The kinetic energy of translation of the mole | vedclass

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Question

(B) The translational kinetic energy of $N$ molecules is given by $K = \frac{3}{2} N k_B T = \frac{3}{2} n R T$,where $n$ is the number of moles.Given

Vedclass · Accepted Answer

$4102.8$

Vedclass · Answer

(B) The translational kinetic energy of $N$ molecules is given by $K = \frac{3}{2} N k_B T = \frac{3}{2} n R T$,where $n$ is the number of moles.Given mass $m = 50 	ext{ g}$,molar mass of $	ext{CO}_2$ $M = 44 	ext{ g/mol}$.Number of moles $n = \frac{m}{M} = \frac{50}{44} \approx 1.136 	ext{ mol}$.Temperature $T = 17 + 273.15 = 290.15 	ext{ K}$.Using $R = 8.314 	ext{ J/(mol K)}$:$K = \frac{3}{2} 	imes \left( \frac{50}{44} ight) 	imes 8.314 	imes 290.15$.$K = 1.5 	imes 1.13636 	imes 8.314 	imes 290.15 \approx 4108.6 	ext{ J}$.Comparing with the given options,the closest value is $4102.8 	ext{ J}$.

The kinetic energy of translation of the molecules in $50 \text{ g}$ of $\text{CO}_2$ gas at $17^{\circ}\text{C}$ is: (in $\text{ J}$)

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