Question

The translational kinetic energy of $1\text{mole}$ of an ideal gas at standard temperature is close to

• A. $3403\text{J}$
• B. $3000\text{J}$
• C. $2342\text{J}$
• D. $1564\text{J}$

Answer 1

The correct option is A $3403\text{J}$

We know that,
Translational kinetic energy, $E=\frac{3}{2}PV$
Using ideal gas equation,
$PV=nRT$
For $n=1$
$PV=RT$
$\therefore E=\frac{3}{2}PV=\frac{3}{2}RT$
$E=\frac{3}{2}\times 8.31\times 273\text{J}$
$\cong 3403\text{J}$