What is the temperature of an ideal gas whose molecules have an average translational kinetic energy of 3.90 10-20 J?
K
(1/2) m v^2 average = (3/2) k T
k = R/Avagadro's number
= 1.38*10^-23 joules/molecule deg K
3.9 *10^-20 = 1.5 (1.38*10^-23) T
T = 188 K
To find the temperature of an ideal gas, we need to use the equation that relates the average translational kinetic energy of the gas molecules to the temperature. This equation is given by:
(3/2) * k * T = 1/2 * m * v^2
Where:
- (3/2) is a constant term that depends on the degrees of freedom of the gas molecules.
- k is the Boltzmann constant, which is equal to 1.38 x 10^-23 J/K.
- T is the temperature in Kelvin.
- m is the mass of a gas molecule.
- v is the root mean square velocity of the gas molecules.
In this case, we are given the average translational kinetic energy of the gas molecules, which is 3.90 x 10^-20 J. Let's solve for T:
(3/2) * k * T = 1/2 * m * v^2
Divide both sides by k:
(3/2) * T = (1/2) * (m * v^2) / k
Substitute the given value for the average translational kinetic energy:
(3/2) * T = (1/2) * (m * v^2) / k
T = (1/2) * (m * v^2) / ((3/2) * k)
T = (m * v^2) / (3 * k)
Since we do not have the values for the mass of the gas molecule and the root mean square velocity, we cannot determine the exact temperature of the gas.