Question

the kinetic energy of one mole of oxygen molecule is calorie /mole at 27degrees
degrees Celsius​

Answer 1

Answer: $900cal/mol$

Explanation:

Kinetic energy is the energy possessed by an object by virtue of its motion.

Average kinetic energy is defined as the average of the kinetic energies of all the particles present in a system.

It is determined by the equation:

$K=\frac{3nRT}{2}$

K= kinetic energy

n= number of moles = 1

R= gas constant  = $2cal/Kmol$

T= temperature =$27^0C=(27+273)K=300K$

$K=\frac{3\times 1\times 2\times 300}{2}=900cal/mol$

Thus the kinetic energy of one mole of oxygen molecule is $900cal/mol$

Answer 2

Answer: the answer comes up to be 900

Explanation:

Since we know that the kinetic energy is equal to 3nRT/2

n is equal to 1

T is taken in absolute scale that is 273 + 27 which is equal to 300.

R valuein calorie per mole is 2.

Now substituting all the values in the formula KE= 3nRT/2

The answer comes up to be 900

Hope you understand

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Shyaja