Determine how many liters 8.73 g of carbon dioxide gas would occupy at the following conditions.

Question

(b) 160°C and 4.00 atm

(c) 249 K and 107 kPa

Answer 1

(b) Mass=8.73g molar mass of CO₂=(1×12)+(16×2)=44gmol^-1

No of moles(n)=Mass/molar mass=8.73/44 =0.198mol

T=160+273=433K P=4atm R=0.082atmL/molK

PV=nRT V=nRT/P

V=(0.198×0.082×433)/4

=1.76L

=1.76litres

(c)T=249K P=107KPa =107×1000 =1.07×10⁵Pa(Concert Pascal to atm)

1.01×10⁵Pa=1atm

1.07×10⁵Pa =1atm

R=0.082atmL/molK n=0.198mol

V=nRT/P

=(0.198×0.082×249)/1

=4.04litres