Pressure Exerted By A Gas

​In a vessel containing gas, the gaseous molecules are moving in all possible directions with very high velocity and are colliding with the walls of the vessel. Due to the continuous bombardment of the gaseous molecules to the walls of the vessel, the gas exerts pressure.

​Consider a cubical vessel of each side of length l containing gas. Let there be n number of gaseous molecules inside the vessel and let m be the mass of each molecule. Let the n molecules be moving with velocities c1, c2, c3, …, cn respectively. Then, the velocity c1 of molecule1 can be resolved into three components u1, v1 and w1 along X, Y and Z-axis respectively as shown in figure.

Pressure Exerted By A Gas

​Then, \[{c_1}^2={u_1}²+{v_1}²+{w_1}²\] Similarly, for other gas molecules, \[{c_2}^2={u_2}²+{v_2}²+{w_2}²\] \[{c_3}^2={u_3}²+{v_3}²+{w_3}²\] \[—————–\] \[{c_n}^2={u_n}²+{v_n}²+{w_n}²\] Let the molecule1 strike the surface of the cube along X-axis. Then, the momentum with which the molecule1 strikes is mu1. Since the collision is perfectly elastic in nature, the molecule1 rebounds back with the same velocity u1 after striking the surface and the momentum becomes -mu1. Change of momentum of the molecule1 along X-axis \[=mu_1-(-mu_1)\] \[=2mu_1\] Here, the molecule1 strikes the surface and covers a distance l. When it rebounds back and strikes the other opposite surface, it again covers the same distance l. So, the molecule1 covers the total distance of 2l. Time between two successive collision \[=\frac{\text{Distance covered}}{Velocity}\] \[=\frac{2l}{u_1}\] Rate of change in momentum of molecule1 along X-axis \[=\frac{\text{Change in Momentum}}{\text{Time Taken}}\] \[=2mu_1×\frac{u_1}{2l}\] \[=\frac{m{u_1}²}{l}\]

​Thus, force exerted by molecule1 along X-axis \[=\frac{m{u_1}²}{l}\] Force exerted by all the gas molecules along X-axis \[F_x=\frac{m{u_1}²}{l}+\frac{m{u_2}²}{l}+…+\frac{m{u_n}²}{l}\] \[F_x=\frac{m}{l}({u_1}²+{u_2}²+…+{u_n}²)\] Pressure exerted by the gas molecules in the direction of X-axis \[P_x=\frac{F_x}{l²}\] \[P_x=\frac{m}{l³}({u_1}²+{u_2}²+…+{u_n}²)\] Similarly, pressure exerted in the direction of Y-axis and Z-axis, \[P_y=\frac{m}{l³}({v_1}²+{v_2}²+…+{v_n}²)\] \[P_z=\frac{m}{l³}({w_1}²+{w_2}²+…+{w_n}²)\]

​Thus, average pressure exerted by all gas molecules, \[P=\frac{1}{3}(P_x+P_y+P_z)\] \[P=\frac{1}{3}\frac{m}{l³}[({u_1}²+{u_2}²+…+{u_n}²)+({v_1}²+{v_2}²+…+{v_n}²)+({w_1}²+{w_2}²+…+{w_n}²)]\] \[P=\frac{1}{3}\frac{m}{l³}[({u_1}²+{v_1}²+{w_1}²)+({u_2}²+{v_2}²+{w_2}²)+…+({u_n}²+{v_n}²+{w_n}²)]\] \[P=\frac{1}{3}\frac{m}{l³}[{c_1}²+{c_2}²+…+{c_n}²]\] \[P=\frac{1}{3}\frac{mn}{l³}\left[\frac{{c_1}²+{c_2}²+…+{c_n}²}{n}\right]\text{___(1)}\] Now, \[\left[\frac{{c_1}²+{c_2}²+…+{c_n}²}{n}\right]=\bar{c}²[\bar{c}²=\text{Mean Square Velocity of gas}]\] \[mn=M [M=\text{Total mass of the gas}]\] \[l³=V [V=\text{Volume of the gas}]\] Then, equation (1) becomes, \[P=\frac{1}{3}\frac{M}{V}\bar{c}²\text{___(2)}\] Also, \[\frac{M}{V}=ρ\text{[Density of the gas]}\] ∴Equation (2) becomes, \[P=\frac{1}{3}ρ\bar{c}²\] This is the pressure exerted by the gas molecules on the walls of the vessel.

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