The rms velocity of oxygen molecule at 27° C is (Atomic weight of oxygen = 16)

Question

The rms velocity of oxygen molecule at 27° C is (Atomic weight of oxygen = 16) (A)  402.5 ms-1  (B)  400.5 text ms-1  (C)  483.5 text ms-1  (D)  532.5 text ms-1

Tardigrade Admin · Accepted Answer

Temperature T=27° C=273+27=300K Atomic weight of oxygen =16 ∴ Molecular weight of oxygen, M=2× 16=32 R=8.31 text J text mol-1K-1 So, the rms velocity, C=√(3RT/M) =√(3× 8.31× 300/32)=483.5 ms-1

Q. The rms velocity of oxygen molecule at $27^{\circ} C$ is (Atomic weight of oxygen = 16)

$402.5 m s^{- 1}$

$400.5 m s^{- 1}$

$483.5 m s^{- 1}$

$532.5 m s^{- 1}$

Temperature $T = 27^{\circ} C = 273 + 27 = 300 K$
Atomic weight of oxygen =16
$∴$ Molecular weight of oxygen, $< b r > M = 2 \times 16 = 32$
$R = 8.31 J m o l^{- 1} K^{- 1}$
So, the rms velocity, $C = \frac{3 RT}{M}$
$= \frac{3 \times 8.31 \times 300}{32} = 483.5 m s^{- 1}$

Q. The rms velocity of oxygen molecule at 27∘C is (Atomic weight of oxygen = 16)

402.5ms−1

400.5ms−1

483.5ms−1

532.5ms−1