Question

What pressure (in atm) would be exerted by a mixture of 1.4 g of Chlorine gas (Mm= 70 g/mol) and 4.8 g of nitrogen gas (Mm= 28 g/mole) in a 200 mL container at 57oC?

Answer 1

Moles of chlorine $\frac{1.4}{70}=0.02moles$

Moles of nitrogen $\frac{4.8}{28}=0.17moles$

PV= nRT

$P×0.2=[0.02+0.17]×0.0821×[57+273]$

$P=\frac{0.19×0.0821×330}{0.2}=25.73atm$