What do you think the pressure is exerted by the gas on the walls of the container in all directions?

Text Solution

Solution : Expression for pressure exerted by a gas: Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l. <br> The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy. but a change in momentum occurs. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_SP_06_E01_043_S01.png" width="80%"> <br> The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision, the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time. <br> A molecule of mass m moving with a velocity `vecv` having components `(v_(x), v_(y), v_(z))` hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. This is shown in the figure. The components of velocity of the molecule after collision are `(v_(x), v_(y), v_(z))` . <br> `"The x-component of momentum of the molecule before collision" = mv_(x)` <br> `"The x-component of momentum of the molecule after collision" = - mv_(x)` <br> The change in momentum of the molecule in x direction <br> `= "Final momentum"- "initial momentum" = -mv_(x)-mv_(x)= - 2mv_(x)` <br> According to law of conservation of linear momentum, the change in momentum of the wall = `2mv_(x)` <br> The number of molecules hitting the right side wall in a small interval of time `Deltat`. <br> The molecules within the distance of `v_(x)Deltat` from the right side wall and moving towards the right will hit the wall in the time interval `Deltat`. The number of molecules that will hit the right side wall in a time interval `Deltat` is equal to the product of volume `(Av_(x)Deltat)` and number density of the molecules (n). Here A is area of the wall and n is number of molecules per unit volume `(N)/(V)` We have assumed that the number density is the same throughout the cube. <br> Not all the n molecules will move to the right, therefore on an average only half of then molecules move to the right and the other half moves towards left side. <br> The number of molecules that hit the right side of the wall in a time interval `Deltat` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_SP_06_E01_043_S02.png" width="80%"> <br> `=(n)/(2)Av_(x)Deltat` ...(1) <br> In the same interval of time `Deltat`, the total momentum transferred by the molecules <br> `DeltaP=(n)/(2)Av_(x)Deltat xx 2mv_(x)=Av_(x)^(2)mnDeltat` ...(2) <br> From Newton.s second law, the change in momentum in a small interval of time rise to force. <br> The force exerted by the molecules on the wall (in magnitude) <br> `F=(DeltaP)/(Deltat)=nmAv_(x)^(2)` ...(3) <br> Pressure, P = force divided by the area of the wall <br> `P=(F)/(A)=nmv_(x)^(2)` ...(4) <br> Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term `v_(x)^(2)` by the average `barv_(x)^(2)` in equation (4). <br> `P=nmbarv_(x)^(2)` ...(5) <br> Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, `barv_(x)^(2)=barv_(y)^(2)=barv_(z)^(2)` . The mean square speed is written as <br> `barv^(2)=barv_(x)^(2)+barv_(y)^(2)+barv_(z)^(2)=3barv_(x)^(2)` <br> `barv_(x)^(2)=(1)/(3)barv^(2)` <br> Using this in equation (5), we get <br> `P=(1)/(3)nmbarv^(2)` or `P=(1)/(3)(N)/(V)mbarv^(2)` as `[n=(N)/(V)]` ...(6)