Answer :
Answer:
a)W=12.62 kJ/mol
b)W=12.59 kJ/mol
Explanation:
At T = 100 °C the second and third virial coefficients are
B = -242.5 cm^3 mol^-1
C = 25200 cm^6 mo1^-2
Now according isothermal work of one mole methyl gas is
W=-[tex]\int\limits^a_b {P} \, dV[/tex]
a=[tex]v_2\\[/tex]
b=[tex]v_1[/tex]
from virial equation
[tex]\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\[/tex]
And
[tex]W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV[/tex]
a=[tex]v_2\\[/tex]
b=[tex]v_1[/tex]
Now calculate V1 and V2 at given condition
[tex]\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}[/tex]
Substitute given values [tex]P_1\\[/tex] = 1 x 10^5 , T = 373.15 and given values of coefficients we get
[tex]10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2[/tex]
Solve for V1 by iterative or alternative cubic equation solver we get
[tex]v_1=30780 cm^3/mol[/tex]
Similarly solve for state 2 at P2 = 50 bar we get
[tex]v_1=241.33 cm^3/mol[/tex]
Now
[tex]W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV[/tex]
a=241.33
b=30780
After performing integration we get work done on the system is
W=12.62 kJ/mol
(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get
dV=RT(-1/p^2+0+C')dP
Hence work done on the system is
[tex]W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP[/tex]
a=[tex]v_2\\[/tex]
b=[tex]v_1[/tex]
by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work
W=12.59 kJ/mol
The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series
The work of mechanically reversible, isothermal compression for the 2 viral equations are;
A) W = 13.74 kJ
B) W = 14.75 kJ
The reason why both equations don't give exactly the same result is; because the 3rd term in the virial equation is different in the two equations.
We are given;
Second virial coefficient; B = −242.5 cm³/mol
Third virial coefficient; C = 25200 cm⁶/mol²
Initial Pressure; P₁ = 1 bar
Final Pressure; P₂ = 55 bar
A) We have the virial equation;
Z = 1 + (B/V) + C/V²
- Now, we know that Z = PV/RT. Thus, our first virial equation can be rewritten as; Z = 1 + (BP/RT) + C(P/RT)²
Thus, the initial volume at initial pressure is given by;
V₁ = (RT/P)(1 + (BP/RT) + C(P/RT)²)
Where;
R is a constant = 0.0821 (atm)L/(mol) K
T = 100°C = 373 k
P = 1 bar = 0.98 atm
- Thus, plugging in the relevant values gives;
V₁ = 31.2 - 0.2425 + 0.0008
V₁ = 30.9583 L/mol
- The final volume at final pressure is given by;
V₂ = (RT/P)(1 + (BP/RT) + C(P/RT)²)
Where;
R is a constant = 0.0821 (atm)L/(mol).K
T = 100°C = 373 k
P = 55 bar = 53.9 atm
Thus;
V₂ = 0.567 - 0.2425 + 0.044
V₂ = 0.3685 L/mol
- Work done for an Isothermal reversal process is given by;
W = -nRT In(V₂/V₁)
Plugging in the relevant values and solving gives;
W = 13.74 kJ
B) We are given the virial equation;
Z = 1 + B'P + C'P²
- This can be rewritten as;
Z = (1 + (BP/RT) + (C - B²)(P/RT)²)
Thus, from Z = PV/RT, we can say that;
V = (RT/P) + B + (C - B²)(P/RT)²)
- Thus, just like in answer A above;
initial volume at initial pressure is;
V₁ = 31.2 - 0.2425 - (0.033*0.032)
V₁ = 30.95 L/atm
- Final volume at final pressure is given by;
V₂ = 0.567 - 0.2425 - (0.033*1.76)
V₂ = 0.266 L/atm
Work done for an Isothermal reversal process is given by;
W = -nRT In(V₂/V₁)
Plugging in the relevant values and solving gives;
W = 14.75 kJ
- In conclusion, The reason why both equations don't give exactly the same result is because the 3rd term in the virial equation is different in the two equations.
Read more about virial equations at; https://brainly.com/question/16177987