Question

Calculate the kinetic energy of translation of 8.5 gm of ammonia gas at 27℃.

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Answer 1

Explanation:

Kinetic energy of the ideal gas can be calculated as

Kinetic energy = {3*n*R*T} / 2

where,

n = number of moles

R = gas constant

T = temperature in kelvin

We have above is, n = 5 moles of N

R = 8/314,

and T = 27 + 273 (to convert celsius to kelvin, add 273)=300.

Kinetic energy = {3*5*8.314*300} / 2

= 18,706.05 joules or 18.706 kj

Hope, it will help you.......

Answer 2

Answer: Total energy of given sample is 1869.75 Joules

Explanation:

Ammonia ⇒ NH₃

Molar mass = 17 gm

Given weight = 8.5 gm

Number of moles = 8.5/17 = 0.5

Number of molecules = 0.5Nₐ     ( Nₐ ⇒ Avagadro number)

Kinetic energy of 1 molecule = $\frac{3kT}{2}$

k = R/Nₐ

Total energy of given sample = 0.5Nₐ × 3RT/2Nₐ  = 3RT/4

R = 8.31

T = 273 + 27 = 300 Kelvin

Total energy of given sample = 3 × 8.31 × 300/4

Total energy of given sample = 1869.75 Joules

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