Question

The average kinetic energy per molecule of an ideal gas at 27 degree c is​

Answer 1

$\huge\underline\green{\sf Answer:-}$

★$\large{\boxed{\sf KE\:per\: molecule=6.21×{10}^{-21}J}}$

$\huge\underline\green{\sf Solution:-}$

★Given :-

Temperature (T) = 27°C

Convert it into Kelvin (K)

Add 273

$\large\implies{\sf 273+27}$

$\large{\sf Temperature (T)= 300K}$

FORMULA TO FIND KINETIC ENERGY (KE) PER MOLECULE IS :-

★$\large{\boxed{\sf KE ={\frac{3RT}{2N_{A}}}}}$

Here ,

R = Gas Constant

Its value is 8.314

$\sf{N_{A}= Avogadro\:no. \implies 6.022×{10}^{23}}$

On Putting value :-

$\large\implies{\sf {\frac{3×8.314×300}{2×6.022×{10}^{23}}}}$

$\large\implies{\sf {\frac{7482.6×{10}^{-23}}{12.044}}}$

$\large\implies{\sf 621.27×{10}^{-23}}$

★$\large\red{\boxed{\sf KE\:per\: Molecule=6.21×{10}^{-21}J}}$