If one gm mole of nitrogen gas occupies `2xx10^(4)` c.c. at a pressure of `10^6` dyne/`cm^(2)`, then the average kineticf energy of nitrogen gas molecule in ergs will be (Avagadro's number `N=6xx10^(23)`)

Options

A. `5xx10^(-14)`
B. `10xx10^(12)`
C. `10^(6)`
D. `2xx10^(6)`

Answer

A

Explanation

K.E./molecule`=(3)/(2)KT=(3)/(2)(RT)/(N_(0))=(3)/(2)(PV)/(N_(0))`