quantity of 2 gram of triatomic gases element was found to occupy volume of 448 ml at 7600 cm 273 Kelvin the mass of its atom is

Dear Student


Mass of triatomic gases element = 2 gm

Volume, V = 448 mL

Pressure, p = 7600 cm

Temperature, T = 273 K



According to the Ideal Gas Law,

pV = nRT


where, n = no. of moles

           V = volume, 0.448 L

           R = Gas constant, 0.0821 L atm/Mol K

           T = Temperature in K

           P = Pressure, 7600 cm = 100 atm (because 76 cm = 1 atm)

So, 7600 x 0.448 L = n x 0.0821 L atm/Mol K x 273 K

100 atm x 0.448 L = n x 0.0821 L atm/Mol K x 273 K

      n = $\frac{100 atm x 0.448 L}{0.0821 l atm/Mol K x 273 K}$ = $\frac{44.8 L atm}{22.4133 L atm/Mol}$ = 2 mol
 
we know, no. of moles = $\frac{mass}{molar mass}$
     n = $\frac{2}{m}$
hence, m = 1 g/mol
    

Mass of 3 atoms of a triatomic gas = 1 g

Mass of 1 atom = $\frac{1}{3 x 6.023 x {10}^{23}}$ = $\frac{1}{18.069 x {10}^{23}}$ = 0.05534 x 10^-23 g or 5.53 x 10^-21 g



Regards