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

1.	Calculate kinetic energy of 5 moles of nitrogen at 27°C
Answer» n = 5 R = 2 cal T = 27° C = 300 k K.E = 3/2 nRT = 3/2 x 5 x 2 x 300 = 4500 cal

Answer»

n = 5

R = 2 cal

T = 27° C = 300 k

K.E = 3/2 nRT

= 3/2 x 5 x 2 x 300

= 4500 cal

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

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