Respuesta :
Answer:
a). P = 688 atm
b). P = 1083.04 atm
c).Δ G = 16.188 J/mol
Explanation:
a). Fugacity 'f' can be calculated from the following equations :
[tex]$\ln \frac{f}{P}=\int^P_0\left(\frac{Z-1}{P}\right) dP$[/tex]
where, P = pressure , Z = compressibility
Now, the virial equation is :
[tex]$PV=R(1+6.4\times 10^{-4} P)$[/tex] ........(1)
Also, PV=ZRT for real gases .......(2)
∴ [tex]$ZRT=RT(1+6.4\times 10^{-4} P)$[/tex]
[tex]$Z=1+6.4\times 10^{-4} P$[/tex]
So from the fugacity equation ,
[tex]$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$[/tex]
[tex]$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$[/tex]
[tex]$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$[/tex]
[tex]$f=Pe^{6.4 \times 10^{-4} P}$[/tex]
Putting the value of P = 500 atm in the above equation, we get,
f = 688 atm
b). Given f = 2P
[tex]$2P=PE^{6.4 \times 10^{-4}P}$[/tex]
[tex]$2=E^{6.4 \times 10^{-4}P}$[/tex]
[tex]$\ln 2 = 6.4 \times 10^{-4}P$[/tex]
∴ P = 1083.04 atm
c). dG = Vdp -S dt at constant temperature, dT = 0
Therefore, dG = V dp
[tex]$\int^{G_2}_{G_1}dG =\int^{P_2}_{P_1}V dp $[/tex]
[tex]$=\int^{P_2}_{P_1}\frac{RT}{P}\left(1+6.4 \times 10^{-4}P\right) dP$[/tex]
[tex]$=\int^{P_2}_{P_1}RT\frac{dp}{P}+\int^{P_2}_{P_1}RT6.4 \times 10^{-4} dP$[/tex]
[tex]$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$[/tex]
[tex]$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$[/tex]
Δ G = 16.188 J/mol
