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The total kinetic energy of 1 mole of \( N_{2} \) at \( 27 \mathrm{C} \) will be approximately

1. 3739.662
2. 1500 calorie
3. 1500 kilo calorie
4. 1500 erg.

Correct answer: 1500 kilo calorie

Solution

For \(N_2\) at 27°C, vibrational modes are not active, so a diatomic gas has 5 active degrees of freedom. Thus the total kinetic energy is \(\frac{5}{2}RT\), which is about \(1.5\times 10^3\) kcal for 1 mole at 300 K.

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