Question

The average K.E. of an ideal gas in calories per mole is approximately equal to.

• A. Three times the absolute temperature
• B. Absolute temperature
• C. Two times the absolute temperature
• D. 1.5 times the absolute temperature

Answer 1

Answer: (A)

Three times the absolute temperature

Answer 2

$K \cdot E . = \frac { 3 } { 2 } \cdot R T = \frac { 3 } { 2 } \cdot 2 \cdot T$ $\because R \approx 2 \mathrm { calK } ^ { - 1 } \mathrm {~mol} ^ { - 1 }$

$K . E . = 3 T$