The average translational kinetic energy of N 2 gas molecules at _ _ _ _ _ _ _ _ _ C ∘ becomes equal to the K . E . of an electron accelerated from rest through a potential difference of 0 . 1 volt . (Given k B = 1 . 38 × 10 - 23

Question

The average translational kinetic energy of N 2 gas molecules at _ _ _ _ _ _ _ _ _ C &#8728; becomes equal to the K . E . of an electron accelerated from rest through a potential difference of 0 . 1 volt . (Given k B = 1 . 38 &#215; 10 - 23 J K - 1 (Fill the nearest integer).

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Given Translation K . E . of N 2 = K . E of electron 3 2 k T = e V 3 2 &#215; 1 . 38 &#215; 10 - 23 &#160; T = 1 . 6 &#215; 10 - 19 &#215; 0 . 1 &#160;&#160; &#8658; T = 773 &#160; K T = 773 - 273 = 500 &#176; C

The average translational kinetic energy of N 2 gas molecules at _ _ _ _ _ _ _ _ _ C ∘ becomes equal to the K . E . of an electron accelerated from rest through a potential difference of 0 . 1 volt . (Given k B = 1 . 38 × 10 - 23 J K - 1 (Fill the nearest integer).

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More from Laws of Motion

The average translational kinetic energy of N 2 gas molecules at _ _ _ _ _ _ _ _ _ C &#8728; becomes equal to the K . E . of an electron accelerated from rest through a potential difference of 0 . 1 volt . (Given k B = 1 . 38 &#215; 10 - 23 J K - 1 (Fill the nearest integer).

Correct answer

Step-by-step solution

More from Laws of Motion

The average translational kinetic energy of N 2 gas molecules at _ _ _ _ _ _ _ _ _ C ∘ becomes equal to the K . E . of an electron accelerated from rest through a potential difference of 0 . 1 volt . (Given k B = 1 . 38 × 10 - 23 J K - 1 (Fill the nearest integer).