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

The correct option is A

Three times the absolute temperature

$K.E=\frac{3}{2}.RT=\frac{3}{2}.2T$ $\because R\approx 2cal{K}^{-1}mo{l}^{-1}$

$K.E.=3T$