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

Answer :

The average kinetic energy is, $7.659\times 10^{-21}J$

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

Explanation:

(a) The formula for average kinetic energy is:

$K.E=\frac{3}{2}kT$

where,

k = Boltzmann’s constant = $1.38\times 10^{-23}J/K$

T = temperature = 370 K

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

$K.E=\frac{3}{2}\times (1.38\times 10^{-23}J/K)\times (370K)$

$K.E=7.659\times 10^{-21}J$

The average kinetic energy is, $7.659\times 10^{-21}J$

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

$\nu_{rms}=\sqrt{\frac{3kN_AT}{M}}$

where,

$\nu_{rms}$ = root mean square speed

k = Boltzmann’s constant = $1.38\times 10^{-23}J/K$

T = temperature = 370 K

M = atomic mass = 0.02 kg/mole

$N_A$ = Avogadro’s number = $6.02\times 10^{23}mol^{-1}$

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

$\nu_{rms}=\sqrt{\frac{3\times (1.38\times 10^{-23}J/K)\times (6.02\times 10^{23}mol^{-1})\times (370K)}{0.02kg/mol}}$

$\nu_{rms}=679.02m/s$

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