The average translational kinetic energy per molecule of an ideal gas at 0ºC (k =1.38 × 10–23 J/K) is –
Question
The average translational kinetic energy per molecule of an ideal gas at 0ºC (k = 1.38 × 10–23 J/K) is –
Solution
The average translational kinetic energy per molecule of an ideal gas is given by the equation:
KE_avg = 3/2 kT
where:
- KE_avg is the average kinetic energy,
- k is the Boltzmann constant (1.38 × 10–23 J/K), and
- T is the temperature in Kelvin.
The temperature given is 0ºC, which is equivalent to 273.15 K in Kelvin.
Substituting these values into the equation gives:
KE_avg = 3/2 * (1.38 × 10–23 J/K) * 273.15 K
Solving this equation gives the average translational kinetic energy per molecule of the ideal gas at 0ºC.
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