R=8.314

I am asking you to solve the equation step by step with all the data listed (please use the formulas above, also the reduced formulas). Calculate the volume of one kilogram of the given substance properties under certain temperature conditions and pressure. Compare the obtained results with the experimental values for gas and liquid phase (Vg and Vc).

I put the data below

Solution (PR):

(p+n2V(V+nb)+nb(Vnb)a)(Vnb)=nRT =(1+(1Tr))2 =0,37464+1,542260,269922 =73(1TbrTbr)log(101300Papk)1 a=pk0,45724R2Tk2b=pk0,077796RTk \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline T [K] & p [MPa] & Tk [K] & pk [MPa] & Tb [K] & M [g/mol] & Vg [dm3/kg] & Vc [dm3/kg] \\ \hline 344 & 3,24964 & 355,31 & 4,0645 & 227,53 & 111,600 & 4,0999 & 1,1051 \\ \hline \end{tabular} modelClapeyronvdWRKRKSPRVc[dm3/kg]1.4695971.3719031.222461Vc[dm3/kg]0.3644970.2668030.117361Vc[%]32.9824.1410.62Vg[dm3/kg]7.8862224.878264.4102914.2742944.057229Vg[dm3/kg]3.7863220.778360.3103910.1743940.042671Vg[%]92.3518.987.574.251.04