show that at absolute temprature T, the quantity KT is roughly equal to the mean energy per degree of freedom?
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• Temperature ~ Average KE of each particle
• Particles have different speeds
• Gas Particles are in constant RANDOM motion
• Average KE of each particle is: 3/2 kT
This subject touches upon statistical mechanics, which is a deep subject. Pressure is force per unit area. Force is rate of change of momentum. That is, the amount of momentum transferred to a wall per second. This is achieved through the gas molecules colliding with the wall. A molecule of mass (m) and velocity (v) facing the wall will change to velocity −v after colliding with the wall (assuming that the collision is elastic). Thus momentum of the molecule changes from to mv to −mv.
The momentum transferred to the wall per molecule collision is therefore = mv−(−mv)=2mv If the wall area is A, the number of collisions on this wallin time Δt is ρAvΔt where ρ=(N/V) is the number density of the gas molecules. We therefore see that the total momentum transferred to the wall of area A in time Δt is equal to 2mv⋅ρAvΔt Dividing by Δt to get the rate change of momentum or force, then dividing by A to get pressure, we have = P=2ρmv²=2(N/V)mv² .
We must next replace by v by vₐv, which means averaging it over the different velocities of the different molecules. So far, vₐv has a direction pointing toward the wall. But actual molecules are moving in all kinds of directions, so in the average, only 1/6 of the molecules are moving in that direction. This cuts down the answer for P by a factor of 6. So finally, we get = P=1/3(N/V)mv²av
• Temperature ~ Average KE of each particle
• Particles have different speeds
• Gas Particles are in constant RANDOM motion
• Average KE of each particle is: 3/2 kT
This subject touches upon statistical mechanics, which is a deep subject. Pressure is force per unit area. Force is rate of change of momentum. That is, the amount of momentum transferred to a wall per second. This is achieved through the gas molecules colliding with the wall. A molecule of mass (m) and velocity (v) facing the wall will change to velocity −v after colliding with the wall (assuming that the collision is elastic). Thus momentum of the molecule changes from to mv to −mv.
The momentum transferred to the wall per molecule collision is therefore = mv−(−mv)=2mv If the wall area is A, the number of collisions on this wallin time Δt is ρAvΔt where ρ=(N/V) is the number density of the gas molecules. We therefore see that the total momentum transferred to the wall of area A in time Δt is equal to 2mv⋅ρAvΔt Dividing by Δt to get the rate change of momentum or force, then dividing by A to get pressure, we have = P=2ρmv²=2(N/V)mv² .
We must next replace by v by vₐv, which means averaging it over the different velocities of the different molecules. So far, vₐv has a direction pointing toward the wall. But actual molecules are moving in all kinds of directions, so in the average, only 1/6 of the molecules are moving in that direction. This cuts down the answer for P by a factor of 6. So finally, we get = P=1/3(N/V)mv²av
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