`color{blue} ✍️`Consider a gas enclosed in a cube of side l. Take the axes to be parallel to the sides of the cube, as shown in Fig. 13.4.
`color{blue} ✍️`A molecule with velocity `(v_x, v_y, v_z )` hits the planar wall parallel to `yz-` plane of area `A (= l^2)`. Since the collision is elastic, the molecule rebounds with the same velocity; its y and z components of velocity do not change in the collision but the x-component reverses sign.
`color{blue} ✍️`That is, the velocity after collision is `(-v_x, v_y, v_z )` . The change in momentum of the molecule is : `–mv_x – (mv_x) = – 2mv_x` . By the principle of conservation of momentum, the momentum imparted to the wall in the collision `= 2mv_x` .
`color{blue} ✍️`To calculate the force (and pressure) on the wall, we need to calculate momentum imparted to the wall per unit time. In a small time interval `Δt`, a molecule with x-component of velocity `v_x` will hit the wall if it is within the distance `v_x Δt` from the wall.
`color{blue} ✍️`That is, all molecules within the volume `Av_x Δt` only can hit the wall in time `Δt`. But, on the average, half of these are moving towards the wall and the other half away from the wall.
`color{blue} ✍️`Thus the number of molecules with velocity `(v_x, v_y, v_z )` hitting the wall in time `Δt` is `½A v_x Δt` n where n is the number of molecules per unit volume. The total momentum transferred to the wall by these molecules in time `Δt` is :
`color{blue} { Q = (2mv_x) ( 1/2 n Av_x Delta t)}`
.........................13.10
`color{blue} ✍️`The force on the wall is the rate of momentum transfer Q/Δt and pressure is force per unit area :
`color{blue} { P = Q // (A Delta t ) = n m_x^2}`
......................(3.11)
`color{blue} ✍️`Actually, all molecules in a gas do not have the same velocity; there is a distribution in velocities. The above equation therefore, stands for pressure due to the group of molecules with speed `vx` in the x-direction and n stands for the number density of that group of molecules. The total pressure is obtained by summing over the contribution due to all groups:
`color{blue} { P = n m bar(v_x^2)}`
...................(13.12)
where `bar(v_x^2)` is the average of `v_x ^2` . Now the gas is isotropic, i.e. there is no preferred direction of velocity of the molecules in the vessel.
`color{blue} ✍️`Therefore, by symmetry,
`color{purple} {bar(v_x^2) = bar(v_y^2) = bar(v_z^2)}`
`color{blue} { = (1/3)} color{blue} { [bar(v_x^2) + bar(v_z^2)] = (1/3) bar(v^2)}`
................(13.13)
`color{blue} ✍️`where `v` is the speed and `bar(v^2)` denotes the mean of the squared speed. Thus
`color{blue} {P = (1//3) nm bar(v^2)} `
.............(13.14)
`color{blue} ✍️`Some remarks on this derivation. First, though we choose the container to be a cube, the shape of the vessel really is immaterial. For a vessel of arbitrary shape, we can always choose a small infinitesimal (planar) area and carry through the steps above.
`color{blue} ✍️`Notice that both A and `Δt ` do not appear in the final result. By Pascal’s law, given in Ch. 10, pressure in one portion of the gas in equilibrium is the same as anywhere else. Second, we have ignored any collisions in the derivation.
`color{blue} ✍️`Though this assumption is difficult to justify rigorously, we can qualitatively see that it will not lead to erroneous results. The number of molecules hitting the wall in time `Δt` was found to be `½ n Av_x Δt`. Now the collisions are random and the gas is in a steady state.
`color{blue} ✍️`Thus, if a molecule with velocity `(v_x, v_y, v_z )` acquires a different velocity due to collision with some molecule, there will always be some other molecule with a different initial velocity which after a collision acquires the velocity `(v_x, v_y, v_z )`.
`color{blue} ✍️` If this were not so, the distribution of velocities would not remain steady. In any case we are finding `bar(v_x^2)` . Thus, on the whole, molecular collisions (if they are not too frequent and the time spent in a collision is negligible compared to time between collisions) will not affect the calculation above.
