The average translational kinetic energy of N₂ 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) (Fill the nearest integer).

Solution

Physics sir · 4 steps

Only translational energy counts here, so use 3/2k_BT even though N₂ is diatomic.

1. Electron energy. An electron accelerated from rest through V = 0.1 V gains eV = 1.6 × 10⁻¹⁹ × 0.1 = 1.6 × 10⁻²⁰ J.
2. Translational part only. Every molecule has 3 translational degrees of freedom, each carrying 1/2k_BT. The 2 rotational degrees of N₂ are not part of translational energy, so use 3/2k_BT, not 5/2k_BT.

   3/2k_BT = eV ⇒ T = 2eV/3k_B

3. Put in numbers. T = (2 × 1.6 × 10⁻²⁰)/(3 × 1.38 × 10⁻²³) = (3.2 × 10⁻²⁰)/(4.14 × 10⁻²³) ≈ 773 K.
4. So. In Celsius, 773 - 273 = 500, so the temperature is 500 °C.

The common slip: 773. Stopping at the kelvin value gives 773, but the blank asks for degrees Celsius, so subtract 273 to get 500.

The idea to keep. Average translational KE is 3/2k_BT for any gas, whether monatomic, diatomic or polyatomic. Extra degrees of freedom matter only when the question asks for total energy.

Answer from the NTA final answer key. Solution by PrepWiser.