Avg. Kinetic Energy: Neon Ratio @ Diff. Temps

[brake4country]

Homework Statement

What is the ratio of the average speed of an atom of neon to another atom of neon at twice the temperature but the same pressure?

Homework Equations

KE = 3/2 RT; v₁/v₂ = sqrt. m₂/sqrt. m₁

The Attempt at a Solution

I first used KE = 3/2RT and substituted 1 and 2 for T and set them equal to each other. I ended up with a ratio of 1:2. Why is the answer 1:1.4?

 

[Quantum Defect]

Homework Statement

What is the ratio of the average speed of an atom of neon to another atom of neon at twice the temperature but the same pressure?

Homework Equations

KE = 3/2 RT; v₁/v₂ = sqrt. m₂/sqrt. m₁

The Attempt at a Solution

I first used KE = 3/2RT and substituted 1 and 2 for T and set them equal to each other. I ended up with a ratio of 1:2. Why is the answer 1:1.4?

Note that you are using equations that are appropriate for root-mean-squared speed. I would take "average" to be "mean", in which case the equations are slightly different. The functional form is the same.

v_mean = SQRT(8*RT/[pi*M]) ==> v2/v1 = ?

 

[brake4country]

Hi, I do not understand what you wrote. I am trying to understand this from a KE=3/2RT and KE=1/2mv^2 point of view. I see that when I set these equal to each other, temp. and velocity are inversely related. T = v^2, thus taking the sqrt. of the temp. I do not know how to use Graham's law for something like this.

 

[Quantum Defect]

Hi, I do not understand what you wrote. I am trying to understand this from a KE=3/2RT and KE=1/2mv^2 point of view. I see that when I set these equal to each other, temp. and velocity are inversely related. T = v^2, thus taking the sqrt. of the temp. I do not know how to use Graham's law for something like this.

You don't need to use Graham's Law.

v_mean = SQRT(8RT/[pi*M])

For the same gas at two different temperatures:

v_2/v_1 = SQRT(8RT_2/(pi*M))/SQRT(8RT_1/(pi*M)) ==> All constants cancel top and bottom ==> v_2/v_1 = SQRT(?)

[Look at the difference between your answer and the book's answer]