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dexar [7]
3 years ago
11

Consider neon, a noble gas whose molecules consist of single atoms of atomic mass 0.02 kg/mol. What is the average kinetic energ

y 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.
Chemistry
1 answer:
Juliette [100K]3 years ago
6 0

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

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