Two molecules of a gas have speeds of $9 \times 10^6 \mathrm{~ms}^{-1}$ and $1 \times 10^6 \mathrm{~ms}^{-1}$, respectively. What is the root mean square speed of these molecules.
For n-molecules, we know that
$$v_{\mathrm{rms}}=\sqrt{\frac{v_1^2+v_2^2+v_3^2+\ldots \ldots+v_n^2}{n}} \quad\left[\begin{array}{c} \left.v_{\mathrm{rms}}=\begin{array}{l} \text { root mean } \\ \text { square velocity } \end{array}\right] \end{array}\right.$$
where $v_1, v_2, v_3 \ldots \ldots \ldots v_n$ are individual velocities of $n$-molecules of the gas. For two molecules,
$$v_{\text {ms }}=\sqrt{\frac{v_1^2+v_2^2}{2}} \quad\left[v_1, v_2, v_3, \ldots \ldots \ldots v_n \text { are individual velocity }\right]$$
Given, $$\quad v_1=9 \times 10^6 \mathrm{~m} / \mathrm{s}$$
and $$\quad v_2=1 \times 10^6 \mathrm{~m} / \mathrm{s}$$
$$\begin{aligned} \therefore \quad v_{r m s} & =\sqrt{\frac{\left(9 \times 10^6\right)^2+\left(1 \times 10^6\right)^2}{2}} \\ & =\sqrt{\frac{81 \times 10^{12}+1 \times 10^{12}}{2}} \\ & =\sqrt{\frac{(81+1) \times 10^{12}}{2}} \\ & =\sqrt{\frac{82 \times 10^{12}}{2}} \\ & =\sqrt{41} \times 10^6 \mathrm{~m} / \mathrm{s} \end{aligned}$$
A gas mixture consists of 2.0 moles of oxygen and 4.0 moles of neon at temperature $T$. Neglecting all vibrational modes, calculate the total internal energy of the system. (0xygen has two rotational modes.)
$\mathrm{O}_2$ is a diatomic gas having 5 degrees of freedom.
Energy (total internal) per mole of the gas $=\frac{5}{2} R T \quad\left[\begin{array}{l}R=\text { Universal gas constant } \\ T=\text { temperature }\end{array}\right]$
For 2 moles of the gas total internal energy $=2 \times \frac{5}{2} R T=5 R T\quad \text{... (i)}$
Neon $(\mathrm{Ne})$ is a monoatomic gas having 3 degrees of freedom.
$\therefore$ Energy per mole $=\frac{3}{2} R T$
We have 4 moles of Ne .
Hence, Energy $=4 \times \frac{3}{2} R T=6 R T\quad$ ...(ii) [Using Eqs. (i) and (ii)]
$$\begin{array}{rlrl} \therefore \quad & \text { Total energy } & =5 R T+6 R T \\ & =11 R T \end{array}$$
Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1$$\mathop A\limits^o $$ and 2$$\mathop A\limits^o $$. The gases may be considered under identical conditions of temperature, pressure and volume.
$$\begin{aligned} &\text { Mean free path of a molecule is given by }\\ &l=\frac{1}{\sqrt{2} d^2 n} \end{aligned}$$
where, $n=$ number of molecules/ volume
$d=$ diameter of the molecule
Now, we can write $l \propto \frac{1}{d^2}$
Given, $$\quad d_1=1\mathop A\limits^o, d_2=2\mathop A\limits^o$$
As $$l_1 \propto \frac{1}{d_1^2} \text { and } l_2 \propto \frac{1}{d_2^2}$$
$$\begin{aligned} \Rightarrow \text{So},\quad\frac{l_1}{l_2} & =\left(\frac{d_2}{d_1}\right)^2=\left(\frac{2}{1}\right)^2=\frac{4}{1} \\ \text{Hence, } \quad l_1: l_2 & =4: 1 \end{aligned}$$
The container shown in figure has two chambers, separated by a partition, of volumes $V_1=2.0 \mathrm{~L}$ and $V_2=3.0 \mathrm{~L}$. The chambers contain $\mu_1=4.0$ and $\mu_2=5.0$ mole of a gas at pressures $p_1=1.00 \mathrm{~atm}$ and $p_2=2.00 \mathrm{~atm}$. Calculate the pressure after the partition is removed and the mixture attains equilibrium.
Consider the diagram,
Given,
$$\begin{aligned} V_1 & =2.0 \mathrm{~L}, V_2=3.0 \mathrm{~L} \\ \mu_1 & =4.0 \mathrm{~mol}, \mu_2=5.0 \mathrm{~mol} \\ p_1 & =1.00 \mathrm{~atm}, p_2=2.00 \mathrm{~atm} \end{aligned}$$
For chamber $1, p_1, V_1=\mu_1 R T_1$
For chamber 2, $p_2, V_2=\mu_2 R T_2$
When the partition is removed the gases get mixed without any loss of energy. The mixture now attains a common equilibrium pressure and the total volume of the system is sum of the volume of individual chambers $V_1$ and $V_2$.
So, $$\mu=\mu_1+\mu_2, V=V_1+V_2$$
From kinetic theory of gases,
$$\begin{aligned} &\text { For } l \text { mole }\\ &p V=\frac{2}{3} E \quad \text{[E = translational kinetic energy]} \end{aligned}$$
$$\begin{aligned} \text{For $\mu_1$ moles,} \quad & p_1 V_1=\frac{2}{3} \mu_1 E_1 \\ \text{For $\mu_2$ moles, } \quad & p_2 V_2=\frac{2}{3} \mu_2 E_2 \end{aligned}$$
$$\begin{aligned} &\text { Total energy is }\\ &\left(\mu_1 E_1+\mu_2 E_2\right)=\frac{3}{2}\left(p_1 V_1+p_2 V_2\right) \end{aligned}$$
$$\text { From the abvne relation, } \quad p V=\frac{2}{3} E_{\text {total }}=\frac{2}{3} \mu E_{\text {per mole }}$$
$$\begin{aligned} p\left(V_1+V_2\right) & =\frac{2}{3} \times \frac{3}{2}\left(p_1 V_1+p_2 V_2\right) \\ p & =\frac{p_1 V_1+p_2 V_2}{V_1+V_2} \\ & =\left(\frac{1.00 \times 2.0+2.00 \times 3.0}{2.0+3.0}\right) \mathrm{atm} \\ & =\frac{8.0}{5.0}=1.60 \mathrm{~atm} \end{aligned}$$
A gas mixture consists of molecules of $A, B$ and $C$ with masses $m_A>m_B>m_C$. Rank the three types of molecules in decreasing order of (a) average KE (b) rms speeds.
The average KE will be the same, as conditions of temperature and pressure are the same.
Now as,
$$\begin{aligned} V_{\text {rms }} & =\sqrt{\frac{3 p V}{M}}=\sqrt{\frac{3 R T}{M}} \\ & =\sqrt{\frac{3 R T}{m N}}=\sqrt{\frac{3 k T}{m}} \end{aligned}$$
where, M = molar mass of the gas
m = mass of each molecular of the gas,
R = gas constant
Clearly, $$v_{\mathrm{rms}} \propto \sqrt{\frac{1}{m}}$$
(b) As $k=$ Boltzmann constant
$T=$ absolute temperature (same for all)
But $m_A>m_B>m_C$
$\left(v_{\text {rms }}\right)_A<\left(v_{\text {rms }}\right)_B<\left(v_{\text {rms }}\right)_C$
$\therefore$ or $\left(v_{\mathrm{rms}}\right)_C>\left(v_{\mathrm{rms}}\right)_B>\left(v_{\mathrm{rms}}\right)_A$