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Chemistry Question on Spontaneity

The average translational kinetic energy of a molecule in a gas becomes equal to 0.69 eV0.69\, eV at a temperature about [Boltzmann constant = 1.38×10−23  J  K−11.38 \times 10^{-23} \; J \; K^{-1}]

A

3370∘C3370 {^{\circ}}C

B

3388∘C3388 {^{\circ}}C

C

5333∘C5333 {^{\circ}}C

D

5060∘C5060 {\circ}C

Answer

5060∘C5060 {\circ}C

Explanation

Solution

Given, average translational kinetic energy
=0.69 eV=0.69×1.6×10−19 V=0.69\, eV =0.69 \times 1.6 \times 10^{-19} \,V
As we know that,
average translational kinetic energy =32kT=\frac{3}{2} k T
0.69×1.6×10−19=32×1.38×10−23T0.69 \times 1.6 \times 10^{-19}=\frac{3}{2} \times 1.38 \times 10^{-23} T
T=0.69×1.6×10−19×23×1.38×10−23T=\frac{0.69 \times 1.6 \times 10^{-19} \times 2}{3 \times 1.38 \times 10^{-23}}
T=5333K=5333−273T=5333 K =5333-273
=5060∘C=5060^{\circ} C