Question

Pressure of an ideal gas is obtained from kinetic gas equation. The kinetic gas equation is: $PV=\frac{1}{3}\left(m{N}_A\right)u^2=\frac{1}{3}M{u}^2$, where, $m{N}_A=M$(molar mass)
$N_A=\text{Avogadro's number}$
$u=\text{root mean square velocity}$
$\text{Translational kinetic energy of n mole}\frac{1}{2}M{u}^2=\frac{3}{2}\left(PV\right)=\frac{3}{2}\left(nRT\right)$
$\text{Average translational kinetic energy per molecule}=\frac{3}{2}\left(\frac{RT}{N_A}\right)=\frac{3}{2}\left(KT\right)$
$u_{rms}=\sqrt{\frac{3PV}{M}}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3P}{d}}$
and $u_{av}=\sqrt{\frac{8RT}{\pi M}}$
What is the rms speed of gas molecule of mass ${10}^{-12}$ gm at room temperature ${27}^{\circ }\text{C}$ according to kinetic theory of gases?
$(\text{Given}R=8J{K}^{-1}mo{l}^{-1},N_A=6\times {10}^{23})$

• A. $0.35\text{cm/s}$
• B. $0.70\text{cm/s}$
• C. $0.70\text{m/s}$
• D. $0.35\text{m/s}$

Answer 1

The correct option is A $0.35\text{cm/s}$

$\text{Kinetic energy of gas per molecule}=\frac{3RT}{2N_A}$
$\text{Therefore,}\frac{1}{2}m{u}^2=\frac{3RT}{2N_A}\text{(where u is rms speed)}$
$\Rightarrow u=\sqrt{\frac{3RT}{m{N}_A}}$
$\Rightarrow u=\sqrt{\frac{3\times 8\times 300}{{10}^{-12}\times {10}^{-3}\times 6\times {10}^{23}}}=0.35\text{cm/s}$