is 2.