Physics · Kinetic theory of gases: assumptions, the concept of pressure, kinetic interpretation of temperature
The total kinetic energy of 1 mole of at will be approximately
The total kinetic energy of 1 mole of \( N_{2} \) at \( 27 \mathrm{C} \) will be approximately
- A. 3739.662
- B. 1500 calorie
- C. 1500 kilo calorie
- D. 1500 erg.
Step-by-step solution
For 1 mole of an ideal gas, the average translational kinetic energy is (3/2)RT. At T = 27°C = 300 K, using R = 8.314 J/(mol·K), we get KE = (3/2) × 8.314 × 300 ≈ 3741.3 J. The given option A, 3739.662 J, is the closest approximation. The other options in different units do not match: 1500 cal ≈ 6276 J, 1500 kcal ≈ 6.276×10^6 J, 1500 erg = 1.5×10^-4 J.
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