Question

find the kinetic energy of 1 molecule
of O2 gas at 127 degree Celsius​

Answer 1

Answer:

K.E of 1 molecule =  3/2 × k×T

where k is boltzman constant and T ( in kelvin) is temperature .

by applying values of k as 1.38 × 10^-23 and temperatue as 400 kelvin

we get 828 × 10 ^-23 joule.

Explanation:

hope it helpfull

Answer 2

The require answer is 4.99kJ.

kinetic energy : To accelerate an object, a force must be applied. To use violence, you have to work. When the object is processed, energy is transferred and the object moves at a new constant speed. The energy transferred is called kinetic energy and depends on the mass and velocity achieved.

The definition of kinetic energy in physics is:

The kinetic energy of an object is a measure of the work an object can do by its motion.

molecule : A molecule is defined as the smallest unit of a compound that contains the chemical properties of the compound.

As per the question we need to find kinetic energy of 1 molecule of O₂.

The given is 1 molecule of O2 gas at 127 degree Celsius​

KE = $\frac{3}{2}$ RT

KE = 3 x 8.314 x $\frac{400}{2}$

    = 49884 J

     = 4.99kJ

Hence the require answer is 4.99kJ.

To learn more kinematics follow the given link

https://brainly.in/question/13977580?

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