The molecules of a given mass of a gas have root mean square speeds of $100 \mathrm{~ms}^{-1}$ at $27^{\circ} \mathrm{C}$ and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at $127^{\circ} \mathrm{C}$ and 2.0 atmospheric pressure?
We know that for a given mass of a gas
$$v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$$
where, $R$ is gas constant $T$ is temperature in kelvin $M$ is molar mass of the gas.
Clearly, $$\quad v_{\mathrm{rms}} \propto \sqrt{T}$$
$$\begin{aligned} &\text { As } R, M \text { are constants, }\\ &\frac{\left(v_{\mathrm{rms}}\right)_1}{\left(v_{\mathrm{rms}}\right)_2}=\sqrt{\frac{T_1}{T_2}} \end{aligned}$$
$$\begin{aligned} &\begin{aligned} \text { Given, }\quad \left(v_{\text {rms }}\right)_1 & =100 \mathrm{~m} / \mathrm{s} \\ T_1=27^{\circ} \mathrm{C} & =27+273=300 \mathrm{~K} \\ T_2=127^{\circ} \mathrm{C} & =127+273=400 \mathrm{~K} \end{aligned} \end{aligned}$$
$$\therefore$$ From Eq. (i)
$$\begin{aligned} & \frac{100}{\left(v_{\mathrm{rms}}\right)_2}=\sqrt{\frac{300}{400}}=\frac{\sqrt{3}}{2} \\ \Rightarrow \quad & \left(v_{\mathrm{rms}}\right)_2=\frac{2 \times 100}{\sqrt{3}}=\frac{200}{\sqrt{3}} \mathrm{~m} / \mathrm{s} \end{aligned}$$
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,
$$\begin{aligned} &\text { For } l \text { mole }\\ &p V=\frac{2}{3} E \quad \text{[E = translational kinetic energy]} \end{aligned}$$
$$\begin{aligned} \text{For $\mu_1$ moles,} \quad & p_1 V_1=\frac{2}{3} \mu_1 E_1 \\ \text{For $\mu_2$ moles, } \quad & p_2 V_2=\frac{2}{3} \mu_2 E_2 \end{aligned}$$
$$\begin{aligned} &\text { Total energy is }\\ &\left(\mu_1 E_1+\mu_2 E_2\right)=\frac{3}{2}\left(p_1 V_1+p_2 V_2\right) \end{aligned}$$
$$\text { From the abvne relation, } \quad p V=\frac{2}{3} E_{\text {total }}=\frac{2}{3} \mu E_{\text {per mole }}$$
$$\begin{aligned} p\left(V_1+V_2\right) & =\frac{2}{3} \times \frac{3}{2}\left(p_1 V_1+p_2 V_2\right) \\ p & =\frac{p_1 V_1+p_2 V_2}{V_1+V_2} \\ & =\left(\frac{1.00 \times 2.0+2.00 \times 3.0}{2.0+3.0}\right) \mathrm{atm} \\ & =\frac{8.0}{5.0}=1.60 \mathrm{~atm} \end{aligned}$$