Question

The kinetic energy for 14 grams of nitrogen gas at 127°C is nearly??

Answer 1

K. E=3/2*n*r*t
Where n= no of moles
R= gas constant
T= temp


Substitue the values u will get
The ans
Hope it helps you

Answer 2

Answer : The kinetic energy of nitrogen gas is, $6.212\times 10^{-21}J$

Solution : Given,

Temperature of gas = $127^oC=273+127=300K$

Formula used for kinetic energy of gas molecule is,

$K.E=\frac{3RT}{2\times N_A}$

where,

R = gas constant = 8.314 J/moleK

T = temperature of gas

$N_A$ = Avogadro's number = $6.022\times 10^{23}mole^{-1}$

Now put all the give values in this formula, we get

$K.E=\frac{3\times (8.314J/moleK)\times (300K)}{2\times 6.022\times 10^{23}mole^{-1}}=6.212\times 10^{-21}J$

Therefore, the kinetic energy of nitrogen gas is, $6.212\times 10^{-21}J$