Which is the average translational kinetic energy of molecules in a gas at temperature 27°C?
Answer: 6.21 × 10⁻²¹
Explanation
The average translational kinetic energy of a molecule in an ideal gas is given by the formula:
E=3/2kT
Where:
-
k=1.38×10^−23J/K (Boltzmann constant)
-
T=27∘C=300K (since 27°C + 273 = 300 K)
Now calculate:
E=3/2×1.38×10^−23×300
E=3/2×4.14×10^−21
6.21×10^−21 J
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