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.
Given that the temperature is 0ºC, we first need to convert this to Kelvin. The Kelvin scale starts at absolute zero (the lowest possible temperature), which is -273.15ºC. So, to convert from Celsius to Kelvin, we add 273.15.
0ºC = 0 + 273.15 = 273.15 K
Now we can substitute the values into the equation:
KE_avg = 3/2 * (1.38 × 10–23 J/K) * 273.15 K
Solving this gives:
KE_avg = 5.65 × 10–21 J
So, the average translational kinetic energy per molecule of an ideal gas at 0ºC is 5.65 × 10–21 J.
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