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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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