calculate the kinetic energy of nitrogen molecule at 27 degree centrigade
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Kinetic enrgy of a gas molecule is given as
K.E = 1.5 kT
[where k = Boltzmann constant = 1.380 × 10^-23 m^2 kg/(s^2 K)]
K.E = 1.5 × 1.380 × 10^-23 × (27 + 273)
= 6.21 × 10^-21 Joules
Kinetic energy of nitrogen molecule at 27°C is 6.21 × 10^-21 Joules
K.E = 1.5 kT
[where k = Boltzmann constant = 1.380 × 10^-23 m^2 kg/(s^2 K)]
K.E = 1.5 × 1.380 × 10^-23 × (27 + 273)
= 6.21 × 10^-21 Joules
Kinetic energy of nitrogen molecule at 27°C is 6.21 × 10^-21 Joules
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