At 27^{circ} C kinetic energy of 4 g of H_2 | vedclass

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(D) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = \frac{3}{2} nRT$. For $H_2$: Molar mass $M(H_2) = 2 \ g/mol$. Number of mo

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$\frac{2 x}{15}$

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(D) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = \frac{3}{2} nRT$. For $H_2$: Molar mass $M(H_2) = 2 \ g/mol$. Number of moles $n_1 = \frac{4 \ g}{2 \ g/mol} = 2 \ mol$. Temperature $T_1 = 27 + 273 = 300 \ K$. $x = \frac{3}{2} 	imes 2 	imes R 	imes 300 = 900R$. For $O_2$: Molar mass $M(O_2) = 32 \ g/mol$. Number of moles $n_2 = \frac{6.4 \ g}{32 \ g/mol} = 0.2 \ mol$. Temperature $T_2 = 127 + 273 = 400 \ K$. $KE_2 = \frac{3}{2} 	imes 0.2 	imes R 	imes 400 = 120R$. Now,find the ratio: $\frac{KE_2}{x} = \frac{120R}{900R} = \frac{12}{90} = \frac{2}{15}$. Therefore,$KE_2 = \frac{2x}{15} \ J$.

At $27^{\circ} C$ kinetic energy of $4 \ g$ of $H_2$ is $x \ J$. What is the kinetic energy (in $J$) of $6.4 \ g$ of oxygen at $127^{\circ} C$?

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