Answer:
[tex]P=35.16[/tex]
[tex]Z=4.6[/tex]
Explanation:
Hello,
In this case, since the VdW equation is:
[tex]P=\frac{nRT}{V-n*b}-a(\frac{n}{V} )^2[/tex]
Since the moles are 10.0 moles, the temperature in K is 300.15 K and the volume is liters is also 4.860 L (1 dm³= 1L), the pressure exerted by the ethane is:
[tex]P=\frac{10.0mol*0.082\frac{atm*L}{mol*K}*300.15K}{4.860mol-10.0mol*0.0651\frac{L}{mol} }-5.507\frac{atm*L^2}{mol^2}(\frac{10.0mol}{4.86L} )^2\\\\P=58.48atm-23.3atm\\\\P=35.16[/tex]
Thus the compression factor turns out:
[tex]Z=\frac{PV}{RT}=\frac{23.3atm*4.86L}{ 0.082\frac{atm*L}{mol*K}*300.15K}\\\\Z=4.6[/tex]
Regards.
