Question

# What is the average translational kinetic energy of a molecule of an ideal gas at temperature of $$27^{\circ}C$$

Solution

## Translational kinetic energy is given by the formula$${ E }_{ T }=\dfrac { 3 }{ 2 } kT$$$${ E }_{ T }=\dfrac { 3 }{ 2 } (1.38X{ 10 }^{ -23 })(300)=6.21X{ 10 }^{ -21 }J$$Answer is $$6.21X{ 10 }^{ -21 }J$$Physics

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