In the question that's attached why can we not use KE=1/2mv^2 to reason that protons have a greater mass than electrons so they would have a higher kinetic energy. When do we know to use the KE=3/2RT? Why isn't KE=1/2mv^2 applicable here?
Kinetic Energy
- Thread starter hibni
- Start date
- Joined
- Aug 2, 2016
- Messages
- 366
- Reaction score
- 559
KE=3/2RT is used in relation to the gas law, and the question states that protons can be assumed to follow the ideal gas law.
- Joined
- Mar 6, 2013
- Messages
- 135
- Reaction score
- 38
Can you show us the passage? I predict that the reason why the kinetic energy is the same is because the input of energy into the system is the same in both cases, so the outgoing kinetic energy must be the same because energy is always conserved. Even though the mass of a proton is higher, we have no reason to assume that the velocity would stay the same. The MCAT loves to trick you into thinking "well a proton is heavier and mass is in KE so KE will be higher" but the more fundamental understanding comes from utilizing conservation of energy.
Your other clue here is that if you think of a proton as an ideal gas, so using KE=3RT/2n or 3kT/2 would work because the ideal gas law makes the assumption that any ideal gas we are talking about has the same properties. Those properties are that the gas particles themselves don't take up any volume, have no intermolecular forces, all collisions are completely elastic, and each particle has a mass that is taken into account by the equation. The equation for the ideal gas law takes the mass into account in the constant, so don't worry about it.
Your other clue here is that if you think of a proton as an ideal gas, so using KE=3RT/2n or 3kT/2 would work because the ideal gas law makes the assumption that any ideal gas we are talking about has the same properties. Those properties are that the gas particles themselves don't take up any volume, have no intermolecular forces, all collisions are completely elastic, and each particle has a mass that is taken into account by the equation. The equation for the ideal gas law takes the mass into account in the constant, so don't worry about it.
- Joined
- Mar 6, 2013
- Messages
- 135
- Reaction score
- 38
For a bit of fun, let's derive the equation to prove that the mass doesn't matter.
Let's imagine a glass cube with a single particle bouncing back and forth between two walls separated by a distance 'x'. Let's find the force of a [perfectly elastic] collision with a wall. Imagine the particle is traveling from the left wall to the right wall with velocity 'v', hits the wall, bounces back the exact same way it came, and travels back to its original point.
Force = ma
a = ∆v/∆t
Force = ∆mv/∆t
m∆v = Impulse (change in momentum) which we refer to as 'P'
Force = ∆P/∆t
What is the ∆P in this case? If the ball hits the wall with velocity v in a perfectly inelastic collision, it'll bounce back with velocity -v. m∆v = m(-v - v) = m*-2v = -2mv. If we had arbitrarily said the initial v was -v and the return was +v, this would just be +2mv. For simplicity's sake, let's just go with 2mv.
How much time did it take to complete one collision and return to the same spot? Well it had to travel 'x' there, and 'x' back, a total distance of 2x, and v = d/t so t = d/v. ∆t = 2x/v
Force = (2mv)/(2x/v)
Force = mv^2/x
Let's find pressure. Pressure = Force/Area. We already found force, and our particle is applying pressure on one side of our cube, so the area of the surface receiving the force is x^2.
Pressure = (mv^2/x) / (x^2)
Pressure of one particle going in the x direction = mv^2/x^3
x^3 is equal to the volume of our cube, so:
Pressure of our one particle between two sides of our cube is = mv^2/V
But what if instead of only talking about one particle going in the x direction, we wanted to talk about all the particles going in the x direction? We could say that we have N particles in the cube, and in our simple model (that actually holds up well under experimentation) they have 3 directions to go (x,y, or z), so really what we've described so far is the pressure generated by one-third of all the particles, which is N. So N/3
Pressure of all particles going in the x direction = P = (mv^2/V) * (n/3)
Let's rearrange a bit:
3PV = mv^2 * N
Let's divide both sides by 2 cuz why not
(3/2)PV = (mv^2)/2 * N
That italicized term is our kinetic energy! To be clear, it is the average kinetic energy for one molecule in our system. mv^2/2 * N is the kinetic energy for all the particles in the cube. This value (KE * N) is what we refer to as internal energy U.
