The kinetic energy of translation of 20 , g o | vedclass

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Question

(A) Given data:Mass of oxygen,$m = 20 \, g$Molar mass of oxygen,$M = 32 \, g/mol$Number of moles,$n = \frac{m}{M} = \frac{20}{32} = 0.625 \, mol$Tempe

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$2490 \, J$

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(A) Given data:Mass of oxygen,$m = 20 \, g$Molar mass of oxygen,$M = 32 \, g/mol$Number of moles,$n = \frac{m}{M} = \frac{20}{32} = 0.625 \, mol$Temperature,$T = 47^{\circ}C = 47 + 273 = 320 \, K$Gas constant,$R = 8.3 \, J/mol \cdot K$The kinetic energy of translation for an ideal gas is given by $E = \frac{3}{2} nRT$.Substituting the values:$E = \frac{3}{2} 	imes 0.625 	imes 8.3 	imes 320$$E = 1.5 	imes 0.625 	imes 8.3 	imes 320$$E = 2490 \, J$Thus,the kinetic energy of translation is $2490 \, J$.

The kinetic energy of translation of $20 \, g$ of oxygen at $47^{\circ}C$ is (molecular weight of oxygen is $32 \, g/mol$ and $R = 8.3 \, J/mol \cdot K$).

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