Question

calculate the kinetic energy of 5moles of nitrogen at27degree celcius

Answer 1

Kinetic energy of the ideal gas can be calculated as 

Kinetic energy = {3*n*R*T} / 2
where, 
n = number of moles
R = gas constant
T = temperature in kelvin

We have above is, n = 5 moles of N
R = 8/314, 
and T = 27 + 273 (to convert celsius to kelvin, add 273)=300. 

Kinetic energy = {3*5*8.314*300} / 2
= 18,706.05 joules or 18.706 kj

Hope this helps. If yes, please mark as Brainliest. 

Answer 2

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\purple{R}}}}$

❤..$\red{\tt{Kinetic}}$$\orange{\tt{Energy}}$=$\frac{3}{2}$

❤.Where n=5 moles;R =8.314J${mol^-1k^-1}$ T= ${27\degree}c$+273=300K

KINETIC ENERGY

${Ek}$=$\frac{3}{2}$×5mol×8.314J${mol^-1k^-1}$=300k=18706.50J.