The average kinetic energy of one molecule of an ideal gas at 27°C and 1 atm pressure is

Question

The average kinetic energy of one molecule of an ideal gas at 27°C and 1 atm pressure is (A) 900 cal K -1 mol -1 (B) 6.21 × 10-21 JK -1 molecule -1 (C) 336.7 JK -1 molecule -1 (D) 3741.3 JK -1 mol -1

Tardigrade Admin · Accepted Answer

Average kinetic energy per molecule =(3/2) k T or =(3/2) (R/N0) T =(3/2) × (8.314/6.023 × 1023) × 300 =6.21 × 10-21 JK -1 molecule -1

Q. The average kinetic energy of one molecule of an ideal gas at $2 7^{\circ}$ C and $1 a t m$ pressure is

$900 c a l K^{- 1} m o l^{- 1}$

$6.21 \times 1 0^{- 21} J K^{- 1}$ molecule $^{- 1}$

$336.7 J K^{- 1}$ molecule $^{- 1}$

$3741.3 J K^{- 1} m o l^{- 1}$

Average kinetic energy per molecule
$= \frac{3}{2} k T$
or $= \frac{3}{2} \frac{R}{N _{0}} T$
$= \frac{3}{2} \times \frac{8.314}{6.023 \times 1 0 ^{23}} \times 300$
$= 6.21 \times 1 0^{- 21} J K^{- 1}$ molecule $^{- 1}$

Q. The average kinetic energy of one molecule of an ideal gas at 27∘C and 1atm pressure is

900calK−1mol−1

6.21×10−21JK−1 molecule −1

336.7JK−1 molecule −1

3741.3JK−1mol−1