3.0 Expression For Pressure

  Kinetic Theory of Gases

Consider a cubical container of edge $l$ in which an ideal gas is kept. Let the number of molecules of gas be $n$ and mass of each gas molecule be $m$.

Consider motion of one gas molecule along $x$-axis. This gas molecule is moving along the $x$-axis with speed ${C_x}$. The initial momentum of the gas molecule is $$\overrightarrow {p_i} = m{C_x}\hat i$$ This gas molecule collides with wall $A$ and returns back with same speed, because the collision is perfectly elastic. Thus the final momentum of the gas molecule is $$\overrightarrow {p_f} = m{C_x}( - \hat i)$$

Thus, momentum change of one gas molecule is $$\overrightarrow {\Delta p} = \overrightarrow {p_f} - \overrightarrow {p_i} = 2m{C_x}( - \hat i)$$

Now, time taken by gas molecule to collide successively with the same wall ($A$) is $$\frac{{C_x}}{{2l}}$$$$t = \frac{{l + l}}{{{C_x}}} = \frac{{2l}}{{{C_x}}}$$

Thus, the number of collision made by gas molecule with the same wall $A$, in unit time is $$ = \frac{{C_x}}{{2l}}$$ Momentum change per second = $(2m{C_x})\frac{{C_x}}{{2l}}( - \hat i)$

$$\therefore \left. {\left| {\overrightarrow {Fx} } \right.} \right| = \frac{{mC_x^2}}{l}$$

$$\therefore \left. {\left| {\overrightarrow {Fx} } \right.} \right| = Pressure\ developed\ ={ \frac{{{F_x}}}{A} = \frac{{mC_x^2}}{{{{\left| x \right|}^2}}} = \frac{{mC_x^2}}{{{l^3}}} } $$$\therefore $ Total pressure developed due to movement of $m$ molecules in $x$-direction is $$\begin{equation} \begin{aligned} \sum {{P_x}} = \frac{m}{{{l^3}}}\left[ {C_{1x}^2 + C_{2x}^2 + .....C_{Nx}^2} \right] \\ \\\end{aligned} \end{equation} $$

Similarly,

$$\begin{equation} \begin{aligned} \sum {{P_y}} = \frac{m}{{{l^3}}}\left[ {C_{1y}^2} \right. + C_{2y}^2 + ......... + \left. {C_{Ny}^2} \right] \\ \sum {{P_z}} = \frac{m}{{{l^3}}}\left[ {C_{1z}^2} \right. + C_{2z}^2 + ......... + \left. {C_{Nz}^2} \right] \\\end{aligned} \end{equation} $$

Thus, average pressure of gas will be

$$\begin{equation} \begin{aligned} P = \frac{{\sum {{P_x}} + \sum {{P_y}} + \sum {{P_z}} }}{3} \\ \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} \therefore P = \frac{m}{{3{l^3}}}\left[ {(C_{1x}^2} \right. + C_{1y}^2 + C_{1z}^2) + (C_{2x}^2 + C_{2y}^2 + C_{2z}^2) + .......\left. {(C_{Nx}^2 + C_{Ny}^2 + C_{Nz}^2)} \right] \\ \\ \therefore P = \frac{m}{{3V}}[C_1^2 + C_2^2 + ...... + C_N^2] \\ \\\end{aligned} \end{equation} $$$$\left[ {\because V = {l^3} } \right]$$

or, $$P = \frac{m}{{3V}}N\overline {{C^2}} $$

where,

$$\overline {{C^2}} =\ mean\ square\ speed\ = \frac{{C_1^2 + C_2^2 + ...... + C_N^2}}{N}$$

Also, ${C_{rms}} = $ Root mean square speed $$\begin{equation} \begin{aligned} = \sqrt {\overline {{C^2}} } = \sqrt {\frac{{C_1^2 + C_2^2 + ...... + C_N^2}}{N}} \\ \\\end{aligned} \end{equation} $$$$\therefore \overline {{C^2}} = {C_{rms}}$$

Thus Expression for pressure is, $$P = \frac{1}{3}\frac{{mN}}{V}C_{rms}^2 $$

The Kinetic equation of gas is, $$PV = \frac{1}{3}mNC_{rms}^2 $$

$$\because PV = \frac{1}{3}mNC_{rms}^2 = \frac{1}{3}MC_{rms}^2$$

where, $M$ = Total mass of gas molecules =$ m \times N$

$$\begin{equation} \begin{aligned} \therefore P = \frac{1}{3}\frac{M}{V}C_{rms}^2 = \frac{1}{3}\rho C_{rms}^2 \\ \\\end{aligned} \end{equation} $$

where, $\rho = \frac{M}{V} = $ Density of gas