Question

calculate kinetic energy of 5 moles of nitrogen at 27°C

Answer 1

kinetic energy of nitrogen=3/2nRT
=3/2x5x25/3x300
=18750 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...!!!

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Sorry baby 'wink'