at what temprature is the average translational k.e of a molecule in an ideal gas equal to 1 ev?

KE=3/2 kT

change 1 ev to joules
k is Boltzmann's constant.
Solve for Temperature T

vander waal b for oxygen is 32 cm3/mol compute the diameter of o2 molecule.

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To determine the temperature at which the average translational kinetic energy (K.E.) of a molecule in an ideal gas is equal to 1 electron volt (1 eV), we can use the equation that relates temperature and kinetic energy. This equation is given by:

K.E. = (3/2) kT

Where:
K.E. is the average translational kinetic energy of a molecule,
k is the Boltzmann constant (approximately 1.38 x 10^-23 J/K),
T is the temperature in Kelvin.

To equate this equation to 1 eV, we need to convert 1 eV to joules (J).

The conversion factor is 1 eV = 1.602 x 10^-19 J.

Now, let's solve for the temperature (T):

1 eV = (3/2) kT

Convert 1 eV to J:

1 eV = 1.602 x 10^-19 J

Substitute the values:

1.602 x 10^-19 J = (3/2) (1.38 x 10^-23 J/K) * T

Simplify:

1.602 x 10^-19 J = (2.07 x 10^-23 J/K) * T

Divide both sides by 2.07 x 10^-23 J/K to solve for T:

T = (1.602 x 10^-19 J) / (2.07 x 10^-23 J/K)

Simplify:

T ≈ 774.88 K

Therefore, the average translational kinetic energy of a molecule in an ideal gas is equal to 1 eV at a temperature of approximately 774.88 Kelvin.