The average translational kinetic energy of the oxygen molecules at a temperature of 127^{} {C} is (. Boltzmann constant =1.38 × 10^{-23} {J} {K}^{-1}) [AP EAPCET 21 May 2025 S2]

Correct Answer is : 8.28×10−21J8.28 × 10^{-21} {J}8.28×10−21J

Correct Answer is : 8.28×10−21J8.28 × 10^{-21} {J}8.28×10−21J

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Heat and Thermodynamics

Section: Physics

Question : 31 of 78

Marks: +1, -0

127^{\circ} \mathrm{C}

(.

=1.38 \times 10^{-23} \mathrm{J} \mathrm{K}^{-1})

Solution:

Given:

Temperature

T=127^{\circ} \mathrm{C}

Boltzmann constant

k=1.38 \times 10^{-23} \text{J/K}

We know that:

Average translational kinetic energy per molecule

=\;\frac{3}{2} k T

Step 1: Convert temperature to kelvins

T=127+273=400 \mathrm{K}

Step 2: Substitute the values

E=\;\frac{3}{2} k T=\;\frac{3}{2} (1.38 \times 10^{-23}) (400)

Step 3: Simplify

\;E=1.5 \times 1.38 \times 400 \times 10^{-23}

\;E=2.07 \times 400 \times 10^{-23}

\;E=8.28 \times 10^{-21} \mathrm{J}

Final Answer:

8.28 \times 10^{-21} \mathrm{J}

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