Calculate the total kinetic energy of 0.5 mole of an ideal gas at 273 Kelvin

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Solution

According to the formula
For average kinetic energy of ideal gases is

$K E = \frac{3}{2} n R T$

Where,

$n = N u m b e r o f m o l e s o f t h e g a s$

$R = G a s c o n s t a n t$

$T = T e m p e r a t u r e i n K .$

S0,

$K E = \frac{3}{2} \times 0.5 \times 8.3145 \times 273$

$= 1702.4 J$
This is the required solution.

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