Question

Calculate the rms speed of an ideal diatomic gas having molecular weight 32gm/mol at 0°c

Answer 1

Answer : The rms speed of an ideal diatomic gas is, 461.287 m/s

Solution :

Formula used :

$v_{rms}=\sqrt{\frac{3RT}{M}}$

where,

$v_{rms}$ = root mean square speed

R = gas constant = $8.314\text{ Kg }m^2s^{-2}K^{-1}mole^{-1}$

T = temperature of gas = $0^oC=0+273=273K$

M = molecular weight = 32 g/mole = 0.032 Kg/mole    (1 Kg = 1000 g)

Now put all the given values in the above formula, we get the rms speed.

$v_{rms}=\sqrt{\frac{3\times (8.314\text{ Kg }m^2s^{-2}K^{-1}mole^{-1})\times (273K)}{(0.032Kg/mole)}}=461.287m/s$

Therefore, the rms speed of an ideal diatomic gas is, 461.287 m/s