At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be 1933 m s-1. The gas is (R = 8.3 J mol-1 K-1 )

Question

At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be 1933 m s-1. The gas is (R = 8.3 J mol-1 K-1 ) (A) H2 (B) F2 (C) Cl2 (D) O2

Tardigrade Admin · Accepted Answer

The velocity of gas at temperature T is given by

υ_{r m s} = \frac{3 RT}{M}

where, R is gas constant and M the molecular weight.
Given,

R = 8.3 J m o l^{- 1} K^{- 1}

T = 2 7^{\circ} C = 27 + 273 = 300 K

υ_{r m s}^{2} = 1933 m s^{- 1}

∴ M = \frac{3 RT}{υ _{r m s}^{2}} = \frac{3 \times 8.3 \times 300}{( 1933 ) ^{2}} ∴ M \approx 2 \times 1 0^{- 3} k g

which is molecular weight of

H_{2} .

Q. At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be $1933 m s^{- 1}$ . The gas is $(R = 8.3 J m o l^{- 1} K^{- 1}$ )

$H_{2}$

$F_{2}$

$C l_{2}$

$O_{2}$

Q. At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be 1933ms−1. The gas is (R=8.3Jmol−1K−1 )

H2​

F2​

Cl2​

O2​

The velocity of gas at temperature T is given by υrms​=M3RT​​ where, R is gas constant and M the molecular weight. Given, R=8.3Jmol−1K−1 T=27∘C=27+273=300K υrms2​=1933ms−1 ∴M=υrms2​3RT​=(1933)23×8.3×300​∴M≈2×10−3kg which is molecular weight of H2​.

Q. At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be $1933 m s^{- 1}$ . The gas is $(R = 8.3 J m o l^{- 1} K^{- 1}$ )

$H_{2}$

$F_{2}$

$C l_{2}$

$O_{2}$