Question

calculate kinetic energy of 5 moles of nitrogen at 27 degree celsius

Answer 1

K.E. of 1 mole N2 gas = (3/2)*8.314*300 J
= 3741.3 J

Thus K.E. of 5 moles N2 gas = 5*3741.3 J
= 18706.5 J

Answer 2

Heya...

Here's your answer.........

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

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✨✨✨✨BY BeWaFa ✨✨✨

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Thanks...!!!

XD

Sorry baby 'wink'