16 g of oxygen occupy `0.02 m^(3) "at" 27^(@) C`. Find the pressure exerted by it. [ Molar mass of oxygen = 32 g/mo. , universal gas constant R = 8.314 /mol.K]

Explanation

T = ` 27^(@) C = (273 + 27) K = 300 K , R = 8.314 ` J/mol .K
pV = nRT , where number of moles ` n= m/n`
The pressure exerted by the gas. ,
` p = (nRT)/V = m/M. (RT)/V = 16/32 xx (8.314 xx 300)/(0.02 )`
= ` 6.236 xx 10^(4) N//m^(2)`