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Q.
Kinetic energy of one mole of an ideal gas at $300 \,K$ is
AMUAMU 2005
Solution:
Key Idea The kinetic energy $(K E)$ of the ideal gas is given by
$K E=\frac{3}{2} n R T$
where, $ n=$ number of moles
$R=g a s$ constant
$T=$ absolute temperature
Here, $n=1$
$T =300 \,K $
$R =8.314 \,J\, mol ^{-1} K ^{-1}$
Now put these value in the above equation
$KE =\frac{3}{2} \times 1 \times 8.314 \times 300 $
$=3741.3\, J $
$=3.74\, kJ$