Question

For gaseous state, if the most probable speed is denoted by $C^{\ast }$, average speed by $\overline{C}$ and root mean square speed by $C$, then for a large number of molecules, the ratios of these speeds are:

• A. $\begin{array}{ccc}{C}^{\ast }& \overline{C}& C\\ 1.225& 1.128& 1\end{array}$
• B. $\begin{array}{ccc}{C}^{\ast }& \overline{C}& C\\ 1& 1.128& 1.225\end{array}$
• C. $\begin{array}{ccc}{C}^{\ast }& \overline{C}& C\\ 1.128& 1.225& 1\end{array}$
• D. $\begin{array}{ccc}{C}^{\ast }& \overline{C}& C\\ 1& 1.125& 1.128\end{array}$

Answer 1

The correct option is B $\begin{array}{ccc}{C}^{\ast }& \overline{C}& C\\ 1& 1.128& 1.225\end{array}$

Using thr formula,
$C^{\ast }=\sqrt{\frac{2RT}{M}};\overline{C}=\sqrt{\frac{8RT}{\pi M}};C=\sqrt{\frac{3RT}{M}}$
$C^{\ast }:\overline{C}:C=\sqrt{2}:\sqrt{\frac{8}{\pi }}:\sqrt{3}=1.414:1.596:1.732$
$=1:1.128:1.225$

Theory:

Kinetic gas equation :

$PV=\frac{1}{3}mN{\overline{u}}^2$

$Volumeofcontainer\left(V\right)=l^3$

Verification of gas laws with kinetic gas equation :

$PV=\frac{1}{3}mN{\overline{u}}^2$.......................(1)

According to Kinetic theory of gases :

$\frac{1}{2}m{\overline{u}}_{rms}^2=\lambda T\left(inkelvin\right)$.................(2)

where $\lambda$ is a proportionality constant.

Multiply and divide equation 1 by 2 on both sides :

$PV=\frac{2}{3}\frac{1}{2}mN{\overline{u}}^2$................(3)

From equation (2) $\frac{1}{2}m{\overline{u}}_{rms}^2=\lambda T$.......(4)

Putting the value of equation 4 in equation 3 :

$PV=\frac{2}{3}\left(\lambda T\right)N$

$P\propto \frac{1}{V}$ which is Boyle's law at constant N and T.

Similarly we can prove charles’ law from kinetic gas equation :

At constant N and Pressure(P):

$V\propto T$ and similarly it can be done at constant N and Volume (V) which is Gay lussac’s law :

$P\propto T$

At constant temperature (T) and Pressure (P) Kinetic gas equation can prove Avogadro’s law :

$V\propto N$ or $V=k_{3}N$

Dividing and multiplying by $N_A$

$V=\frac{k_3}{N_A}\left(\frac{N}{N_A}\right)$ or $V=k_4\left(\frac{N}{N_A}\right)$

Calculation of $\lambda$ :

From ideal gases :

$PV=nRT=\frac{2}{3}\lambda NT$........(1)

Where $n=numberofmolesofgasand{N}_A=Avogadr{o}^{\prime }snumber$

Since $numberofmoles\left(n\right)=\frac{Givennumberofmolecules\left(N\right)}{Avogadr{o}^{\prime }snumber\left(N_A\right)}$

$N=n{N}_A$...........(2)

Putting the value of N from 2 in 1 :

$PV=nRT=\frac{2}{3}\left(n{N}_A\right)\lambda T$

$\lambda =\frac{3}{2}\frac{R}{N_A}$ or $\lambda =\frac{3}{2}k$ where k is Boltzmann constant

$k\left(Boltzmannconstant\right)=\frac{R}{N_A}$

Average kinetic energy of molecules :

$\frac{1}{2}m{\overline{u}}_{rms}^2\propto T\left(inkelvin\right)$or $\frac{1}{2}m{\overline{u}}_{rms}^2=\lambda T\left(inkelvin\right)$ where $\lambda$ is a proportionality constant.

$\lambda T=\frac{3}{2}kT$ where k is Boltzmann constant

$k=1.38\times {10}^{-23}J{K}^{-}1$

Total Kinetic energy per mole :

$Totalkineticenergy=N_A\left(\frac{1}{2}m{\overline{u}}_{rms}^2\right)=N_A\left(\lambda T\right)$

$=\frac{3}{2}\left(\frac{R}{N_A}\right)N_A=\frac{3}{2}R$

Root mean square speed $V_{rms}$ :

$Totalkineticenergy=\frac{1}{2}M{\overline{u}}_{rms}^2=\frac{3}{2}RT$

${\overline{u}}_{rms}^2=\frac{3RT}{M}$ or ${\overline{u}}_{rms}=\sqrt[2]{2\frac{3RT}{M}}$

Unit of $u_{rms}$:

$\text{Unit of}{u}_{rms}=\sqrt{\frac{\left(J{K}^{-1}mo{l}^{-1}\right)\left(K\right)}{kgmo{l}^{-1}}}$

$\text{Unit of}{u}_{rms}=\sqrt{\frac{J}{kg}}$

$\text{Unit of}{u}_{rms}=\sqrt{\frac{kg{m}^2{s}^{-2}}{kg}}$

$\text{Unit of}{u}_{rms}=m{s}^{-1}$

Note: Always take $R$ in $J{K}^{-1}mo{l}^{-1}$ and $M$ in $kgmo{l}^{-1}$