The average translational kinetic energy of $N_{2}$ gas molecules at $_________{}^{\circ}C$ becomes equal to the $K . E .$ of an electron accelerated from rest through a potential difference of $0.1 volt .$
(Given $k_{B} = 1.38 \times 10^{- 23} J K^{- 1})$ (Fill the nearest integer).

Question

The average translational kinetic energy of $N_{2}$ gas molecules at $_________{}^{\circ}C$ becomes equal to the $K . E .$ of an electron accelerated from rest through a potential difference of $0.1 volt .$

(Given $k_{B} = 1.38 \times 10^{- 23} J K^{- 1})$ (Fill the nearest integer).

TestHub · Accepted Answer

Given Translation K . E . of N 2 = K . E of electron 3 2 k T = e V 3 2 × 1 . 38 × 10 - 23 T = 1 . 6 × 10 - 19 × 0 . 1 ⇒ T = 773 K T = 773 - 273 = 500 ° C

Physics - Heat & Thermo Question with Solution | TestHub

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