Question

Calculate Kinetic energy of 5 moles of nitrogen at 27°C ?​

Answer 1

Explanation:

v²=3RT/m

T is in kelvin

Kinetic energy=1/2mv²

then you can calculate by putting value

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.