The average translational kinetic energy of a molecule in a gas becomes equal to 0.69 eV at a temperature about [Boltzmann constant = 1.38 × 10-23 J K-1]

Question

The average translational kinetic energy of a molecule in a gas becomes equal to 0.69 eV at a temperature about [Boltzmann constant = 1.38 × 10-23 J K-1] (A) 3370 °C (B) 3388 °C (C) 5333 °C (D) 5060 °C

Tardigrade Admin · Accepted Answer

Given, average translational kinetic energy

= 0.69 e V = 0.69 \times 1.6 \times 1 0^{- 19} V

As we know that,
average translational kinetic energy

= \frac{3}{2} k T

0.69 \times 1.6 \times 1 0^{- 19} = \frac{3}{2} \times 1.38 \times 1 0^{- 23} T

T = \frac{0.69 \times 1.6 \times 1 0 ^{- 19} \times 2}{3 \times 1.38 \times 1 0 ^{- 23}}

T = 5333 K = 5333 - 273

= 506 0^{\circ} C

Q. The average translational kinetic energy of a molecule in a gas becomes equal to $0.69 e V$ at a temperature about
[Boltzmann constant = $1.38 \times 1 0^{- 23} J K^{- 1}$ ]

$3370^{\circ} C$

$3388^{\circ} C$

$5333^{\circ} C$

$5060 \circ C$

Q. The average translational kinetic energy of a molecule in a gas becomes equal to 0.69eV at a temperature about [Boltzmann constant = 1.38×10−23JK−1]

3370∘C

3388∘C

5333∘C

5060∘C

Given, average translational kinetic energy =0.69eV=0.69×1.6×10−19V As we know that, average translational kinetic energy =23​kT 0.69×1.6×10−19=23​×1.38×10−23T T=3×1.38×10−230.69×1.6×10−19×2​ T=5333K=5333−273 =5060∘C

Q. The average translational kinetic energy of a molecule in a gas becomes equal to $0.69 e V$ at a temperature about
[Boltzmann constant = $1.38 \times 1 0^{- 23} J K^{- 1}$ ]

$3370^{\circ} C$

$3388^{\circ} C$

$5333^{\circ} C$

$5060 \circ C$