JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Heat & Thermodynamics

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 1 0^{- 23} J / K

) (Fill the nearest integer).

Answer: 500

Solution

The electron gains

E = e V = 1.6 \times 1 0^{- 19} \times 0.1 = 1.6 \times 1 0^{- 20} J

. Setting this equal to the average translational energy

\frac{3}{2} k_{B} T

gives

T = \frac{2 \times 1.6 \times 1 0 ^{- 20}}{3 \times 1.38 \times 1 0 ^{- 23}} = 773 K

. In Celsius this is

773 - 273 = 500^{\circ} C

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Integer answer

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