Kinetic Theory of Gases
5.0 Ideal Gas Laws
5.0 Ideal Gas Laws
1. Boyle's Law:
For a given mass of a gas at constant temperature, the volume of gas is inversely proportional to its temperature.
$$\begin{equation} \begin{aligned} V \propto \frac{1}{P} \\ \\\end{aligned} \end{equation} $$ or $$PV = constant$$
2. Charle's Law:
For a given mass of a gas at constant pressure the volume of gas is directly proportional to its temperature.
$$V \propto T$$ or $$\frac{V}{T} = \;constant$$
3. Avagadro's Law:
At same temperature and pressure equal volumes of all gases contain equal number of molecules.
if $P$, $V$ and $T$ are same $${N_1} = {N_2}$$
4. Dalton's Law:
The pressure exerted by a mixture of several gases is equal to the sum of pressure exerted by each component of gas present in mixture.
i.e., $${P_{mix}} = {P_1} + {P_2} + {P_3}$$
5. Graham's Law of Diffusion:
At same temperature and pressure the rate of diffusion of gas is inversely proportional to the square root of the density of gas.
Rate of diffusion $$\begin{equation} \begin{aligned} {r_d} \propto \frac{1}{{\sqrt P }} \\ \\\end{aligned} \end{equation} $$
Question 1. A container has volume 10 L. It is filled with air at room temperature and atmospheric pressure. Given that the air contains 21% oxygen, find out the number of moles of oxygen in the container.
Solution: Assume that the air is an ideal gas. The formula $PV = N RT$ gives,
$$\begin{equation} \begin{aligned} N = \frac{{PV}}{{RT}} = \frac{{(1.01 \times {{10}^5})(10 \times {{10}^{ - 3}})}}{{8.3 \times 298}} \\ \\\end{aligned} \end{equation} $$
where $N=0.408$ moles of air
Hence, there are $0.408 \times \frac{{21}}{{100}} = 0.0857$ moles of oxygen gas in the container.
Question 2. An ideal gas obeying the equation is at STP. If volume of gas changes from $V$ to $8V$, find out the final temperature of the gas.
Solution: Given, $$T\sqrt P = \;constant$$
$$ \Rightarrow {T^2}P = \;constant\ \ \ ...(i)$$
According to ideal gas equation $PV = N RT\ \ \ ...(ii)$
$$P = \frac{{NRT}}{V}\ \ \ ...(iii)$$
From (i) & (ii)
$$\begin{equation} \begin{aligned} {T^2} \times \frac{{N RT}}{V} = constant \Rightarrow \frac{{{T^3}}}{V} = constant \\ \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} \therefore \frac{{T_1^3}}{{{V_1}}} = \frac{{T_2^3}}{{{V_2}}} \Rightarrow \frac{{{{273}^3}}}{V} = \frac{{T_f^3}}{{8V}} \Rightarrow T_f^3 = 8{(273)^3} \\ \\\end{aligned} \end{equation} $$$${T_f} = 2 \times 273 = 546K$$
Question 3. The P-V diagram for one mole of an ideal gas is shown in figure. Find the pressure and volume of the gas when its temperature is maximum.
Solution: The equation of line $AB$ can be written as
$$\begin{equation} \begin{aligned} \frac{{P - {P_A}}}{{{P_B} - {P_A}}} = \frac{{V - {V_A}}}{{{V_B} - {V_A}}} \\ \\ \Rightarrow \frac{{P - 3{P_0}}}{{{P_0} - 3{P_0}}} = \frac{{V - {V_0}}}{{2{V_0} - {V_0}}} \Rightarrow \frac{{P - 3{P_0}}}{{ - 2{P_0}}} = \frac{{V - {V_0}}}{{{V_0}}} \\ \\\end{aligned} \end{equation} $$$$\Rightarrow P = 3{P_0} - \frac{{2{P_0}}}{{{V_0}}}(V - {V_0}) = 3{P_0} - \frac{{2{P_0}V}}{{{V_0}}} + 2{P_0} $$$$ \Rightarrow P = 5{P_0} - \frac{{2{P_0}V}}{{{V_0}}}\ \ \ ...(i)$$
$\therefore $ from equation (i)
$$\begin{equation} \begin{aligned} P = \frac{{RT}}{V} = 5{P_0} - \frac{{2{P_0}V}}{{{V_0}}} \\ \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} \Rightarrow T = \left[ {5{P_0}V - \frac{{2{P_0}{V^2}}}{{{V_0}}}} \right]\frac{1}{R} \\ \\ \Rightarrow T = \frac{{{P_0}}}{R}\left[ {5V - \frac{{2{V^2}}}{{{V_0}}}} \right] \\ \\\end{aligned} \end{equation} $$
For $T$ to be maximum, $$\begin{equation} \begin{aligned} \left[ {\frac{{dT}}{{dV}} = 0,\;\frac{{{d^2}T}}{{{d^2}V}} < 0} \right] \\ \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} \frac{{dT}}{{dV}} = 0 \Rightarrow \frac{{{P_0}}}{R}\left[ {5 - \frac{{4V}}{{{V_0}}}} \right] \\ \\ {\text{ }} \Rightarrow V = \frac{{5{V_0}}}{4} \\ \\ \frac{{{d^2}T}}{{{d^2}V}} = \frac{{{P_0}}}{R}\left[ {\frac{{ - 4}}{{{V_0}}}} \right] \\ \\\end{aligned} \end{equation} $$
Thus maximum temperature occur when $$\begin{equation} \begin{aligned} V = \frac{{5{V_0}}}{4} \\ \\\end{aligned} \end{equation} $$
Now, $$\begin{equation} \begin{aligned} {T_{\max }} = \frac{{{P_0}}}{R}\left[ {\frac{{5 \times 5{V_0}}}{4} - \frac{2}{{{V_0}}} \times \frac{{25V_0^2}}{{16}}} \right] = \frac{{25{P_0}{V_0}}}{R}\left[ {\frac{1}{4} - \frac{1}{8}} \right] \\ \\ {T_{\max}} = \frac{{25}}{8}\frac{{{P_0}{V_0}}}{R} \\ \\ \therefore P = \frac{{N RT}}{V} = \frac{{R \times 25{P_0}{V_0} \times 4}}{{8R \times 5{V_0}}} = \frac{5}{2}{P_0} \\ \\\end{aligned} \end{equation} $$
Thus $P = 2.5{P_0}$ and $V = 1.25{V_0}$
