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(D) The kinetic energy $(E)$ of one mole of an ideal gas is given by the formula $E = \frac{3}{2}RT$.At normal temperature and pressure $(NTP)$,the st

Vedclass · Accepted Answer

$3.4 	imes 10^3 	ext{ J}$

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(D) The kinetic energy $(E)$ of one mole of an ideal gas is given by the formula $E = \frac{3}{2}RT$.At normal temperature and pressure $(NTP)$,the standard temperature is $T = 273 	ext{ K}$.Given the gas constant $R = 8.31 	ext{ J/mol-K}$.Substituting these values into the formula:$E = \frac{3}{2} 	imes 8.31 	imes 273$$E = 1.5 	imes 8.31 	imes 273$$E = 12.465 	imes 273$$E \approx 3402.945 	ext{ J}$Rounding to the appropriate significant figures,we get $E \approx 3.4 	imes 10^3 	ext{ J}$.

The kinetic energy of one gm-mole of a gas at normal temperature and pressure is $(R = 8.31 \text{ J/mol-K})$.

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