calculate kinetic energy of 1 molecule at 27 degree Celsius
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The total kinetic energy of 1 mole of N2 at 27° Celsius will be approximately??
answer:
KE of ideal gas=3/2nRT
n=1 mole
R=8.314 jk^-1mol^-1
T=300k
hence KE=3/2*1*8.314*300=3741.3 J
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