`color{blue} ✍️`Consider a gas enclosed in a cube of side l. Take the axes to be parallel to the sides of the cube, as shown in Fig. 13.4.
`color{blue} ✍️`A molecule with velocity `(v_x, v_y, v_z )` hits the planar wall parallel to `yz-` plane of area `A (= l^2)`. Since the collision is elastic, the molecule rebounds with the same velocity; its y and z components of velocity do not change in the collision but the x-component reverses sign.
`color{blue} ✍️`That is, the velocity after collision is `(-v_x, v_y, v_z )` . The change in momentum of the molecule is : `–mv_x – (mv_x) = – 2mv_x` . By the principle of conservation of momentum, the momentum imparted to the wall in the collision `= 2mv_x` .
`color{blue} ✍️`To calculate the force (and pressure) on the wall, we need to calculate momentum imparted to the wall per unit time. In a small time interval `Δt`, a molecule with x-component of velocity `v_x` will hit the wall if it is within the distance `v_x Δt` from the wall.
`color{blue} ✍️`That is, all molecules within the volume `Av_x Δt` only can hit the wall in time `Δt`. But, on the average, half of these are moving towards the wall and the other half away from the wall.
`color{blue} ✍️`Thus the number of molecules with velocity `(v_x, v_y, v_z )` hitting the wall in time `Δt` is `½A v_x Δt` n where n is the number of molecules per unit volume. The total momentum transferred to the wall by these molecules in time `Δt` is :
`color{blue} { Q = (2mv_x) ( 1/2 n Av_x Delta t)}`
.........................13.10
`color{blue} ✍️`The force on the wall is the rate of momentum transfer Q/Δt and pressure is force per unit area :
`color{blue} { P = Q // (A Delta t ) = n m_x^2}`
......................(3.11)
`color{blue} ✍️`Actually, all molecules in a gas do not have the same velocity; there is a distribution in velocities. The above equation therefore, stands for pressure due to the group of molecules with speed `vx` in the x-direction and n stands for the number density of that group of molecules. The total pressure is obtained by summing over the contribution due to all groups:
`color{blue} { P = n m bar(v_x^2)}`
...................(13.12)
where `bar(v_x^2)` is the average of `v_x ^2` . Now the gas is isotropic, i.e. there is no preferred direction of velocity of the molecules in the vessel.
`color{blue} ✍️`Therefore, by symmetry,
`color{purple} {bar(v_x^2) = bar(v_y^2) = bar(v_z^2)}`
`color{blue} { = (1/3)} color{blue} { [bar(v_x^2) + bar(v_z^2)] = (1/3) bar(v^2)}`
................(13.13)
`color{blue} ✍️`where `v` is the speed and `bar(v^2)` denotes the mean of the squared speed. Thus
`color{blue} {P = (1//3) nm bar(v^2)} `
.............(13.14)
`color{blue} ✍️`Some remarks on this derivation. First, though we choose the container to be a cube, the shape of the vessel really is immaterial. For a vessel of arbitrary shape, we can always choose a small infinitesimal (planar) area and carry through the steps above.
`color{blue} ✍️`Notice that both A and `Δt ` do not appear in the final result. By Pascal’s law, given in Ch. 10, pressure in one portion of the gas in equilibrium is the same as anywhere else. Second, we have ignored any collisions in the derivation.
`color{blue} ✍️`Though this assumption is difficult to justify rigorously, we can qualitatively see that it will not lead to erroneous results. The number of molecules hitting the wall in time `Δt` was found to be `½ n Av_x Δt`. Now the collisions are random and the gas is in a steady state.
`color{blue} ✍️`Thus, if a molecule with velocity `(v_x, v_y, v_z )` acquires a different velocity due to collision with some molecule, there will always be some other molecule with a different initial velocity which after a collision acquires the velocity `(v_x, v_y, v_z )`.
`color{blue} ✍️` If this were not so, the distribution of velocities would not remain steady. In any case we are finding `bar(v_x^2)` . Thus, on the whole, molecular collisions (if they are not too frequent and the time spent in a collision is negligible compared to time between collisions) will not affect the calculation above.