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 $\left(v_{\mathrm{rms}}\right)_C>\left(v_{\mathrm{rms}}\right)_B>\left(v_{\mathrm{rms}}\right)_A$
We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm . Is one justified in assuming the ideal gas law, in the final state? (Hydrogen molecules can be consider as spheres of radius $1 \mathop A\limits^o$ ).
Assuming hydrogen molecules as spheres of radius $1 \mathop A\limits^o$.
So, $r=1 \mathop A\limits^o=$ radius.
$$\begin{aligned} \text { The volume of hydrogen molecules } & =\frac{4}{3} \pi r^3 \\ & =\frac{4}{3}(3.14)\left(10^{-10}\right)^3 \\ & \approx 4 \times 10^{-30} \mathrm{~m}^3 \end{aligned}$$
$$\begin{aligned} \text { Number of moles of } \quad \mathrm{H}_2 & =\frac{\text { Mass }}{\text { Molecular mass }} \\ & =\frac{0.5}{2}=0.25 \end{aligned}$$
$$\begin{aligned} & \text { Molecules of } \mathrm{H}_2 \text { present }=\text { Number of moles of } \mathrm{H}_2 \text { present } \times 6.023 \times 10^{23} \\ &=0.25 \times 6.023 \times 10^{23} \\ & \therefore \text { Volume of molecules present }=\text { Molecules number } \times \text { volume of each molecule } \\ &=0.25 \times 6.023 \times 10^{23} \times 4 \times 10^{-30} \\ &=6.023 \times 10^{23} \times 10^{-30} \\ & \approx 6 \times 10^{-7} \mathrm{~m}^3 \quad \text{... (i)} \end{aligned}$$
$$\begin{aligned} &\text { Now, if ideal gas law is considered to be constant. }\\ &\begin{aligned} p_i V_i & =p_f V_f \\ V_f & =\left(\frac{p_i}{p_f}\right) V_i=\left(\frac{1}{100}\right)\left(3 \times 10^{-2}\right)^3 \\ & =\frac{27 \times 10^{-6}}{10^2} \\ & =2.7 \times 10^{-7} \mathrm{~m}^3\quad \text{... (ii)} \end{aligned} \end{aligned}$$
Hence, on compression the volume of the gas is of the order of the molecular volume [form Eq.(i) and Eq.(ii)]. The intermolecular forces will play role and the gas will deviate from ideal gas behaviour.
When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased. What about Boyle's law in this case?
When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules remains constant. In this case, as the number of air molecules keep increasing. Hence, this is a case of variable mass. Boyle's law (and even Charle's law) is only applicable in situations, where number of gas molecules remains fixed. Hence, in this case Boyle's law is not applicable.
A balloon has 5.0 mole of helium at $7^{\circ} \mathrm{C}$. Calculate
(a) the number of atoms of helium in the balloon.
(b) the total internal energy of the system.
$$\begin{aligned} &\text { Given, number of moles of helium }=5\\ &T=7^{\circ} \mathrm{C}=7+273=280 \mathrm{~K} \end{aligned}$$
$$\begin{aligned} &\text { (a) Hence, number of atoms (He is monoatomic) }\\ &\begin{aligned} & =\text { Number of moles } \times \text { Avogadro's number } \\ & =5 \times 6.023 \times 10^{23} \\ & =30.015 \times 10^{23} \\ & =3.0 \times 10^{24} \text { atoms } \end{aligned} \end{aligned}$$
$$\text { (b) Now, average kinetic energy per molecule }=\frac{3}{2} k_B T$$
Here, $k_B=$ Boltzmann constant. $$\quad$$ (It has only 3 degrees of freedom)
$$\begin{aligned} &\therefore \text { Total energy of all the atoms }\\ &\begin{aligned} & =\text { Total internal energy } \\ & =\frac{3}{2} k_b T \times \text { number of atoms } \\ & =\frac{3}{2} \times 1.38 \times 10^{-23} \times 280 \times 3.0 \times 10^{24} \\ & =1.74 \times 10^4 \mathrm{~J} \end{aligned} \end{aligned}$$