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Question

(D) The average translational kinetic energy of a gas molecule is given by the formula $E = \frac{3}{2} k T$,where $k$ is the Boltzmann constant and $

Vedclass · Accepted Answer

$5060$

Vedclass · Answer

(D) The average translational kinetic energy of a gas molecule is given by the formula $E = \frac{3}{2} k T$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.Given,$E = 0.69 \ eV = 0.69 	imes 1.6 	imes 10^{-19} \ J$.Substituting the values into the formula:$0.69 	imes 1.6 	imes 10^{-19} = \frac{3}{2} 	imes 1.38 	imes 10^{-23} 	imes T$$T = \frac{0.69 	imes 1.6 	imes 10^{-19} 	imes 2}{3 	imes 1.38 	imes 10^{-23}}$$T = \frac{2.208 	imes 10^{-19}}{4.14 	imes 10^{-23}} \approx 5333 \ K$To convert the temperature to Celsius,we use $T(^{\circ}C) = T(K) - 273.15$.$T(^{\circ}C) = 5333 - 273 = 5060^{\circ} C$.Therefore,the correct option is $D$.

The average translational kinetic energy of a molecule in a gas becomes equal to $0.69 \ eV$ at a temperature of about,[Boltzmann's constant $= 1.38 \times 10^{-23} \ J \ K^{-1}$] (in $^{\circ} C$)

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