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JEE Main 202427 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
27 January
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2024 · 27 January · Evening Shift · Physics · Kinetic Theory of Gases

The total kinetic energy of 1 mole of oxygen at 27 ° C is : [Use...

The total kinetic energy of 11 mole of oxygen at 27°C27 ° C is :
[Use universal gas constant (R)=8.31Jmol−1K−1( R ) = 8 . 31J{mol}^{- 1}{K}^{- 1} ]

Options

  1. A6845.5J6845 . 5J
  2. B5942.0J5942 . 0J
  3. C6232.5J6232 . 5J
  4. D5670.5J5670 . 5J

Correct answer

C. 6232.5J6232 . 5J

Step-by-step solution

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{cases}K & = & \dfrac{5}{2} \times 1 \times 8 . 31 \times 300J \\ & = & 6232 . 5J\end{cases}

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