The total kinetic energy of 1 mole of oxygen at 27° C is : [Use universal gas constant (R)=8.31 J / mole K]

Solution

Physics sir · 4 steps

Oxygen is diatomic, so give it 5 degrees of freedom, each worth 1/2RT per mole.

1. Count degrees of freedom. O₂ is a diatomic molecule. At room temperature it has 3 translational and 2 rotational degrees of freedom, so f = 5. Vibration is not excited at this temperature.
2. Use equipartition. Each degree of freedom carries 1/2RT of energy per mole. Multiply by the number of degrees of freedom to get the total kinetic energy.

   E = f/2RT = 5/2RT

3. Put in numbers. Convert to kelvin: T = 27 + 273 = 300 K. Then E = 5/2 × 8.31 × 300 = 2.5 × 2493 = 6232.5 J.
4. So. The total kinetic energy of 1 mole of oxygen at 27°C is 6232.5 J, which is option (a).

The idea to keep. Total kinetic energy of a mole is f/2RT. Monatomic has f = 3, diatomic has f = 5. Count all degrees of freedom, not only the translational ones.

Answer from the NTA final answer key. Solution by PrepWiser.