Question

8.2L of an ideal gas weigh 9.0g at 300k and 1atm pressure. The molecular gas is

Answer 1

p=1atm,

v=8.2 L,

temp=300k,

mass=9g

pV=nRT... where n is no.of moles

n=mass÷molarmass

pV=(mass÷molarmass)RT

1×8.2=(9÷molarmass)0.082×300

8.2÷8.2×3=9÷molarmass

molarmass=27