Question

The av. K.E./mole of an ideal monoatomic gas at 27°C is

• A. 900 cal
• B. 1800 cal
• C. 300 cal
• D. None

Answer 1

Answer: (A)

900 cal

Answer 2

The temperature is converted to Kelvin: $T = 27^\circ\text{C} + 273 = 300 \text{ K}$. For an ideal monoatomic gas, the degrees of freedom $f=3$. Using the equipartition theorem, the average kinetic energy per mole is $\text{KE}_{\text{mole}} = \frac{f}{2}RT$. Using $R \approx 2 \text{ cal/mol·K}$, $\text{KE}_{\text{mole}} = \frac{3}{2} \times (2 \text{ cal/mol·K}) \times (300 \text{ K}) = 900 \text{ cal/mol}$.