Question

Calculate the kinetic energy of $10$ gram of Argon molecules at ${127}^{\circ }C$.
[Universal gas constant $R=8320J/molK$. Atomic weight of Argon $=40kg/mol$]

Answer 1

Given : $m=10g=10\times {10}^{-3}kg$

$T=273+127=400K$
$R=8320J/mol.K$
$M=40kg/mol$
Kinetic energy per unit mass $=\frac{3}{2}\frac{RT}{M}$

$=\frac{3\times 8320\times 400}{2\times 40}$
$=15\times 8320$

$=1.248\times {10}^{5}J/kg$

The kinetic energy of $10\times {10}^{-3}kg$ of Argon at ${127}^{\circ }C$
$=1.248\times {10}^5\times 10\times {10}^{-3}$
$=1248J$