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

1.	The average K.E. of one mole of an ideal gas in calories is equal to
Answer» 3 times of its ABSOLUTE TEMP 2 times of its absolute temp 4 times of its absolute temp 2/3 times of its absolute temp Solution :KE = `3/2 XX RT = 3/2 xx 2T = 3T`

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