The average kinetic energy of a monoatomic gas molecule kept at temperature 27°C is (Boltzmann constant k = 1.3 × 10-23 JK-1)

Question

The average kinetic energy of a monoatomic gas molecule kept at temperature 27°C is (Boltzmann constant k = 1.3 × 10-23 JK-1) (A) 5.85 × 10-21 J (B) 4.12 × 10-21 J (C) 3.75 × 10-21 J (D) 2.85 × 10-21 J

Tardigrade Admin · Accepted Answer

K . E=(3/2) kβ ⋅ T=(3/2) × 1.39 × 10-23 × 300 J

Q. The average kinetic energy of a monoatomic gas molecule kept at temperature $2 7^{\circ} C$ is
(Boltzmann constant $k = 1.3 \times 1 0^{- 23} J K^{- 1}$ )

$5.85 \times 1 0^{- 21} J$

$4.12 \times 1 0^{- 21} J$

$3.75 \times 1 0^{- 21} J$

$2.85 \times 1 0^{- 21} J$

$7.55 \times 1 0^{- 2} J$

$K . E = \frac{3}{2} k_{β} \cdot T = \frac{3}{2} \times 1.39 \times 1 0^{- 23} \times 300 J$

Q. The average kinetic energy of a monoatomic gas molecule kept at temperature 27∘C is (Boltzmann constant k=1.3×10−23JK−1)

5.85×10−21J

4.12×10−21J

3.75×10−21J

2.85×10−21J

7.55×10−2J

K.E=23​kβ​⋅T=23​×1.39×10−23×300J