And so the total kinetic energy U of the system is equal to (3/2)*PV. But wait! PV = nRT. So, (3/2)*PV = (3/2)*nRT.
Where n is the number of moles. We also know that R = Kb (boltzmann's constant) * avagadro's number (6.022E23 particles/mole). So U is also = (3/2)*NKbT
In conclusion, for an ideal gas:
U = KEavg * N
= (3/2)*PV
= (3/2)*nRT
= (3/2)*NKbT
Taking this a step further:
KEavg * N = (3/2) * KbT * N
Cancel out Ns:
Average Kinetic Energy = (3/2)KbT
And Kb is a constant. This proves that for an ideal gas, kinetic energy is really just dependent on temperature.
So you can see that as long as the number of particles and the temperature remains constant, the kinetic energy of the system (internal energy) remains constant.
Let's imagine a glass cube with a single particle bouncing back and forth between two walls separated by a distance 'x'. Let's find the force of a [perfectly elastic] collision with a wall. Imagine the particle is traveling from the left wall to the right wall with velocity 'v', hits the wall, bounces back the exact same way it came, and travels back to its original point.
Force = ma
a = ∆v/∆t
Force = ∆mv/∆t
m∆v = Impulse (change in momentum) which we refer to as 'P'
Force = ∆P/∆t
What is the ∆P in this case? If the ball hits the wall with velocity v in a perfectly inelastic collision, it'll bounce back with velocity -v. m∆v = m(-v - v) = m*-2v = -2mv. If we had arbitrarily said the initial v was -v and the return was +v, this would just be +2mv. For simplicity's sake, let's just go with 2mv.
How much time did it take to complete one collision and return to the same spot? Well it had to travel 'x' there, and 'x' back, a total distance of 2x, and v = d/t so t = d/v. ∆t = 2x/v
Force = (2mv)/(2x/v)
Force = mv^2/x
Let's find pressure. Pressure = Force/Area. We already found force, and our particle is applying pressure on one side of our cube, so the area of the surface receiving the force is x^2.
Pressure = (mv^2/x) / (x^2)
Pressure of one particle going in the x direction = mv^2/x^3
x^3 is equal to the volume of our cube, so:
Pressure of our one particle between two sides of our cube is = mv^2/V
But what if instead of only talking about one particle going in the x direction, we wanted to talk about all the particles going in the x direction? We could say that we have N particles in the cube, and in our simple model (that actually holds up well under experimentation) they have 3 directions to go (x,y, or z), so really what we've described so far is the pressure generated by one-third of all the particles, which is N. So N/3
Pressure of all particles going in the x direction = P = (mv^2/V) * (n/3)
Let's rearrange a bit:
3PV = mv^2 * N
Let's divide both sides by 2 cuz why not
(3/2)PV = (mv^2)/2 * N
That italicized term is our kinetic energy! To be clear, it is the average kinetic energy for one molecule in our system. mv^2/2 * N is the kinetic energy for all the particles in the cube. This value (KE * N) is what we refer to as internal energy U.
And so the total kinetic energy U of the system is equal to (3/2)*PV. But wait! PV = nRT. So, (3/2)*PV = (3/2)*nRT.
Where n is the number of moles. We also know that R = Kb (boltzmann's constant) * avagadro's number (6.022E23 particles/mole). So U is also = (3/2)*NKbT
In conclusion, for an ideal gas:
U = KEavg * N
= (3/2)*PV
= (3/2)*nRT
= (3/2)*NKbT
Taking this a step further:
KEavg * N = (3/2) * KbT * N
Cancel out Ns:
Average Kinetic Energy = (3/2)KbT
And Kb is a constant. This proves that for an ideal gas, kinetic energy is really just dependent on temperature.
So you can see that as long as the number of particles and the temperature remains constant, the kinetic energy of the system (internal energy) remains constant.
Last edited:
