The total kinetic energy of 1 mole of oxygen at 27°C is : [Take R = (25/3) J/mol-K]

The correct option is 6250 J.

To find the total kinetic energy of 1 mole of oxygen gas at $27°\text{C}$, we can use the kinetic theory of gases. Let's break down the calculation step-by-step.

Step 1: Identify the given values
Number of moles ($n$) = $1\text{mol}$
Temperature in Celsius ($t$) = $27°\text{C}$
Universal gas constant ($R$) = $\frac{25}{3}\text{J/mol-K}$

Step 2: Convert the temperature to Kelvin
To use the gas equations, we must express the temperature in the absolute scale (Kelvin):
$T=t+273$
$T=27+273=300\text{K}$

Step 3: Determine the degrees of freedom for oxygen
Oxygen (${\text{O}}_2$) is a diatomic gas. At normal temperatures (like $27°\text{C}$), a diatomic molecule has 5 degrees of freedom ($f=5$), consisting of 3 translational degrees of freedom and 2 rotational degrees of freedom.

Step 4: Write the formula for the total kinetic energy
According to the law of equipartition of energy, the total internal kinetic energy of $n$ moles of a gas is given by:
$E=\frac{f}{2}nRT$

Step 5: Substitute the values into the formula
Plugging the values into the formula:
$E=\frac{5}{2}\times 1\times \left(\frac{25}{3}\right)\times 300$

Step 6: Simplify the expression
First, divide 300 by 3:
$E=\frac{5}{2}\times 25\times 100$
Next, divide 100 by 2:
$E=5\times 25\times 50$
Now multiply the numbers:
$E=125\times 50$
$E=6250\text{J}$

Thus, the total kinetic energy of 1 mole of oxygen at $27°\text{C}$ is $6250\text{J}$.