Kinetic Theory

JEE Main level, single correct, medium

131 of 168

Question 131 of 168

The total kinetic energy of 11 mole of oxygen at 27∘C27^{\circ}C is :
[Use universal gas constant (R)=8.31Jmol−1K−1(R)=8.31J{mol}^{-1}{K}^{-1} ]

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Answer: (C) 6232.5J6232.5J

The formula to calculate the kinetic energy can be written as
K=f2nRT...(1)K=\dfrac{f}{2}nRT...\left(1\right)
Since oxygen is a diatomic gas, the number of degrees of freedom is 55.
Thus, from equation (1), it follows that
K=52×1×8.31×300J=6232.5J\begin{matrix}K & = & \dfrac{5}{2}\times 1\times 8.31\times 300J \\ & = & 6232.5J\end{matrix}

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