Question

Consider neon, a noble gas whose molecules consist of single atoms of atomic mass 0.02 kg/mol. What is the average kinetic energy of a neon atom when the gas is at a temperature of 370 K? Avogadro’s number is 6.02 × 1023 mol−1 and Boltzmann’s constant is 1.38 × 10−23 J/K. Answer in units of J. Question 5, chap 19, sect 4. part 2 of 3 10 points What is the root mean square speed of a neon atom under such conditions? Answer in units of m/s.

Answer 1

The average kinetic energy is,
`7.659* 10^(-21)J`

The root mean square speed is,
`679.02m/s`

Step-by-step explanation:

(a) The formula for average kinetic energy is:

`K.E=(3)/(2)kT`

where,

k = Boltzmann’s constant =
`1.38* 10^(-23)J/K`

T = temperature = 370 K

Now put all the given values in the above average kinetic energy formula, we get:

`K.E=(3)/(2)* (1.38* 10^(-23)J/K)* (370K)`

`K.E=7.659* 10^(-21)J`

The average kinetic energy is,
`7.659* 10^(-21)J`

(b) The formula used for root mean square speed is:

`\\u_(rms)=\sqrt{(3kN_AT)/(M)}`

where,

`\\u_(rms)` = root mean square speed

k = Boltzmann’s constant =
`1.38* 10^(-23)J/K`

T = temperature = 370 K

M = atomic mass = 0.02 kg/mole

`N_A` = Avogadro’s number =
`6.02* 10^(23)mol^(-1)`

Now put all the given values in the above root mean square speed formula, we get:

`\\u_(rms)=\sqrt{(3* (1.38* 10^(-23)J/K)* (6.02* 10^(23)mol^(-1))* (370K))/(0.02kg/mol)}`

`\\u_(rms)=679.02m/s`

The root mean square speed is,
`679.02m/s`